https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Alexlikemath&feedformat=atom AoPS Wiki - User contributions [en] 2020-12-02T22:59:59Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14&diff=134670 1984 AIME Problems/Problem 14 2020-10-06T02:38:25Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> What is the largest even integer that cannot be written as the sum of two odd composite numbers?<br /> <br /> == Solution 1 ==<br /> <br /> Take an even positive integer &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is either &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;2 \bmod{6}&lt;/math&gt;, or &lt;math&gt;4 \bmod{6}&lt;/math&gt;. Notice that the numbers &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;21&lt;/math&gt;, ... , and in general &lt;math&gt;9 + 6n&lt;/math&gt; for nonnegative &lt;math&gt;n&lt;/math&gt; are odd composites. We now have 3 cases:<br /> <br /> If &lt;math&gt;x \ge 18&lt;/math&gt; and is &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;9 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 44&lt;/math&gt; and is &lt;math&gt;2 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;35 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;35&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 34&lt;/math&gt; and is &lt;math&gt;4 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;25 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;25&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites.<br /> <br /> <br /> Clearly, if &lt;math&gt;x \ge 44&lt;/math&gt;, it can be expressed as a sum of 2 odd composites. However, if &lt;math&gt;x = 42&lt;/math&gt;, it can also be expressed using case 1, and if &lt;math&gt;x = 40&lt;/math&gt;, using case 3. &lt;math&gt;38&lt;/math&gt; is the largest even integer that our cases do not cover. If we examine the possible ways of splitting &lt;math&gt;38&lt;/math&gt; into two addends, we see that no pair of odd composites add to &lt;math&gt;38&lt;/math&gt;. Therefore, &lt;math&gt;\boxed{038}&lt;/math&gt; is the largest possible number that is not expressible as the sum of two odd composite numbers.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be an integer that cannot be written as the sum of two odd composite numbers. If &lt;math&gt;n&gt;33&lt;/math&gt;, then &lt;math&gt;n-9,n-15,n-21,n-25,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; must all be prime (or &lt;math&gt;n-33=1&lt;/math&gt;, which yields &lt;math&gt;n=34=9+25&lt;/math&gt; which does not work). Thus &lt;math&gt;n-9,n-15,n-21,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is &lt;math&gt;5,11,17,23,&lt;/math&gt; and &lt;math&gt;29&lt;/math&gt;, yielding a maximal answer of 38. Since &lt;math&gt;38-25=13&lt;/math&gt;, which is prime, the answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> == Solution 3 (bash) ==<br /> Let &lt;math&gt;2n&lt;/math&gt; be an even integer. Using the [[Chicken McNugget Theorem]] on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as &lt;math&gt;n+n&lt;/math&gt;. We bash each case until we find one that works.<br /> <br /> == Solution 4 (easiest) ==<br /> The easiest method is to notice that any odd number that ends is a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...<br /> <br /> For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9. Hence the largest number that ends with a 4 that satisfies the conditions is 14. <br /> <br /> If you list out all the numbers, you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and anything else that ends with a 3 is bad, so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be &lt;math&gt;\boxed{38}&lt;/math&gt;<br /> <br /> == Solution 5 ==<br /> Claim: The answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> Proof:<br /> It is fairly easy to show 38 can't be split into 2 odd composites.<br /> <br /> Assume there exists an even integer &lt;math&gt;m &gt; 38&lt;/math&gt; that m can't be split into 2 odd composites.<br /> <br /> Then, we can consider m modulo 5.<br /> <br /> If m = 0 mod 5, we can express m = 15 + 5k for some integer k. Since &lt;math&gt;m &gt; 20&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 15 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 1 mod 5, we can express m = 21 + 5k for some integer k. Since &lt;math&gt;m &gt; 26&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 21 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 2 mod 5, we can express m = 27 + 5k for some integer k. Since &lt;math&gt;m &gt; 32&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 27 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 3 mod 5, we can express m = 33 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 4 mod 5, we can express m = 9 + 5k for some integer k. Since &lt;math&gt;m &gt; 14&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 9 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> Thus, in all cases we can split m into 2 odd composites, and we get at a contradiction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=13|num-a=15}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14&diff=134669 1984 AIME Problems/Problem 14 2020-10-06T02:35:49Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> What is the largest even integer that cannot be written as the sum of two odd composite numbers?<br /> <br /> == Solution 1 ==<br /> <br /> Take an even positive integer &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is either &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;2 \bmod{6}&lt;/math&gt;, or &lt;math&gt;4 \bmod{6}&lt;/math&gt;. Notice that the numbers &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;21&lt;/math&gt;, ... , and in general &lt;math&gt;9 + 6n&lt;/math&gt; for nonnegative &lt;math&gt;n&lt;/math&gt; are odd composites. We now have 3 cases:<br /> <br /> If &lt;math&gt;x \ge 18&lt;/math&gt; and is &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;9 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 44&lt;/math&gt; and is &lt;math&gt;2 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;35 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;35&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 34&lt;/math&gt; and is &lt;math&gt;4 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;25 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;25&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites.<br /> <br /> <br /> Clearly, if &lt;math&gt;x \ge 44&lt;/math&gt;, it can be expressed as a sum of 2 odd composites. However, if &lt;math&gt;x = 42&lt;/math&gt;, it can also be expressed using case 1, and if &lt;math&gt;x = 40&lt;/math&gt;, using case 3. &lt;math&gt;38&lt;/math&gt; is the largest even integer that our cases do not cover. If we examine the possible ways of splitting &lt;math&gt;38&lt;/math&gt; into two addends, we see that no pair of odd composites add to &lt;math&gt;38&lt;/math&gt;. Therefore, &lt;math&gt;\boxed{038}&lt;/math&gt; is the largest possible number that is not expressible as the sum of two odd composite numbers.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be an integer that cannot be written as the sum of two odd composite numbers. If &lt;math&gt;n&gt;33&lt;/math&gt;, then &lt;math&gt;n-9,n-15,n-21,n-25,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; must all be prime (or &lt;math&gt;n-33=1&lt;/math&gt;, which yields &lt;math&gt;n=34=9+25&lt;/math&gt; which does not work). Thus &lt;math&gt;n-9,n-15,n-21,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is &lt;math&gt;5,11,17,23,&lt;/math&gt; and &lt;math&gt;29&lt;/math&gt;, yielding a maximal answer of 38. Since &lt;math&gt;38-25=13&lt;/math&gt;, which is prime, the answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> == Solution 3 (bash) ==<br /> Let &lt;math&gt;2n&lt;/math&gt; be an even integer. Using the [[Chicken McNugget Theorem]] on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as &lt;math&gt;n+n&lt;/math&gt;. We bash each case until we find one that works.<br /> <br /> == Solution 4 (easiest) ==<br /> The easiest method is to notice that any odd number that ends is a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...<br /> <br /> For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9. Hence the largest number that ends with a 4 that satisfies the conditions is 14. <br /> <br /> If you list out all the numbers, you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and anything else that ends with a 3 is bad, so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be &lt;math&gt;\boxed{38}&lt;/math&gt;<br /> <br /> == Solution 5 ==<br /> Claim: The answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> Proof:<br /> It is fairly easy to show 38 can't be split into 2 odd composites.<br /> <br /> Assume there exists an even integer &lt;math&gt;m &gt; 38&lt;/math&gt; that m can't be split into 2 odd composites.<br /> <br /> Then, we can consider m modulo 5.<br /> <br /> If m = 0 mod 5, we can express m = 15 + 5k for some integer k. Since &lt;math&gt;m &gt; 20&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 15 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 1 mod 5, we can express m = 21 + 5k for some integer k. Since &lt;math&gt;m &gt; 26&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 21 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 2 mod 5, we can express m = 27 + 5k for some integer k. Since &lt;math&gt;m &gt; 32&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 27 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 3 mod 5, we can express m = 33 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 4 mod 5, we can express m = 9 + 5k for some integer k. Since &lt;math&gt;m &gt; 14&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> Thus, in all cases we can split m into 2 odd composites, and we arrive at a contradiction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=13|num-a=15}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14&diff=134668 1984 AIME Problems/Problem 14 2020-10-06T02:34:51Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> What is the largest even integer that cannot be written as the sum of two odd composite numbers?<br /> <br /> == Solution 1 ==<br /> <br /> Take an even positive integer &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is either &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;2 \bmod{6}&lt;/math&gt;, or &lt;math&gt;4 \bmod{6}&lt;/math&gt;. Notice that the numbers &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;21&lt;/math&gt;, ... , and in general &lt;math&gt;9 + 6n&lt;/math&gt; for nonnegative &lt;math&gt;n&lt;/math&gt; are odd composites. We now have 3 cases:<br /> <br /> If &lt;math&gt;x \ge 18&lt;/math&gt; and is &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;9 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 44&lt;/math&gt; and is &lt;math&gt;2 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;35 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;35&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 34&lt;/math&gt; and is &lt;math&gt;4 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;25 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;25&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites.<br /> <br /> <br /> Clearly, if &lt;math&gt;x \ge 44&lt;/math&gt;, it can be expressed as a sum of 2 odd composites. However, if &lt;math&gt;x = 42&lt;/math&gt;, it can also be expressed using case 1, and if &lt;math&gt;x = 40&lt;/math&gt;, using case 3. &lt;math&gt;38&lt;/math&gt; is the largest even integer that our cases do not cover. If we examine the possible ways of splitting &lt;math&gt;38&lt;/math&gt; into two addends, we see that no pair of odd composites add to &lt;math&gt;38&lt;/math&gt;. Therefore, &lt;math&gt;\boxed{038}&lt;/math&gt; is the largest possible number that is not expressible as the sum of two odd composite numbers.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be an integer that cannot be written as the sum of two odd composite numbers. If &lt;math&gt;n&gt;33&lt;/math&gt;, then &lt;math&gt;n-9,n-15,n-21,n-25,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; must all be prime (or &lt;math&gt;n-33=1&lt;/math&gt;, which yields &lt;math&gt;n=34=9+25&lt;/math&gt; which does not work). Thus &lt;math&gt;n-9,n-15,n-21,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is &lt;math&gt;5,11,17,23,&lt;/math&gt; and &lt;math&gt;29&lt;/math&gt;, yielding a maximal answer of 38. Since &lt;math&gt;38-25=13&lt;/math&gt;, which is prime, the answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> == Solution 3 (bash) ==<br /> Let &lt;math&gt;2n&lt;/math&gt; be an even integer. Using the [[Chicken McNugget Theorem]] on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as &lt;math&gt;n+n&lt;/math&gt;. We bash each case until we find one that works.<br /> <br /> == Solution 4 (easiest) ==<br /> The easiest method is to notice that any odd number that ends is a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...<br /> <br /> For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9. Hence the largest number that ends with a 4 that satisfies the conditions is 14. <br /> <br /> If you list out all the numbers, you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and anything else that ends with a 3 is bad, so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be &lt;math&gt;\boxed{38}&lt;/math&gt;<br /> <br /> == Solution 5 ==<br /> Claim: The answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> Proof:<br /> It is fairly easy to show 38 can't be split into 2 odd composites.<br /> <br /> Assume there exists an even integer &lt;math&gt;m &gt; 38&lt;/math&gt; that m can't be split into 2 odd composites.<br /> <br /> Then, we can consider m modulo 5.<br /> <br /> If m = 0 mod 5, we can express m = 15 + 5k for some integer k. Since &lt;math&gt;m &gt; 20&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 15 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 1 mod 5, we can express m = 21 + 5k for some integer k. Since &lt;math&gt;m &gt; 26&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 21 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 2 mod 5, we can express m = 27 + 5k for some integer k. Since &lt;math&gt;m &gt; 32&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 27 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 3 mod 5, we can express m = 33 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 4 mod 5, we can express m = 9 + 5k for some integer k. Since &lt;math&gt;m &gt; 14&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; so &lt;math&gt;k \geq 2&lt;/math&gt;. Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> Thus, in all cases we can split m into 2 odd composites, and we arrive at a contradiction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=13|num-a=15}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14&diff=134667 1984 AIME Problems/Problem 14 2020-10-06T02:31:55Z <p>Alexlikemath: Solution 5</p> <hr /> <div>== Problem ==<br /> What is the largest even integer that cannot be written as the sum of two odd composite numbers?<br /> <br /> == Solution 1 ==<br /> <br /> Take an even positive integer &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is either &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;2 \bmod{6}&lt;/math&gt;, or &lt;math&gt;4 \bmod{6}&lt;/math&gt;. Notice that the numbers &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;21&lt;/math&gt;, ... , and in general &lt;math&gt;9 + 6n&lt;/math&gt; for nonnegative &lt;math&gt;n&lt;/math&gt; are odd composites. We now have 3 cases:<br /> <br /> If &lt;math&gt;x \ge 18&lt;/math&gt; and is &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;9 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 44&lt;/math&gt; and is &lt;math&gt;2 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;35 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;35&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 34&lt;/math&gt; and is &lt;math&gt;4 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;25 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;25&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites.<br /> <br /> <br /> Clearly, if &lt;math&gt;x \ge 44&lt;/math&gt;, it can be expressed as a sum of 2 odd composites. However, if &lt;math&gt;x = 42&lt;/math&gt;, it can also be expressed using case 1, and if &lt;math&gt;x = 40&lt;/math&gt;, using case 3. &lt;math&gt;38&lt;/math&gt; is the largest even integer that our cases do not cover. If we examine the possible ways of splitting &lt;math&gt;38&lt;/math&gt; into two addends, we see that no pair of odd composites add to &lt;math&gt;38&lt;/math&gt;. Therefore, &lt;math&gt;\boxed{038}&lt;/math&gt; is the largest possible number that is not expressible as the sum of two odd composite numbers.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be an integer that cannot be written as the sum of two odd composite numbers. If &lt;math&gt;n&gt;33&lt;/math&gt;, then &lt;math&gt;n-9,n-15,n-21,n-25,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; must all be prime (or &lt;math&gt;n-33=1&lt;/math&gt;, which yields &lt;math&gt;n=34=9+25&lt;/math&gt; which does not work). Thus &lt;math&gt;n-9,n-15,n-21,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is &lt;math&gt;5,11,17,23,&lt;/math&gt; and &lt;math&gt;29&lt;/math&gt;, yielding a maximal answer of 38. Since &lt;math&gt;38-25=13&lt;/math&gt;, which is prime, the answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> == Solution 3 (bash) ==<br /> Let &lt;math&gt;2n&lt;/math&gt; be an even integer. Using the [[Chicken McNugget Theorem]] on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as &lt;math&gt;n+n&lt;/math&gt;. We bash each case until we find one that works.<br /> <br /> == Solution 4 (easiest) ==<br /> The easiest method is to notice that any odd number that ends is a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...<br /> <br /> For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9. Hence the largest number that ends with a 4 that satisfies the conditions is 14. <br /> <br /> If you list out all the numbers, you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and anything else that ends with a 3 is bad, so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be &lt;math&gt;\boxed{38}&lt;/math&gt;<br /> <br /> == Solution 5 ==<br /> Claim: The answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> Proof:<br /> It is fairly easy to show 38 can't be split into 2 odd composites.<br /> <br /> Assume there exists an even integer &lt;math&gt;m &gt; 38&lt;/math&gt; that m can't be split into 2 odd composites.<br /> <br /> Then, we can consider m modulo 5.<br /> <br /> If m = 0 mod 5, we can express m = 15 + 5k for some integer k. Since &lt;math&gt;m &gt; 20&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 15 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 1 mod 5, we can express m = 21 + 5k for some integer k. Since &lt;math&gt;m &gt; 26&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 21 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 2 mod 5, we can express m = 27 + 5k for some integer k. Since &lt;math&gt;m &gt; 32&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 27 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 3 mod 5, we can express m = 33 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 4 mod 5, we can express m = 9 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> Thus, in all cases we can split m into 2 odd composites, and we arrive at a contradiction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=13|num-a=15}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14&diff=134666 1984 AIME Problems/Problem 14 2020-10-06T02:27:09Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> What is the largest even integer that cannot be written as the sum of two odd composite numbers?<br /> <br /> == Solution 1 ==<br /> <br /> Take an even positive integer &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is either &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;2 \bmod{6}&lt;/math&gt;, or &lt;math&gt;4 \bmod{6}&lt;/math&gt;. Notice that the numbers &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;21&lt;/math&gt;, ... , and in general &lt;math&gt;9 + 6n&lt;/math&gt; for nonnegative &lt;math&gt;n&lt;/math&gt; are odd composites. We now have 3 cases:<br /> <br /> If &lt;math&gt;x \ge 18&lt;/math&gt; and is &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;9 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 44&lt;/math&gt; and is &lt;math&gt;2 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;35 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;35&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 34&lt;/math&gt; and is &lt;math&gt;4 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;25 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;25&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites.<br /> <br /> <br /> Clearly, if &lt;math&gt;x \ge 44&lt;/math&gt;, it can be expressed as a sum of 2 odd composites. However, if &lt;math&gt;x = 42&lt;/math&gt;, it can also be expressed using case 1, and if &lt;math&gt;x = 40&lt;/math&gt;, using case 3. &lt;math&gt;38&lt;/math&gt; is the largest even integer that our cases do not cover. If we examine the possible ways of splitting &lt;math&gt;38&lt;/math&gt; into two addends, we see that no pair of odd composites add to &lt;math&gt;38&lt;/math&gt;. Therefore, &lt;math&gt;\boxed{038}&lt;/math&gt; is the largest possible number that is not expressible as the sum of two odd composite numbers.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be an integer that cannot be written as the sum of two odd composite numbers. If &lt;math&gt;n&gt;33&lt;/math&gt;, then &lt;math&gt;n-9,n-15,n-21,n-25,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; must all be prime (or &lt;math&gt;n-33=1&lt;/math&gt;, which yields &lt;math&gt;n=34=9+25&lt;/math&gt; which does not work). Thus &lt;math&gt;n-9,n-15,n-21,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is &lt;math&gt;5,11,17,23,&lt;/math&gt; and &lt;math&gt;29&lt;/math&gt;, yielding a maximal answer of 38. Since &lt;math&gt;38-25=13&lt;/math&gt;, which is prime, the answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> == Solution 3 (bash) ==<br /> Let &lt;math&gt;2n&lt;/math&gt; be an even integer. Using the [[Chicken McNugget Theorem]] on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as &lt;math&gt;n+n&lt;/math&gt;. We bash each case until we find one that works.<br /> <br /> == Solution 4 (easiest) ==<br /> The easiest method is to notice that any odd number that ends is a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...<br /> <br /> For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9. Hence the largest number that ends with a 4 that satisfies the conditions is 14. <br /> <br /> If you list out all the numbers, you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and anything else that ends with a 3 is bad, so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be &lt;math&gt;\boxed{38}&lt;/math&gt;<br /> <br /> == Solution 5 ==<br /> Claim: The answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> Proof:<br /> It is fairly easy to show 38 can't be split into 2 odd composites.<br /> <br /> Assume there exists a even integer &lt;math&gt;m &gt; 38&lt;/math&gt; that m can't be split into 2 odd composites.<br /> <br /> Then, we can consider m modulo 5.<br /> <br /> If m = 0 mod 5, we can express m = 15 + 5k for some integer k. Since &lt;math&gt;m &gt; 20&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 15 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 1 mod 5, we can express m = 21 + 5k for some integer k. Since &lt;math&gt;m &gt; 26&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 21 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 2 mod 5, we can express m = 27 + 5k for some integer k. Since &lt;math&gt;m &gt; 32&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 27 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 3 mod 5, we can express m = 33 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 4 mod 5, we can express m = 9 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> Thus, in all cases we can split m into 2 odd composites, and we arrive at a contradiction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> <br /> == See also ==<br /> {{AIME box|year=1984|num-b=13|num-a=15}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14&diff=134665 1984 AIME Problems/Problem 14 2020-10-06T02:26:43Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> What is the largest even integer that cannot be written as the sum of two odd composite numbers?<br /> <br /> == Solution 1 ==<br /> <br /> Take an even positive integer &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is either &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;2 \bmod{6}&lt;/math&gt;, or &lt;math&gt;4 \bmod{6}&lt;/math&gt;. Notice that the numbers &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;21&lt;/math&gt;, ... , and in general &lt;math&gt;9 + 6n&lt;/math&gt; for nonnegative &lt;math&gt;n&lt;/math&gt; are odd composites. We now have 3 cases:<br /> <br /> If &lt;math&gt;x \ge 18&lt;/math&gt; and is &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;9 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 44&lt;/math&gt; and is &lt;math&gt;2 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;35 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;35&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 34&lt;/math&gt; and is &lt;math&gt;4 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;25 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;25&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites.<br /> <br /> <br /> Clearly, if &lt;math&gt;x \ge 44&lt;/math&gt;, it can be expressed as a sum of 2 odd composites. However, if &lt;math&gt;x = 42&lt;/math&gt;, it can also be expressed using case 1, and if &lt;math&gt;x = 40&lt;/math&gt;, using case 3. &lt;math&gt;38&lt;/math&gt; is the largest even integer that our cases do not cover. If we examine the possible ways of splitting &lt;math&gt;38&lt;/math&gt; into two addends, we see that no pair of odd composites add to &lt;math&gt;38&lt;/math&gt;. Therefore, &lt;math&gt;\boxed{038}&lt;/math&gt; is the largest possible number that is not expressible as the sum of two odd composite numbers.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be an integer that cannot be written as the sum of two odd composite numbers. If &lt;math&gt;n&gt;33&lt;/math&gt;, then &lt;math&gt;n-9,n-15,n-21,n-25,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; must all be prime (or &lt;math&gt;n-33=1&lt;/math&gt;, which yields &lt;math&gt;n=34=9+25&lt;/math&gt; which does not work). Thus &lt;math&gt;n-9,n-15,n-21,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is &lt;math&gt;5,11,17,23,&lt;/math&gt; and &lt;math&gt;29&lt;/math&gt;, yielding a maximal answer of 38. Since &lt;math&gt;38-25=13&lt;/math&gt;, which is prime, the answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> == Solution 3 (bash) ==<br /> Let &lt;math&gt;2n&lt;/math&gt; be an even integer. Using the [[Chicken McNugget Theorem]] on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as &lt;math&gt;n+n&lt;/math&gt;. We bash each case until we find one that works.<br /> <br /> == Solution 4 (easiest) ==<br /> The easiest method is to notice that any odd number that ends is a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...<br /> <br /> For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9. Hence the largest number that ends with a 4 that satisfies the conditions is 14. <br /> <br /> If you list out all the numbers, you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and anything else that ends with a 3 is bad, so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be &lt;math&gt;\boxed{38}&lt;/math&gt;<br /> <br /> == Solution 5 ==<br /> Claim: The answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> Proof:<br /> It is fairly easy to show 38 can't be split into 2 odd composites.<br /> <br /> Assume there exists a even integer &lt;math&gt;m &gt; 38&lt;/math&gt; that m can't be split into 2 odd composites.<br /> <br /> Then, we can consider m modulo 5.<br /> <br /> If m = 0 mod 5, we can express m = 15 + 5k for some integer k. Since &lt;math&gt;m &gt; 20&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 15 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 1 mod 5, we can express m = 21 + 5k for some integer k. Since &lt;math&gt;m &gt; 26&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 21 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 2 mod 5, we can express m = 27 + 5k for some integer k. Since &lt;math&gt;m &gt; 32&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 27 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 3 mod 5, we can express m = 33 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 4 mod 5, we can express m = 9 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> Thus, in all cases we can split m into 2 odd composites, and we arrive at a contradiction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> == See also ==<br /> {{AIME box|year=1984|num-b=13|num-a=15}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_14&diff=134664 1984 AIME Problems/Problem 14 2020-10-06T02:26:25Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> What is the largest even integer that cannot be written as the sum of two odd composite numbers?<br /> <br /> == Solution 1 ==<br /> <br /> Take an even positive integer &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;x&lt;/math&gt; is either &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;2 \bmod{6}&lt;/math&gt;, or &lt;math&gt;4 \bmod{6}&lt;/math&gt;. Notice that the numbers &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;15&lt;/math&gt;, &lt;math&gt;21&lt;/math&gt;, ... , and in general &lt;math&gt;9 + 6n&lt;/math&gt; for nonnegative &lt;math&gt;n&lt;/math&gt; are odd composites. We now have 3 cases:<br /> <br /> If &lt;math&gt;x \ge 18&lt;/math&gt; and is &lt;math&gt;0 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;9 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;9&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 44&lt;/math&gt; and is &lt;math&gt;2 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;35 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;35&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites. <br /> <br /> If &lt;math&gt;x\ge 34&lt;/math&gt; and is &lt;math&gt;4 \bmod{6}&lt;/math&gt;, &lt;math&gt;x&lt;/math&gt; can be expressed as &lt;math&gt;25 + (9+6n)&lt;/math&gt; for some nonnegative &lt;math&gt;n&lt;/math&gt;. Note that &lt;math&gt;25&lt;/math&gt; and &lt;math&gt;9+6n&lt;/math&gt; are both odd composites.<br /> <br /> <br /> Clearly, if &lt;math&gt;x \ge 44&lt;/math&gt;, it can be expressed as a sum of 2 odd composites. However, if &lt;math&gt;x = 42&lt;/math&gt;, it can also be expressed using case 1, and if &lt;math&gt;x = 40&lt;/math&gt;, using case 3. &lt;math&gt;38&lt;/math&gt; is the largest even integer that our cases do not cover. If we examine the possible ways of splitting &lt;math&gt;38&lt;/math&gt; into two addends, we see that no pair of odd composites add to &lt;math&gt;38&lt;/math&gt;. Therefore, &lt;math&gt;\boxed{038}&lt;/math&gt; is the largest possible number that is not expressible as the sum of two odd composite numbers.<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;n&lt;/math&gt; be an integer that cannot be written as the sum of two odd composite numbers. If &lt;math&gt;n&gt;33&lt;/math&gt;, then &lt;math&gt;n-9,n-15,n-21,n-25,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; must all be prime (or &lt;math&gt;n-33=1&lt;/math&gt;, which yields &lt;math&gt;n=34=9+25&lt;/math&gt; which does not work). Thus &lt;math&gt;n-9,n-15,n-21,n-27,&lt;/math&gt; and &lt;math&gt;n-33&lt;/math&gt; form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is &lt;math&gt;5,11,17,23,&lt;/math&gt; and &lt;math&gt;29&lt;/math&gt;, yielding a maximal answer of 38. Since &lt;math&gt;38-25=13&lt;/math&gt;, which is prime, the answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> == Solution 3 (bash) ==<br /> Let &lt;math&gt;2n&lt;/math&gt; be an even integer. Using the [[Chicken McNugget Theorem]] on the two smallest odd composite integers that are relatively prime from each other, 9 and 25, show that the maximum is 191, and the maximum even integer is 190 or lower. We use the fact that sufficiently high multiples of 6, 10, 14, 22, etc. can be represented as &lt;math&gt;n+n&lt;/math&gt;. We bash each case until we find one that works.<br /> <br /> == Solution 4 (easiest) ==<br /> The easiest method is to notice that any odd number that ends is a 5 is a composite (except for 5 itself). This means that we will have 15, 25, 35, etc... no matter what. What it also means is that if we look at the end digit, if 15 plus another number will equal that number, then any number that has that same end digit can be added by that same number plus a version of 15, 25, 35...<br /> <br /> For example, let's say we assume our end digit of the number is 4. If we have 5 as one of our end digits, then 9 must be the end digit of the other number. If we go down our list of numbers that end with a 9 and is composite, we will stumble upon the number 9 itself. That means that the number 15+9 is able to be written in a composite form, but also anything that ends with a 4 and is above 15+9. Hence the largest number that ends with a 4 that satisfies the conditions is 14. <br /> <br /> If you list out all the numbers, you will notice that 33 is the largest number where the last digit is not repeated (13 and 23 are not composite). That means that 33+15 and anything else that ends with a 3 is bad, so the largest number that satisfies the conditions is the largest number that ends with a 8 and is below 48. That number would be &lt;math&gt;\boxed{38}&lt;/math&gt;<br /> <br /> == Solution 5 ==<br /> Claim: The answer is &lt;math&gt;\boxed{038}&lt;/math&gt;.<br /> <br /> Proof:<br /> It is fairly easy to show 38 can't be split into 2 odd composites.<br /> <br /> Assume there exists a even integer &lt;math&gt;m &gt; 38&lt;/math&gt; that m can't be split into 2 odd composites.<br /> <br /> Then, we can consider m modulo 5.<br /> <br /> If m = 0 mod 5, we can express m = 15 + 5k for some integer k. Since &lt;math&gt;m &gt; 20&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 15 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 1 mod 5, we can express m = 21 + 5k for some integer k. Since &lt;math&gt;m &gt; 26&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 21 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 2 mod 5, we can express m = 27 + 5k for some integer k. Since &lt;math&gt;m &gt; 32&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt;. Thus, 5k is composite. Since 27 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 3 mod 5, we can express m = 33 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> If m = 4 mod 5, we can express m = 9 + 5k for some integer k. Since &lt;math&gt;m &gt; 38&lt;/math&gt;, &lt;math&gt;k &gt; 1&lt;/math&gt; Thus, 5k is composite. Since 33 is odd and composite, m is even, 5k should be odd as well. <br /> <br /> Thus, in all cases we can split m into 2 odd composites, and we arrive at a contradiction. &lt;cmath&gt;\blacksquare&lt;/cmath&gt;<br /> <br /> -Alexlikemath<br /> == See also ==<br /> {{AIME box|year=1984|num-b=13|num-a=15}}<br /> * [[AIME Problems and Solutions]]<br /> * [[American Invitational Mathematics Examination]]<br /> * [[Mathematics competition resources]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_15&diff=134649 1983 AIME Problems/Problem 15 2020-10-05T21:33:42Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> The adjoining figure shows two intersecting chords in a circle, with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the sine of the central angle of minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this number is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> <br /> &lt;asy&gt;size(100);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=1;<br /> pair O1=(0,0);<br /> pair A=(-0.91,-0.41);<br /> pair B=(-0.99,0.13);<br /> pair C=(0.688,0.728);<br /> pair D=(-0.25,0.97);<br /> path C1=Circle(O1,1);<br /> draw(C1);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> draw(A--D);<br /> draw(B--C);<br /> pair F=intersectionpoint(A--D,B--C);<br /> add(pathticks(A--F,1,0.5,0,3.5));<br /> add(pathticks(F--D,1,0.5,0,3.5));<br /> &lt;/asy&gt;<br /> &lt;!-- [[Image:1983_AIME-15.png|200px]] --&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> -Credit to Adamz for diagram-<br /> &lt;asy&gt;<br /> size(10cm);<br /> import olympiad;<br /> pair O = (0,0);dot(O);label(&quot;$O$&quot;,O,SW);<br /> pair M = (4,0);dot(M);label(&quot;$M$&quot;,M,SE);<br /> pair N = (4,2);dot(N);label(&quot;$N$&quot;,N,NE);<br /> draw(circle(O,5));<br /> pair B = (4,3);dot(B);label(&quot;$B$&quot;,B,NE);<br /> pair C = (4,-3);dot(C);label(&quot;$C$&quot;,C,SE);<br /> draw(B--C);draw(O--M);<br /> pair P = (1.5,2);dot(P);label(&quot;$P$&quot;,P,W);<br /> draw(circle(P,2.5));<br /> pair A=(3,4);dot(A);label(&quot;$A$&quot;,A,NE);<br /> draw(O--A);<br /> draw(O--B);<br /> pair Q = (1.5,0); dot(Q); label(&quot;$Q$&quot;,Q,S);<br /> pair R = (3,0); dot(R); label(&quot;$R$&quot;,R,S);<br /> draw(P--Q,dotted); draw(A--R,dotted);<br /> pair D=(5,0); dot(D); label(&quot;$D$&quot;,D,E);<br /> draw(A--D);<br /> &lt;/asy&gt;Let &lt;math&gt;A&lt;/math&gt; be any fixed point on circle &lt;math&gt;O&lt;/math&gt;, and let &lt;math&gt;AD&lt;/math&gt; be a chord of circle &lt;math&gt;O&lt;/math&gt;. The [[locus]] of midpoints &lt;math&gt;N&lt;/math&gt; of the chord &lt;math&gt;AD&lt;/math&gt; is a circle &lt;math&gt;P&lt;/math&gt;, with diameter &lt;math&gt;AO&lt;/math&gt;. Generally, the circle &lt;math&gt;P&lt;/math&gt; can intersect the chord &lt;math&gt;BC&lt;/math&gt; at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle &lt;math&gt;P&lt;/math&gt; is tangent to &lt;math&gt;BC&lt;/math&gt; at point &lt;math&gt;N&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;M&lt;/math&gt; be the midpoint of the chord &lt;math&gt;BC&lt;/math&gt;. From right triangle &lt;math&gt;OMB&lt;/math&gt;, we have &lt;math&gt;OM = \sqrt{OB^2 - BM^2} =4&lt;/math&gt;. This gives &lt;math&gt;\tan \angle BOM = \frac{BM}{OM} = \frac 3 4&lt;/math&gt;.<br /> <br /> Notice that the distance &lt;math&gt;OM&lt;/math&gt; equals &lt;math&gt;PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the radius of circle &lt;math&gt;P&lt;/math&gt;. <br /> <br /> Hence &lt;cmath&gt;\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5&lt;/cmath&gt; (where &lt;math&gt;R&lt;/math&gt; represents the radius, &lt;math&gt;5&lt;/math&gt;, of the large circle given in the question). Therefore, since &lt;math&gt;\angle AOM&lt;/math&gt; is clearly acute, we see that &lt;cmath&gt;\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3&lt;/cmath&gt;<br /> <br /> Next, notice that &lt;math&gt;\angle AOB = \angle AOM - \angle BOM&lt;/math&gt;. We can therefore apply the subtraction formula for &lt;math&gt;\tan&lt;/math&gt; to obtain &lt;cmath&gt;\tan \angle AOB =\frac{\tan \angle AOM - \tan \angle BOM}{1 + \tan \angle AOM \cdot \tan \angle BOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}&lt;/cmath&gt; It follows that &lt;math&gt;\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}&lt;/math&gt;, such that the answer is &lt;math&gt;7 \cdot 25=\boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> This solution, while similar to Solution 1, is far more motivated and less contrived.<br /> <br /> Firstly, we note the statement in the problem that &quot;&lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; and bisected by &lt;math&gt;BC&lt;/math&gt;&quot; &amp;ndash; what is its significance? What is the criterion for this statement to be true?<br /> <br /> We consider the locus of midpoints of the chords from &lt;math&gt;A&lt;/math&gt;. It is well-known that this is the circle with diameter &lt;math&gt;AO&lt;/math&gt;, where &lt;math&gt;O&lt;/math&gt; is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor &lt;math&gt;\frac{1}{2}&lt;/math&gt; and center &lt;math&gt;A&lt;/math&gt;. Thus, the locus is the result of the dilation with scale factor &lt;math&gt;\frac{1}{2}&lt;/math&gt; and centre &lt;math&gt;A&lt;/math&gt; of circle &lt;math&gt;O&lt;/math&gt;. Let the center of this circle be &lt;math&gt;P&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt; if they cross at some point &lt;math&gt;N&lt;/math&gt; on the circle. Moreover, since &lt;math&gt;AD&lt;/math&gt; is the only chord, &lt;math&gt;BC&lt;/math&gt; must be tangent to the circle &lt;math&gt;P&lt;/math&gt;.<br /> <br /> The rest of this problem is straightforward.<br /> <br /> Our goal is to find &lt;math&gt;\sin \angle AOB = \sin{\left(\angle AOM - \angle BOM\right)}&lt;/math&gt;, where &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;. We have &lt;math&gt;BM=3&lt;/math&gt; and &lt;math&gt;OM=4&lt;/math&gt;.<br /> Let &lt;math&gt;R&lt;/math&gt; be the projection of &lt;math&gt;A&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;, and similarly let &lt;math&gt;Q&lt;/math&gt; be the projection of &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;. Then it remains to find &lt;math&gt;AR&lt;/math&gt; so that we can use the addition formula for &lt;math&gt;\sin&lt;/math&gt;.<br /> <br /> As &lt;math&gt;PN&lt;/math&gt; is a radius of circle &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;PN=2.5&lt;/math&gt;, and similarly, &lt;math&gt;PO=2.5&lt;/math&gt;. Since &lt;math&gt;OM=4&lt;/math&gt;, we have &lt;math&gt;OQ=OM-QM=OM-PN=4-2.5=1.5&lt;/math&gt;. Thus &lt;math&gt;PQ=\sqrt{2.5^2-1.5^2}=2&lt;/math&gt;.<br /> <br /> Further, we see that &lt;math&gt;\triangle OAR&lt;/math&gt; is a dilation of &lt;math&gt;\triangle OPQ&lt;/math&gt; about center &lt;math&gt;O&lt;/math&gt; with scale factor &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;AR=2PQ=4&lt;/math&gt;.<br /> <br /> Lastly, we apply the formula:<br /> &lt;cmath&gt; \sin{\left(\angle AOM - \angle BOM\right)} = \sin \angle AOM \cos \angle BOM - \sin \angle BOM \cos \angle AOM = \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)=\frac{7}{25}&lt;/cmath&gt;<br /> Thus the answer is &lt;math&gt;7\cdot25=\boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 3 (coordinate geometry) ===<br /> [[File:Aime1983p15s2.png|500px|link=]]<br /> <br /> Let the circle have equation &lt;math&gt;x^2 + y^2 = 25&lt;/math&gt;, with centre &lt;math&gt;O(0,0)&lt;/math&gt;. Since &lt;math&gt;BC=6&lt;/math&gt;, we can calculate (by the Pythagorean Theorem) that the distance from &lt;math&gt;O&lt;/math&gt; to the line &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. Therefore, we can let &lt;math&gt;B=(3,4)&lt;/math&gt; and &lt;math&gt;C=(-3,4)&lt;/math&gt;. Now, assume that &lt;math&gt;A&lt;/math&gt; is any point on the major arc BC, and &lt;math&gt;D&lt;/math&gt; any point on the minor arc BC. We can write &lt;math&gt;A=(5 \cos \alpha, 5 \sin \alpha)&lt;/math&gt;, where &lt;math&gt;\alpha&lt;/math&gt; is the angle measured from the positive &lt;math&gt;x&lt;/math&gt; axis to the ray &lt;math&gt;OA&lt;/math&gt;. It will also be convenient to define &lt;math&gt;\angle XOB = \alpha_0&lt;/math&gt;. <br /> <br /> Firstly, since &lt;math&gt;B&lt;/math&gt; must lie in the minor arc &lt;math&gt;AD&lt;/math&gt;, we see that &lt;math&gt;\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)&lt;/math&gt;. However, since the midpoint of &lt;math&gt;AD&lt;/math&gt; must lie on &lt;math&gt;BC&lt;/math&gt;, and the highest possible &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;D&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt;, we see that the &lt;math&gt;y&lt;/math&gt;-coordinate cannot be lower than &lt;math&gt;3&lt;/math&gt;, that is, &lt;math&gt;\alpha \in \left[\sin^{-1}\frac{3}{5},\alpha_0\right)&lt;/math&gt;.<br /> <br /> Secondly, there is a theorem that says that, in a circle, if a chord is bisected by a radius, then they must be perpendicular. Therefore, suppose that &lt;math&gt;P&lt;/math&gt; is the intersection point of &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, so that by the theorem, &lt;math&gt;OP&lt;/math&gt; is perpendicular to &lt;math&gt;AD&lt;/math&gt;. So, if &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;, this means that &lt;math&gt;P&lt;/math&gt; is the only point on the chord &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;OP&lt;/math&gt; is perpendicular to &lt;math&gt;AD&lt;/math&gt;. Now suppose that &lt;math&gt;P=(p,4)&lt;/math&gt;, where &lt;math&gt;p \in (-3,3)&lt;/math&gt;. The fact that &lt;math&gt;OP&lt;/math&gt; must be perpendicular to &lt;math&gt;AD&lt;/math&gt; is equivalent to the following equation:<br /> <br /> &lt;cmath&gt; -1 = \left(\text{slope of } OP\right)\left(\text{slope of } AP\right)&lt;/cmath&gt;<br /> which becomes<br /> &lt;cmath&gt; -1 = \frac{4}{p} \cdot \frac{5\sin \alpha - 4}{5\cos \alpha - p}&lt;/cmath&gt;<br /> <br /> This rearranges to<br /> <br /> &lt;cmath&gt; p^2 - (5\cos \alpha)p + 16 - 20 \sin \alpha = 0&lt;/cmath&gt;<br /> <br /> Given that this equation must have only one real root &lt;math&gt;p\in (-3,3)&lt;/math&gt;, we study the following function:<br /> <br /> &lt;cmath&gt;f(x) = x^2 - (5\cos \alpha)x + 16 - 20 \sin \alpha&lt;/cmath&gt;<br /> <br /> First, by the fact that the equation &lt;math&gt;f(x)=0&lt;/math&gt; has real solutions, its discriminant &lt;math&gt;\Delta&lt;/math&gt; must be non-negative, so we calculate<br /> <br /> &lt;cmath&gt; \begin{split}\Delta &amp; = (5\cos \alpha)^2 - 4(16-20\sin \alpha) \\<br /> &amp; = 25 (1- \sin^2 \alpha) - 64 + 80 \sin \alpha \\<br /> &amp; = -25 \sin^2 \alpha + 80\sin \alpha - 39 \\<br /> &amp; = (13 - 5\sin \alpha)(5\sin \alpha - 3)\end{split}&lt;/cmath&gt;<br /> <br /> It is obvious that this is in fact non-negative. If it is actually zero, then &lt;math&gt;\sin \alpha = \frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\cos \alpha = \frac{4}{5}&lt;/math&gt;. In this case, &lt;math&gt;p = \frac{5\cos \alpha}{2} = 2 \in (-3,3)&lt;/math&gt;, so we have found a possible solution. We thus calculate &lt;math&gt;\sin(\text{central angle of minor arc } AB) = \sin (\alpha_0 - \alpha) = \frac{4}{5}\cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{5} = \frac{7}{25}&lt;/math&gt; by the subtraction formula for &lt;math&gt;\sin&lt;/math&gt;. This means that the answer is &lt;math&gt;7 \cdot 25 = 175&lt;/math&gt;.<br /> <br /> === Addendum to Solution 3 ===<br /> Since this is an AIME problem, we can assume that we are done since we have found one possible case. However, in a full-solution contest, we could not assume that the answer is unique, but would need to prove that this is the unique solution. This can be proven as follows.<br /> <br /> Suppose that &lt;math&gt;\Delta &gt; 0&lt;/math&gt;, which would mean that there could be two real roots of &lt;math&gt;f(x)&lt;/math&gt;, one lying in the interval &lt;math&gt;(-3,3)&lt;/math&gt;, and another outside of it. We also see, by [[Vieta's Formulas]], that the average of the two roots is &lt;math&gt;\frac{5\cos \alpha}{2}&lt;/math&gt;, which is non-negative, so the root outside of &lt;math&gt;(-3,3)&lt;/math&gt; must be no less than &lt;math&gt;3&lt;/math&gt;. By considering the graph of &lt;math&gt;y=f(x)&lt;/math&gt;, which is a &quot;U-shaped&quot; parabola, it is now evident that &lt;math&gt;f(-3) &gt; 0&lt;/math&gt; and &lt;math&gt;f(3)\leq 0&lt;/math&gt;. We can just use the second inequality:<br /> <br /> &lt;cmath&gt;0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt; 3\cos \alpha + 4 \sin \alpha \geq 5 &lt;/cmath&gt;<br /> <br /> The only way for this inequality to be satisfied is when &lt;math&gt;A=B&lt;/math&gt; (by applying the Cauchy-Schwarz inequality, or just plotting the line &lt;math&gt;3x+4y=5&lt;/math&gt; to see that point &lt;math&gt;A&lt;/math&gt; can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point &lt;math&gt;A&lt;/math&gt; lies in the half-plane above the line &lt;math&gt;3x+4y=5&lt;/math&gt;, inclusive, and the half-plane below the line &lt;math&gt;-3x+4y=5&lt;/math&gt;, exclusive. This can be seen to be impossible by drawing the lines and observing that the intersection of the two half-planes does not share any point with the circle.)<br /> ===Solution 4===<br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. Fix &lt;math&gt;B,C,&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt;. Then, as &lt;math&gt;D&lt;/math&gt; moves around the circle, the locus of the midpoints of &lt;math&gt;AD&lt;/math&gt; is clearly a circle. Since the problems gives that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; bisected by &lt;math&gt;BC&lt;/math&gt;, it follows that the circle with diameter &lt;math&gt;DO&lt;/math&gt; and &lt;math&gt;AO&lt;/math&gt; is tangent to &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> Now, let the intersection of &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt; and let the midpoint of &lt;math&gt;AO&lt;/math&gt; (the center of the circle tangent to &lt;math&gt;BC&lt;/math&gt; that we described beforehand) be &lt;math&gt;F&lt;/math&gt;. Drop the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call its intersection with &lt;math&gt;BC&lt;/math&gt; &lt;math&gt;K&lt;/math&gt;. Drop the perpendicular from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;KO&lt;/math&gt; and call its intersection with &lt;math&gt;KO&lt;/math&gt; &lt;math&gt;L&lt;/math&gt;. Clearly, &lt;math&gt;KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4&lt;/math&gt; and since &lt;math&gt;EF&lt;/math&gt; is radius, it equals &lt;math&gt;\frac{5}{2}&lt;/math&gt;. The same applies for &lt;math&gt;FO&lt;/math&gt;, which also equals &lt;math&gt;\frac{5}{2}&lt;/math&gt;. By the Pythagorean theorem, we deduce that &lt;math&gt;FL = 2&lt;/math&gt;, so &lt;math&gt;EK = 2&lt;/math&gt;. This is very important information! Now we know that &lt;math&gt;BE = 1&lt;/math&gt;, so by Power of a Point, &lt;math&gt;AE = ED = \sqrt{5}&lt;/math&gt;.<br /> <br /> We’re almost there! Since by the Pythagorean theorem, &lt;math&gt;ED^2 + EO^2 = 5&lt;/math&gt;, we deduce that &lt;math&gt;EO = 2\sqrt{5}&lt;/math&gt;. &lt;math&gt;EC=OC=5&lt;/math&gt;, so &lt;math&gt;\sin (CEO) = \frac{2\sqrt{5}}{5}&lt;/math&gt;. Furthermore, since &lt;math&gt;\sin (CEO) = \cos(DEC)&lt;/math&gt;, we know that &lt;math&gt;\cos (DEC) = \frac{2\sqrt{5}}{5}&lt;/math&gt;. By the law of cosines,<br /> &lt;cmath&gt;DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10&lt;/cmath&gt;Therefore, &lt;math&gt;DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}&lt;/math&gt;. Now, drop the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BA&lt;/math&gt; and call its intersection with &lt;math&gt;BA&lt;/math&gt; &lt;math&gt;Z&lt;/math&gt;. Then, by the Pythagorean theorem, &lt;math&gt;OZ = \frac{7\sqrt{2}}{2}&lt;/math&gt;. Thus, &lt;math&gt;\sin (BOZ) = \frac{\sqrt{2}}{10}&lt;/math&gt; and &lt;math&gt;\cos (BOZ) = \frac{7\sqrt{2}}{10}&lt;/math&gt;. As a result, &lt;math&gt;\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}&lt;/math&gt;. &lt;math&gt;7 \cdot 25 = \boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> [[Image:Dgram.png|thumb|none|800px]]<br /> <br /> Let I be the intersection of AD and BC.<br /> <br /> Lemma: &lt;math&gt;AI = ID&lt;/math&gt; if and only if &lt;math&gt;\angle AIO = 90&lt;/math&gt;.<br /> <br /> Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If &lt;math&gt;\angle AIO = 90&lt;/math&gt;, We can get &lt;math&gt;\triangle AIO \cong \triangle OID&lt;/math&gt;<br /> <br /> Let be this the circle with diameter AO.<br /> <br /> Thus, we get &lt;math&gt;\angle AIO = 90&lt;/math&gt;, implying I must lie on &lt;math&gt;\omega&lt;/math&gt;. I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC.<br /> <br /> Now, create a coordinate system such that the origin is O and the y-axis is perpendicular to BC. The positive x direction is from B to C.<br /> <br /> Let Z be (0,5).<br /> Let Y be (-5,0).<br /> Let X be the center of &lt;math&gt;\omega&lt;/math&gt;. Since &lt;math&gt;\omega&lt;/math&gt;'s radius is &lt;math&gt;\frac{5}{2}&lt;/math&gt;, the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so &lt;math&gt;sin(XOY) = sin(AOY) = \frac{3}{5}&lt;/math&gt;. &lt;math&gt;sin(BOZ) = \frac{3}{5}&lt;/math&gt;. If we let &lt;math&gt;sin(\theta) = \frac{3}{5}&lt;/math&gt;, we can find that what we are looking for is &lt;math&gt;sin(90 - 2\theta)&lt;/math&gt;, which we can evaluate and get &lt;math&gt;\frac{7}{25} \implies \boxed{175}&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_22&diff=132802 2007 AMC 12A Problems/Problem 22 2020-08-31T04:42:09Z <p>Alexlikemath: </p> <hr /> <div>{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #22]] and [[2007 AMC 10A Problems/Problem 25|2007 AMC 10A #25]]}}<br /> == Problem ==<br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, let &lt;math&gt;S(n)&lt;/math&gt; denote the sum of the digits of &lt;math&gt;n.&lt;/math&gt; For how many values of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;n + S(n) + S(S(n)) = 2007?&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5&lt;/math&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> For the sake of notation let &lt;math&gt;T(n) = n + S(n) + S(S(n))&lt;/math&gt;. Obviously &lt;math&gt;n&lt;2007&lt;/math&gt;. Then the maximum value of &lt;math&gt;S(n) + S(S(n))&lt;/math&gt; is when &lt;math&gt;n = 1999&lt;/math&gt;, and the sum becomes &lt;math&gt;28 + 10 = 38&lt;/math&gt;. So the minimum bound is &lt;math&gt;1969&lt;/math&gt;. We do [[casework]] upon the tens digit:<br /> <br /> Case 1: &lt;math&gt;196u \Longrightarrow u = 9&lt;/math&gt;. Easy to directly disprove.<br /> <br /> Case 2: &lt;math&gt;197u&lt;/math&gt;. &lt;math&gt;S(n) = 1 + 9 + 7 + u = 17 + u&lt;/math&gt;, and &lt;math&gt;S(S(n)) = 8+u&lt;/math&gt; if &lt;math&gt;u \le 2&lt;/math&gt; and &lt;math&gt;S(S(n)) = 2 + (u-3) = u-1&lt;/math&gt; otherwise. <br /> <br /> :Subcase a: &lt;math&gt;T(n) = 1970 + u + 17 + u + 8 + u = 1995 + 3u = 2007 \Longrightarrow u = 4&lt;/math&gt;. This exceeds our bounds, so no solution here.<br /> :Subcase b: &lt;math&gt;T(n) = 1970 + u + 17 + u + u - 1 = 1986 + 3u = 2007 \Longrightarrow u = 7&lt;/math&gt;. First solution.<br /> <br /> Case 3: &lt;math&gt;198u&lt;/math&gt;. &lt;math&gt;S(n) = 18 + u&lt;/math&gt;, and &lt;math&gt;S(S(n)) = 9 + u&lt;/math&gt; if &lt;math&gt;u \le 1&lt;/math&gt; and &lt;math&gt;2 + (u-2) = u&lt;/math&gt; otherwise.<br /> <br /> :Subcase a: &lt;math&gt;T(n) = 1980 + u + 18 + u + 9 + u = 2007 + 3u = 2007 \Longrightarrow u = 0&lt;/math&gt;. Second solution.<br /> :Subcase b: &lt;math&gt;T(n) = 1980 + u + 18 + u + u = 1998 + 3u = 2007 \Longrightarrow u = 3&lt;/math&gt;. Third solution. <br /> <br /> Case 4: &lt;math&gt;199u&lt;/math&gt;. But &lt;math&gt;S(n) &gt; 19&lt;/math&gt;, and &lt;math&gt;n + S(n)&lt;/math&gt; clearly sum to &lt;math&gt;&gt; 2007&lt;/math&gt;.<br /> <br /> Case 5: &lt;math&gt;200u&lt;/math&gt;. So &lt;math&gt;S(n) = 2 + u&lt;/math&gt; and &lt;math&gt;S(S(n)) = 2 + u&lt;/math&gt; (recall that &lt;math&gt;n &lt; 2007 &lt;/math&gt;), and &lt;math&gt;2000 + u + 2 + u + 2 + u = 2004 + 3u = 2007 \Longrightarrow u = 1&lt;/math&gt;. Fourth solution.<br /> <br /> In total we have &lt;math&gt;4 \mathrm{(D)}&lt;/math&gt; solutions, which are &lt;math&gt;1977, 1980, 1983, &lt;/math&gt; and &lt;math&gt;2001&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Clearly, &lt;math&gt;n &gt; 1950&lt;/math&gt;. We can break this into three cases:<br /> <br /> Case 1: &lt;math&gt;n \geq 2000&lt;/math&gt;<br /> <br /> :Inspection gives &lt;math&gt;n = 2001&lt;/math&gt;.<br /> <br /> Case 2: &lt;math&gt;n &lt; 2000&lt;/math&gt;, &lt;math&gt;n = 19xy&lt;/math&gt; (not to be confused with &lt;math&gt;19*x*y&lt;/math&gt;), &lt;math&gt;x + y &lt; 10 &lt;/math&gt;<br /> <br /> :If you set up an equation, it reduces to<br /> <br /> &lt;math&gt;4x + y = 32&lt;/math&gt;<br /> <br /> :which has as its only solution satisfying the constraints &lt;math&gt;x = 8&lt;/math&gt;, &lt;math&gt;y = 0&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;n &lt; 2000&lt;/math&gt;, &lt;math&gt;n = 19xy&lt;/math&gt;, &lt;math&gt;x + y \geq 10&lt;/math&gt;<br /> <br /> :This reduces to<br /> <br /> :&lt;math&gt;4x + y = 35&lt;/math&gt;. The only two solutions satisfying the constraints for this equation are &lt;math&gt;x = 7&lt;/math&gt;, &lt;math&gt;y = 7&lt;/math&gt; and &lt;math&gt;x = 8&lt;/math&gt;, &lt;math&gt;y = 3&lt;/math&gt;. <br /> <br /> The solutions are thus &lt;math&gt;1977, 1980, 1983, 2001&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(D)}\ 4&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> As in Solution 1, we note that &lt;math&gt;S(n)\leq 28&lt;/math&gt; and &lt;math&gt;S(S(n))\leq 10&lt;/math&gt;. &lt;br/&gt;<br /> Obviously, &lt;math&gt;n\equiv S(n)\equiv S(S(n)) \pmod 9&lt;/math&gt;. &lt;br/&gt;<br /> As &lt;math&gt;2007\equiv 0 \pmod 9&lt;/math&gt;, this means that &lt;math&gt;n\bmod 9 \in\{0,3,6\}&lt;/math&gt;, or equivalently that &lt;math&gt;n\equiv S(n)\equiv S(S(n))\equiv 0 \pmod 3&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;S(S(n))\in\{3,6,9\}&lt;/math&gt;. For each possible &lt;math&gt;S(S(n))&lt;/math&gt; we get three possible &lt;math&gt;S(n)&lt;/math&gt;. &lt;br/&gt;<br /> (E. g., if &lt;math&gt;S(S(n))=6&lt;/math&gt;, then &lt;math&gt;S(n)=x&lt;/math&gt; is a number such that &lt;math&gt;x\leq 28&lt;/math&gt; and &lt;math&gt;S(x)=6&lt;/math&gt;, therefore &lt;math&gt;S(n)\in\{6,15,24\}&lt;/math&gt;.)<br /> <br /> For each of these nine possibilities we compute &lt;math&gt;n_{?}&lt;/math&gt; as &lt;math&gt;2007-S(n)-S(S(n))&lt;/math&gt; and check whether &lt;math&gt;S(n_{?})=S(n)&lt;/math&gt;. &lt;br/&gt;<br /> We'll find out that out of the 9 cases, in 4 the value &lt;math&gt;n_{?}&lt;/math&gt; has the correct sum of digits. &lt;br/&gt;<br /> This happens for &lt;math&gt;n_{?}\in \{ 1977, 1980, 1983, 2001 \}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> *This solution is not a good solution, but is viable for in contest situations<br /> Clearly &lt;math&gt;n\equiv S(n) \pmod 9&lt;/math&gt;. Thus, &lt;cmath&gt;n+S(n)+S(S(n))\equiv 0 \pmod 9 \implies n\equiv 0\pmod 3.&lt;/cmath&gt;<br /> Now we need a bound for &lt;math&gt;n&lt;/math&gt;. It is clear that the maximum for &lt;math&gt;S(n)=36&lt;/math&gt; (from &lt;math&gt;n=9999&lt;/math&gt;) which means the maximum for &lt;math&gt;S(n)+S(S(n))&lt;/math&gt; is &lt;math&gt;45&lt;/math&gt;. This means that &lt;math&gt;n\geq 1962&lt;/math&gt;.<br /> *Warning: This is where you will cringe badly<br /> Now check all multiples of &lt;math&gt;3&lt;/math&gt; from &lt;math&gt;1962&lt;/math&gt; to &lt;math&gt;2007&lt;/math&gt; and we find that only &lt;math&gt;n=1977, 1980, 1983, 2001&lt;/math&gt; work, so our answer is &lt;math&gt;\mathrm{(D)}\ 4&lt;/math&gt;.<br /> <br /> Remark: this may seem time consuming, but in reality, calculating &lt;math&gt;n+S(n)+S(S(n))&lt;/math&gt; for &lt;math&gt;16&lt;/math&gt; values is actually very quick, so this solution would only take approximately 3-5 minutes, helpful in a contest.<br /> <br /> ==Solution 5 (Rigorous)==<br /> Let the number of digits of &lt;math&gt;n&lt;/math&gt; be &lt;math&gt;m&lt;/math&gt;. If &lt;math&gt;m = 5&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; will already be greater than &lt;math&gt;2007&lt;/math&gt;. Notice that &lt;math&gt;S(n)&lt;/math&gt; is always at most &lt;math&gt;9m&lt;/math&gt;. Then if &lt;math&gt;m = 3&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; will be at most &lt;math&gt;999&lt;/math&gt;, &lt;math&gt;S(n)&lt;/math&gt; will be at most &lt;math&gt;27&lt;/math&gt;, and &lt;math&gt;S(S(n))&lt;/math&gt; will be even smaller than &lt;math&gt;27&lt;/math&gt;. Clearly we cannot reach a sum of &lt;math&gt;2007&lt;/math&gt;, unless &lt;math&gt;m = 4&lt;/math&gt; (i.e. &lt;math&gt;n&lt;/math&gt; has &lt;math&gt;4&lt;/math&gt; digits).<br /> <br /> Then, let &lt;math&gt;n&lt;/math&gt; be a four digit number in the form &lt;math&gt;1000a + 100b + 10c + d&lt;/math&gt;. Then &lt;math&gt;S(n) = a + b + c + d&lt;/math&gt;.<br /> <br /> &lt;math&gt;S(S(n))&lt;/math&gt; is the sum of the digits of &lt;math&gt;a + b + c + d&lt;/math&gt;. We can represent &lt;math&gt;S(S(n))&lt;/math&gt; as the sum of the tens digit and the ones digit of &lt;math&gt;S(n)&lt;/math&gt;. The tens digit in the form of a decimal is<br /> <br /> <br /> &lt;math&gt;\frac{a + b + c + d}{10}&lt;/math&gt;.<br /> <br /> <br /> To remove the decimal portion, we can simply take the floor of the expression,<br /> <br /> <br /> &lt;math&gt;\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> Now that we have expressed the tens digit, we can express the ones digit as &lt;math&gt;S(S(n)) -10&lt;/math&gt; times the above expression, or<br /> <br /> <br /> &lt;math&gt;a + b + c + d - 10\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> Adding the two expressions yields the value of &lt;math&gt;S(S(n))&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt; = a + b + c + d - 9\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> Combining this expression to the ones for &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;S(n)&lt;/math&gt; yields<br /> <br /> <br /> &lt;math&gt;1002a + 102b + 12c + 3d - 9\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> Setting this equal to &lt;math&gt;2007&lt;/math&gt; and rearranging a bit yields<br /> <br /> <br /> &lt;math&gt;12c + 3d = 2007 - 1002a - 102b + 9\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;4c + d = 669 - 334a - 34b + 3\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> (The reason for this slightly weird arrangement will soon become evident)<br /> <br /> <br /> Now we examine the possible values of &lt;math&gt;a&lt;/math&gt;. If &lt;math&gt;a \ge 3&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; is already too large. &lt;math&gt;a&lt;/math&gt; must also be greater than &lt;math&gt;0&lt;/math&gt;, or &lt;math&gt;n&lt;/math&gt; would be a &lt;math&gt;3&lt;/math&gt;-digit number. Therefore, &lt;math&gt;a = 1 \, \text{or} \, 2&lt;/math&gt;. Now we examine by case.<br /> <br /> If &lt;math&gt;a = 2&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must both be &lt;math&gt;0&lt;/math&gt; (otherwise &lt;math&gt;n&lt;/math&gt; would already be greater than &lt;math&gt;2007&lt;/math&gt;). Substituting these values into the equation yields<br /> <br /> <br /> &lt;math&gt;d = 1 + 3\lfloor\frac{2 + d}{10}\rfloor&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;d=1&lt;/math&gt;.<br /> <br /> <br /> Sure enough, &lt;math&gt;2001 + (2+1) + 3=2007&lt;/math&gt;.<br /> <br /> Now we move onto the case where &lt;math&gt;a = 1&lt;/math&gt;. Then our initial equation simplifies to<br /> <br /> <br /> &lt;math&gt;4c + d = 335 - 34b + 3\lfloor\frac{1 + b + c + d}{10}\rfloor&lt;/math&gt;<br /> <br /> <br /> Since &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; can each be at most &lt;math&gt;9&lt;/math&gt;, we substitute that value to find the lower bound of &lt;math&gt;b&lt;/math&gt;. Doing so yields<br /> <br /> <br /> &lt;math&gt;34b \ge 290 + 3\lfloor\frac{19 + b}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> The floor expression is at least &lt;math&gt;3\lfloor\frac{19}{10}\rfloor=3&lt;/math&gt; , so the right-hand side is at least &lt;math&gt;293&lt;/math&gt;. Solving for &lt;math&gt;b&lt;/math&gt;, we see that &lt;math&gt;b \ge 9 &lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;b=9&lt;/math&gt;. Again, we substitute for &lt;math&gt;b&lt;/math&gt; and the equation becomes<br /> <br /> <br /> &lt;math&gt;4c + d = 29 + 3\lfloor\frac{10 + c + d}{10}\rfloor&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;4c + d = 32 + 3\lfloor\frac{c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> Just like we did for &lt;math&gt;b&lt;/math&gt;, we can find the lower bound of &lt;math&gt;c&lt;/math&gt; by assuming &lt;math&gt;d = 9&lt;/math&gt; and solving:<br /> <br /> <br /> &lt;math&gt;4c + 9 \ge 29 + 3\lfloor\frac{c + 9}{10}\rfloor&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;4c \ge 20 + 3\lfloor\frac{c + 9}{10}\rfloor&lt;/math&gt;<br /> <br /> <br /> The right hand side is &lt;math&gt;20&lt;/math&gt; for &lt;math&gt;c=0&lt;/math&gt; and &lt;math&gt;23&lt;/math&gt; for &lt;math&gt;c \ge 1&lt;/math&gt;. Solving for c yields &lt;math&gt;c \ge 6&lt;/math&gt;. Looking back at the previous equation, the floor expression is &lt;math&gt;0&lt;/math&gt; for &lt;math&gt;c+d \le 9&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; for &lt;math&gt;c+d \ge 10&lt;/math&gt;. Thus, the right-hand side is &lt;math&gt;32&lt;/math&gt; for &lt;math&gt;c+d \le 9&lt;/math&gt; and &lt;math&gt;35&lt;/math&gt; for &lt;math&gt;c+d \ge 10&lt;/math&gt;. We can solve these two scenarios as systems of equations/inequalities:<br /> <br /> &lt;math&gt;4c+d = 32&lt;/math&gt;<br /> <br /> &lt;math&gt;c+d \le 9&lt;/math&gt;<br /> <br /> and<br /> <br /> &lt;math&gt;4c+d=35&lt;/math&gt;<br /> <br /> &lt;math&gt;c+d \ge 10&lt;/math&gt;<br /> <br /> Solving yields three pairs &lt;math&gt;(c, d):&lt;/math&gt; &lt;math&gt;(8, 0)&lt;/math&gt;; &lt;math&gt;(8, 3)&lt;/math&gt;; and &lt;math&gt;(7, 7)&lt;/math&gt;. Checking the numbers &lt;math&gt;1980&lt;/math&gt;, &lt;math&gt;1983&lt;/math&gt;, and &lt;math&gt;1977&lt;/math&gt;; we find that all three work. Therefore there are a total of &lt;math&gt;4&lt;/math&gt; possibilities for &lt;math&gt;n&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;.<br /> <br /> Note: Although this solution takes a while to read (as well as to write) the actual time it takes to think through the process above is very short in comparison to the solution length.<br /> <br /> ==Solution 6==<br /> <br /> <br /> Rearranging, we get <br /> 2007 - S(n) - S(S(n)) = n.<br /> <br /> We can now try S(n) = 1-28, to find values of n. Then, you need to check whether S(n) = sum digits of n. If so, you found a solution!<br /> <br /> The bash involves making a table with 3 columns: S(n), n, S(n). We write the numbers 1-28 in the S(n) column, we can find n using the above equation, and the third column we can just find the sum of the digits of n, and if the first column/3rd column match, we have a solution.<br /> <br /> You will in the end find S(n) = 3, 18, 21, 24 yield solutions, which corrispond to n = 2001, 1980, 1983, 1977. There are 4 solutions, so the answer is &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> == See also ==<br /> {{AMC12 box|year=2007|ab=A|num-b=21|num-a=23}}<br /> {{AMC10 box|year=2007|ab=A|num-b=24|after=Last question}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12A_Problems/Problem_22&diff=132801 2007 AMC 12A Problems/Problem 22 2020-08-31T04:39:56Z <p>Alexlikemath: </p> <hr /> <div>{{duplicate|[[2007 AMC 12A Problems|2007 AMC 12A #22]] and [[2007 AMC 10A Problems/Problem 25|2007 AMC 10A #25]]}}<br /> == Problem ==<br /> For each positive integer &lt;math&gt;n&lt;/math&gt;, let &lt;math&gt;S(n)&lt;/math&gt; denote the sum of the digits of &lt;math&gt;n.&lt;/math&gt; For how many values of &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;n + S(n) + S(S(n)) = 2007?&lt;/math&gt; <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 1 \qquad \mathrm{(B)}\ 2 \qquad \mathrm{(C)}\ 3 \qquad \mathrm{(D)}\ 4 \qquad \mathrm{(E)}\ 5&lt;/math&gt;<br /> <br /> == Solution ==<br /> === Solution 1 ===<br /> For the sake of notation let &lt;math&gt;T(n) = n + S(n) + S(S(n))&lt;/math&gt;. Obviously &lt;math&gt;n&lt;2007&lt;/math&gt;. Then the maximum value of &lt;math&gt;S(n) + S(S(n))&lt;/math&gt; is when &lt;math&gt;n = 1999&lt;/math&gt;, and the sum becomes &lt;math&gt;28 + 10 = 38&lt;/math&gt;. So the minimum bound is &lt;math&gt;1969&lt;/math&gt;. We do [[casework]] upon the tens digit:<br /> <br /> Case 1: &lt;math&gt;196u \Longrightarrow u = 9&lt;/math&gt;. Easy to directly disprove.<br /> <br /> Case 2: &lt;math&gt;197u&lt;/math&gt;. &lt;math&gt;S(n) = 1 + 9 + 7 + u = 17 + u&lt;/math&gt;, and &lt;math&gt;S(S(n)) = 8+u&lt;/math&gt; if &lt;math&gt;u \le 2&lt;/math&gt; and &lt;math&gt;S(S(n)) = 2 + (u-3) = u-1&lt;/math&gt; otherwise. <br /> <br /> :Subcase a: &lt;math&gt;T(n) = 1970 + u + 17 + u + 8 + u = 1995 + 3u = 2007 \Longrightarrow u = 4&lt;/math&gt;. This exceeds our bounds, so no solution here.<br /> :Subcase b: &lt;math&gt;T(n) = 1970 + u + 17 + u + u - 1 = 1986 + 3u = 2007 \Longrightarrow u = 7&lt;/math&gt;. First solution.<br /> <br /> Case 3: &lt;math&gt;198u&lt;/math&gt;. &lt;math&gt;S(n) = 18 + u&lt;/math&gt;, and &lt;math&gt;S(S(n)) = 9 + u&lt;/math&gt; if &lt;math&gt;u \le 1&lt;/math&gt; and &lt;math&gt;2 + (u-2) = u&lt;/math&gt; otherwise.<br /> <br /> :Subcase a: &lt;math&gt;T(n) = 1980 + u + 18 + u + 9 + u = 2007 + 3u = 2007 \Longrightarrow u = 0&lt;/math&gt;. Second solution.<br /> :Subcase b: &lt;math&gt;T(n) = 1980 + u + 18 + u + u = 1998 + 3u = 2007 \Longrightarrow u = 3&lt;/math&gt;. Third solution. <br /> <br /> Case 4: &lt;math&gt;199u&lt;/math&gt;. But &lt;math&gt;S(n) &gt; 19&lt;/math&gt;, and &lt;math&gt;n + S(n)&lt;/math&gt; clearly sum to &lt;math&gt;&gt; 2007&lt;/math&gt;.<br /> <br /> Case 5: &lt;math&gt;200u&lt;/math&gt;. So &lt;math&gt;S(n) = 2 + u&lt;/math&gt; and &lt;math&gt;S(S(n)) = 2 + u&lt;/math&gt; (recall that &lt;math&gt;n &lt; 2007 &lt;/math&gt;), and &lt;math&gt;2000 + u + 2 + u + 2 + u = 2004 + 3u = 2007 \Longrightarrow u = 1&lt;/math&gt;. Fourth solution.<br /> <br /> In total we have &lt;math&gt;4 \mathrm{(D)}&lt;/math&gt; solutions, which are &lt;math&gt;1977, 1980, 1983, &lt;/math&gt; and &lt;math&gt;2001&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Clearly, &lt;math&gt;n &gt; 1950&lt;/math&gt;. We can break this into three cases:<br /> <br /> Case 1: &lt;math&gt;n \geq 2000&lt;/math&gt;<br /> <br /> :Inspection gives &lt;math&gt;n = 2001&lt;/math&gt;.<br /> <br /> Case 2: &lt;math&gt;n &lt; 2000&lt;/math&gt;, &lt;math&gt;n = 19xy&lt;/math&gt; (not to be confused with &lt;math&gt;19*x*y&lt;/math&gt;), &lt;math&gt;x + y &lt; 10 &lt;/math&gt;<br /> <br /> :If you set up an equation, it reduces to<br /> <br /> &lt;math&gt;4x + y = 32&lt;/math&gt;<br /> <br /> :which has as its only solution satisfying the constraints &lt;math&gt;x = 8&lt;/math&gt;, &lt;math&gt;y = 0&lt;/math&gt;.<br /> <br /> Case 3: &lt;math&gt;n &lt; 2000&lt;/math&gt;, &lt;math&gt;n = 19xy&lt;/math&gt;, &lt;math&gt;x + y \geq 10&lt;/math&gt;<br /> <br /> :This reduces to<br /> <br /> :&lt;math&gt;4x + y = 35&lt;/math&gt;. The only two solutions satisfying the constraints for this equation are &lt;math&gt;x = 7&lt;/math&gt;, &lt;math&gt;y = 7&lt;/math&gt; and &lt;math&gt;x = 8&lt;/math&gt;, &lt;math&gt;y = 3&lt;/math&gt;. <br /> <br /> The solutions are thus &lt;math&gt;1977, 1980, 1983, 2001&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(D)}\ 4&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> As in Solution 1, we note that &lt;math&gt;S(n)\leq 28&lt;/math&gt; and &lt;math&gt;S(S(n))\leq 10&lt;/math&gt;. &lt;br/&gt;<br /> Obviously, &lt;math&gt;n\equiv S(n)\equiv S(S(n)) \pmod 9&lt;/math&gt;. &lt;br/&gt;<br /> As &lt;math&gt;2007\equiv 0 \pmod 9&lt;/math&gt;, this means that &lt;math&gt;n\bmod 9 \in\{0,3,6\}&lt;/math&gt;, or equivalently that &lt;math&gt;n\equiv S(n)\equiv S(S(n))\equiv 0 \pmod 3&lt;/math&gt;.<br /> <br /> Thus &lt;math&gt;S(S(n))\in\{3,6,9\}&lt;/math&gt;. For each possible &lt;math&gt;S(S(n))&lt;/math&gt; we get three possible &lt;math&gt;S(n)&lt;/math&gt;. &lt;br/&gt;<br /> (E. g., if &lt;math&gt;S(S(n))=6&lt;/math&gt;, then &lt;math&gt;S(n)=x&lt;/math&gt; is a number such that &lt;math&gt;x\leq 28&lt;/math&gt; and &lt;math&gt;S(x)=6&lt;/math&gt;, therefore &lt;math&gt;S(n)\in\{6,15,24\}&lt;/math&gt;.)<br /> <br /> For each of these nine possibilities we compute &lt;math&gt;n_{?}&lt;/math&gt; as &lt;math&gt;2007-S(n)-S(S(n))&lt;/math&gt; and check whether &lt;math&gt;S(n_{?})=S(n)&lt;/math&gt;. &lt;br/&gt;<br /> We'll find out that out of the 9 cases, in 4 the value &lt;math&gt;n_{?}&lt;/math&gt; has the correct sum of digits. &lt;br/&gt;<br /> This happens for &lt;math&gt;n_{?}\in \{ 1977, 1980, 1983, 2001 \}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> *This solution is not a good solution, but is viable for in contest situations<br /> Clearly &lt;math&gt;n\equiv S(n) \pmod 9&lt;/math&gt;. Thus, &lt;cmath&gt;n+S(n)+S(S(n))\equiv 0 \pmod 9 \implies n\equiv 0\pmod 3.&lt;/cmath&gt;<br /> Now we need a bound for &lt;math&gt;n&lt;/math&gt;. It is clear that the maximum for &lt;math&gt;S(n)=36&lt;/math&gt; (from &lt;math&gt;n=9999&lt;/math&gt;) which means the maximum for &lt;math&gt;S(n)+S(S(n))&lt;/math&gt; is &lt;math&gt;45&lt;/math&gt;. This means that &lt;math&gt;n\geq 1962&lt;/math&gt;.<br /> *Warning: This is where you will cringe badly<br /> Now check all multiples of &lt;math&gt;3&lt;/math&gt; from &lt;math&gt;1962&lt;/math&gt; to &lt;math&gt;2007&lt;/math&gt; and we find that only &lt;math&gt;n=1977, 1980, 1983, 2001&lt;/math&gt; work, so our answer is &lt;math&gt;\mathrm{(D)}\ 4&lt;/math&gt;.<br /> <br /> Remark: this may seem time consuming, but in reality, calculating &lt;math&gt;n+S(n)+S(S(n))&lt;/math&gt; for &lt;math&gt;16&lt;/math&gt; values is actually very quick, so this solution would only take approximately 3-5 minutes, helpful in a contest.<br /> <br /> ==Solution 5 (Rigorous)==<br /> Let the number of digits of &lt;math&gt;n&lt;/math&gt; be &lt;math&gt;m&lt;/math&gt;. If &lt;math&gt;m = 5&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; will already be greater than &lt;math&gt;2007&lt;/math&gt;. Notice that &lt;math&gt;S(n)&lt;/math&gt; is always at most &lt;math&gt;9m&lt;/math&gt;. Then if &lt;math&gt;m = 3&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; will be at most &lt;math&gt;999&lt;/math&gt;, &lt;math&gt;S(n)&lt;/math&gt; will be at most &lt;math&gt;27&lt;/math&gt;, and &lt;math&gt;S(S(n))&lt;/math&gt; will be even smaller than &lt;math&gt;27&lt;/math&gt;. Clearly we cannot reach a sum of &lt;math&gt;2007&lt;/math&gt;, unless &lt;math&gt;m = 4&lt;/math&gt; (i.e. &lt;math&gt;n&lt;/math&gt; has &lt;math&gt;4&lt;/math&gt; digits).<br /> <br /> Then, let &lt;math&gt;n&lt;/math&gt; be a four digit number in the form &lt;math&gt;1000a + 100b + 10c + d&lt;/math&gt;. Then &lt;math&gt;S(n) = a + b + c + d&lt;/math&gt;.<br /> <br /> &lt;math&gt;S(S(n))&lt;/math&gt; is the sum of the digits of &lt;math&gt;a + b + c + d&lt;/math&gt;. We can represent &lt;math&gt;S(S(n))&lt;/math&gt; as the sum of the tens digit and the ones digit of &lt;math&gt;S(n)&lt;/math&gt;. The tens digit in the form of a decimal is<br /> <br /> <br /> &lt;math&gt;\frac{a + b + c + d}{10}&lt;/math&gt;.<br /> <br /> <br /> To remove the decimal portion, we can simply take the floor of the expression,<br /> <br /> <br /> &lt;math&gt;\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> Now that we have expressed the tens digit, we can express the ones digit as &lt;math&gt;S(S(n)) -10&lt;/math&gt; times the above expression, or<br /> <br /> <br /> &lt;math&gt;a + b + c + d - 10\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> Adding the two expressions yields the value of &lt;math&gt;S(S(n))&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt; = a + b + c + d - 9\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> Combining this expression to the ones for &lt;math&gt;n&lt;/math&gt; and &lt;math&gt;S(n)&lt;/math&gt; yields<br /> <br /> <br /> &lt;math&gt;1002a + 102b + 12c + 3d - 9\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> Setting this equal to &lt;math&gt;2007&lt;/math&gt; and rearranging a bit yields<br /> <br /> <br /> &lt;math&gt;12c + 3d = 2007 - 1002a - 102b + 9\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;4c + d = 669 - 334a - 34b + 3\lfloor\frac{a + b + c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> (The reason for this slightly weird arrangement will soon become evident)<br /> <br /> <br /> Now we examine the possible values of &lt;math&gt;a&lt;/math&gt;. If &lt;math&gt;a \ge 3&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; is already too large. &lt;math&gt;a&lt;/math&gt; must also be greater than &lt;math&gt;0&lt;/math&gt;, or &lt;math&gt;n&lt;/math&gt; would be a &lt;math&gt;3&lt;/math&gt;-digit number. Therefore, &lt;math&gt;a = 1 \, \text{or} \, 2&lt;/math&gt;. Now we examine by case.<br /> <br /> If &lt;math&gt;a = 2&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; must both be &lt;math&gt;0&lt;/math&gt; (otherwise &lt;math&gt;n&lt;/math&gt; would already be greater than &lt;math&gt;2007&lt;/math&gt;). Substituting these values into the equation yields<br /> <br /> <br /> &lt;math&gt;d = 1 + 3\lfloor\frac{2 + d}{10}\rfloor&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;d=1&lt;/math&gt;.<br /> <br /> <br /> Sure enough, &lt;math&gt;2001 + (2+1) + 3=2007&lt;/math&gt;.<br /> <br /> Now we move onto the case where &lt;math&gt;a = 1&lt;/math&gt;. Then our initial equation simplifies to<br /> <br /> <br /> &lt;math&gt;4c + d = 335 - 34b + 3\lfloor\frac{1 + b + c + d}{10}\rfloor&lt;/math&gt;<br /> <br /> <br /> Since &lt;math&gt;c&lt;/math&gt; and &lt;math&gt;d&lt;/math&gt; can each be at most &lt;math&gt;9&lt;/math&gt;, we substitute that value to find the lower bound of &lt;math&gt;b&lt;/math&gt;. Doing so yields<br /> <br /> <br /> &lt;math&gt;34b \ge 290 + 3\lfloor\frac{19 + b}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> The floor expression is at least &lt;math&gt;3\lfloor\frac{19}{10}\rfloor=3&lt;/math&gt; , so the right-hand side is at least &lt;math&gt;293&lt;/math&gt;. Solving for &lt;math&gt;b&lt;/math&gt;, we see that &lt;math&gt;b \ge 9 &lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;b=9&lt;/math&gt;. Again, we substitute for &lt;math&gt;b&lt;/math&gt; and the equation becomes<br /> <br /> <br /> &lt;math&gt;4c + d = 29 + 3\lfloor\frac{10 + c + d}{10}\rfloor&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;4c + d = 32 + 3\lfloor\frac{c + d}{10}\rfloor&lt;/math&gt;.<br /> <br /> <br /> Just like we did for &lt;math&gt;b&lt;/math&gt;, we can find the lower bound of &lt;math&gt;c&lt;/math&gt; by assuming &lt;math&gt;d = 9&lt;/math&gt; and solving:<br /> <br /> <br /> &lt;math&gt;4c + 9 \ge 29 + 3\lfloor\frac{c + 9}{10}\rfloor&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;4c \ge 20 + 3\lfloor\frac{c + 9}{10}\rfloor&lt;/math&gt;<br /> <br /> <br /> The right hand side is &lt;math&gt;20&lt;/math&gt; for &lt;math&gt;c=0&lt;/math&gt; and &lt;math&gt;23&lt;/math&gt; for &lt;math&gt;c \ge 1&lt;/math&gt;. Solving for c yields &lt;math&gt;c \ge 6&lt;/math&gt;. Looking back at the previous equation, the floor expression is &lt;math&gt;0&lt;/math&gt; for &lt;math&gt;c+d \le 9&lt;/math&gt; and &lt;math&gt;3&lt;/math&gt; for &lt;math&gt;c+d \ge 10&lt;/math&gt;. Thus, the right-hand side is &lt;math&gt;32&lt;/math&gt; for &lt;math&gt;c+d \le 9&lt;/math&gt; and &lt;math&gt;35&lt;/math&gt; for &lt;math&gt;c+d \ge 10&lt;/math&gt;. We can solve these two scenarios as systems of equations/inequalities:<br /> <br /> &lt;math&gt;4c+d = 32&lt;/math&gt;<br /> <br /> &lt;math&gt;c+d \le 9&lt;/math&gt;<br /> <br /> and<br /> <br /> &lt;math&gt;4c+d=35&lt;/math&gt;<br /> <br /> &lt;math&gt;c+d \ge 10&lt;/math&gt;<br /> <br /> Solving yields three pairs &lt;math&gt;(c, d):&lt;/math&gt; &lt;math&gt;(8, 0)&lt;/math&gt;; &lt;math&gt;(8, 3)&lt;/math&gt;; and &lt;math&gt;(7, 7)&lt;/math&gt;. Checking the numbers &lt;math&gt;1980&lt;/math&gt;, &lt;math&gt;1983&lt;/math&gt;, and &lt;math&gt;1977&lt;/math&gt;; we find that all three work. Therefore there are a total of &lt;math&gt;4&lt;/math&gt; possibilities for &lt;math&gt;n&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;.<br /> <br /> Note: Although this solution takes a while to read (as well as to write) the actual time it takes to think through the process above is very short in comparison to the solution length.<br /> <br /> ==Solution 6==<br /> <br /> <br /> Rearranging, we get <br /> 2007 - S(n) - S(S(n)) = n.<br /> <br /> We can now try S(n) = 1-28, to find values of n. Then, you need to check whether S(n) = sum digits of n. If so, you found a solution!<br /> <br /> The bash involves making a table with 3 columns: S(n), n, S(n). We write the numbers 1-28 in the S(n) column, we can find n using the above equation, and the third column we can just find the sum of the digits of n, and if the first column/3rd column match, we have a solution.<br /> <br /> You will in the end find S(n) = 3, 18, 21, 24 yield solutions, which corrispond to n = 2001, 1980, 1983, 1977. There are 4 solutions, so the answer is &lt;math&gt;\boxed{\text{D}}&lt;/math&gt;<br /> <br /> <br /> <br /> == See also ==<br /> {{AMC12 box|year=2007|ab=A|num-b=21|num-a=23}}<br /> {{AMC10 box|year=2007|ab=A|num-b=24|after=Last question}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_15&diff=132435 1983 AIME Problems/Problem 15 2020-08-24T01:27:17Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> The adjoining figure shows two intersecting chords in a circle, with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the sine of the central angle of minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this number is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> <br /> &lt;asy&gt;size(100);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=1;<br /> pair O1=(0,0);<br /> pair A=(-0.91,-0.41);<br /> pair B=(-0.99,0.13);<br /> pair C=(0.688,0.728);<br /> pair D=(-0.25,0.97);<br /> path C1=Circle(O1,1);<br /> draw(C1);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> draw(A--D);<br /> draw(B--C);<br /> pair F=intersectionpoint(A--D,B--C);<br /> add(pathticks(A--F,1,0.5,0,3.5));<br /> add(pathticks(F--D,1,0.5,0,3.5));<br /> &lt;/asy&gt;<br /> &lt;!-- [[Image:1983_AIME-15.png|200px]] --&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> -Credit to Adamz for diagram-<br /> &lt;asy&gt;<br /> size(10cm);<br /> import olympiad;<br /> pair O = (0,0);dot(O);label(&quot;$O$&quot;,O,SW);<br /> pair M = (4,0);dot(M);label(&quot;$M$&quot;,M,SE);<br /> pair N = (4,2);dot(N);label(&quot;$N$&quot;,N,NE);<br /> draw(circle(O,5));<br /> pair B = (4,3);dot(B);label(&quot;$B$&quot;,B,NE);<br /> pair C = (4,-3);dot(C);label(&quot;$C$&quot;,C,SE);<br /> draw(B--C);draw(O--M);<br /> pair P = (1.5,2);dot(P);label(&quot;$P$&quot;,P,W);<br /> draw(circle(P,2.5));<br /> pair A=(3,4);dot(A);label(&quot;$A$&quot;,A,NE);<br /> draw(O--A);<br /> draw(O--B);<br /> pair Q = (1.5,0); dot(Q); label(&quot;$Q$&quot;,Q,S);<br /> pair R = (3,0); dot(R); label(&quot;$R$&quot;,R,S);<br /> draw(P--Q,dotted); draw(A--R,dotted);<br /> pair D=(5,0); dot(D); label(&quot;$D$&quot;,D,E);<br /> draw(A--D);<br /> &lt;/asy&gt;Let &lt;math&gt;A&lt;/math&gt; be any fixed point on circle &lt;math&gt;O&lt;/math&gt;, and let &lt;math&gt;AD&lt;/math&gt; be a chord of circle &lt;math&gt;O&lt;/math&gt;. The [[locus]] of midpoints &lt;math&gt;N&lt;/math&gt; of the chord &lt;math&gt;AD&lt;/math&gt; is a circle &lt;math&gt;P&lt;/math&gt;, with diameter &lt;math&gt;AO&lt;/math&gt;. Generally, the circle &lt;math&gt;P&lt;/math&gt; can intersect the chord &lt;math&gt;BC&lt;/math&gt; at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle &lt;math&gt;P&lt;/math&gt; is tangent to &lt;math&gt;BC&lt;/math&gt; at point &lt;math&gt;N&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;M&lt;/math&gt; be the midpoint of the chord &lt;math&gt;BC&lt;/math&gt;. From right triangle &lt;math&gt;OMB&lt;/math&gt;, we have &lt;math&gt;OM = \sqrt{OB^2 - BM^2} =4&lt;/math&gt;. This gives &lt;math&gt;\tan \angle BOM = \frac{BM}{OM} = \frac 3 4&lt;/math&gt;.<br /> <br /> Notice that the distance &lt;math&gt;OM&lt;/math&gt; equals &lt;math&gt;PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the radius of circle &lt;math&gt;P&lt;/math&gt;. <br /> <br /> Hence &lt;cmath&gt;\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5&lt;/cmath&gt; (where &lt;math&gt;R&lt;/math&gt; represents the radius, &lt;math&gt;5&lt;/math&gt;, of the large circle given in the question). Therefore, since &lt;math&gt;\angle AOM&lt;/math&gt; is clearly acute, we see that &lt;cmath&gt;\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3&lt;/cmath&gt;<br /> <br /> Next, notice that &lt;math&gt;\angle AOB = \angle AOM - \angle BOM&lt;/math&gt;. We can therefore apply the subtraction formula for &lt;math&gt;\tan&lt;/math&gt; to obtain &lt;cmath&gt;\tan \angle AOB =\frac{\tan \angle AOM - \tan \angle BOM}{1 + \tan \angle AOM \cdot \tan \angle BOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}&lt;/cmath&gt; It follows that &lt;math&gt;\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}&lt;/math&gt;, such that the answer is &lt;math&gt;7 \cdot 25=\boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> This solution, while similar to Solution 1, is far more motivated and less contrived.<br /> <br /> Firstly, we note the statement in the problem that &quot;&lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; and bisected by &lt;math&gt;BC&lt;/math&gt;&quot; &amp;ndash; what is its significance? What is the criterion for this statement to be true?<br /> <br /> We consider the locus of midpoints of the chords from &lt;math&gt;A&lt;/math&gt;. It is well-known that this is the circle with diameter &lt;math&gt;AO&lt;/math&gt;, where &lt;math&gt;O&lt;/math&gt; is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor &lt;math&gt;\frac{1}{2}&lt;/math&gt; and center &lt;math&gt;A&lt;/math&gt;. Thus, the locus is the result of the dilation with scale factor &lt;math&gt;\frac{1}{2}&lt;/math&gt; and centre &lt;math&gt;A&lt;/math&gt; of circle &lt;math&gt;O&lt;/math&gt;. Let the center of this circle be &lt;math&gt;P&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt; if they cross at some point &lt;math&gt;N&lt;/math&gt; on the circle. Moreover, since &lt;math&gt;AD&lt;/math&gt; is the only chord, &lt;math&gt;BC&lt;/math&gt; must be tangent to the circle &lt;math&gt;P&lt;/math&gt;.<br /> <br /> The rest of this problem is straightforward.<br /> <br /> Our goal is to find &lt;math&gt;\sin \angle AOB = \sin{\left(\angle AOM - \angle BOM\right)}&lt;/math&gt;, where &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;. We have &lt;math&gt;BM=3&lt;/math&gt; and &lt;math&gt;OM=4&lt;/math&gt;.<br /> Let &lt;math&gt;R&lt;/math&gt; be the projection of &lt;math&gt;A&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;, and similarly let &lt;math&gt;Q&lt;/math&gt; be the projection of &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;. Then it remains to find &lt;math&gt;AR&lt;/math&gt; so that we can use the addition formula for &lt;math&gt;\sin&lt;/math&gt;.<br /> <br /> As &lt;math&gt;PN&lt;/math&gt; is a radius of circle &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;PN=2.5&lt;/math&gt;, and similarly, &lt;math&gt;PO=2.5&lt;/math&gt;. Since &lt;math&gt;OM=4&lt;/math&gt;, we have &lt;math&gt;OQ=OM-QM=OM-PN=4-2.5=1.5&lt;/math&gt;. Thus &lt;math&gt;PQ=\sqrt{2.5^2-1.5^2}=2&lt;/math&gt;.<br /> <br /> Further, we see that &lt;math&gt;\triangle OAR&lt;/math&gt; is a dilation of &lt;math&gt;\triangle OPQ&lt;/math&gt; about center &lt;math&gt;O&lt;/math&gt; with scale factor &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;AR=2PQ=4&lt;/math&gt;.<br /> <br /> Lastly, we apply the formula:<br /> &lt;cmath&gt; \sin{\left(\angle AOM - \angle BOM\right)} = \sin \angle AOM \cos \angle BOM - \sin \angle BOM \cos \angle AOM = \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)=\frac{7}{25}&lt;/cmath&gt;<br /> Thus the answer is &lt;math&gt;7\cdot25=\boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 3 (coordinate geometry) ===<br /> [[File:Aime1983p15s2.png|500px|link=]]<br /> <br /> Let the circle have equation &lt;math&gt;x^2 + y^2 = 25&lt;/math&gt;, with centre &lt;math&gt;O(0,0)&lt;/math&gt;. Since &lt;math&gt;BC=6&lt;/math&gt;, we can calculate (by the Pythagorean Theorem) that the distance from &lt;math&gt;O&lt;/math&gt; to the line &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. Therefore, we can let &lt;math&gt;B=(3,4)&lt;/math&gt; and &lt;math&gt;C=(-3,4)&lt;/math&gt;. Now, assume that &lt;math&gt;A&lt;/math&gt; is any point on the major arc BC, and &lt;math&gt;D&lt;/math&gt; any point on the minor arc BC. We can write &lt;math&gt;A=(5 \cos \alpha, 5 \sin \alpha)&lt;/math&gt;, where &lt;math&gt;\alpha&lt;/math&gt; is the angle measured from the positive &lt;math&gt;x&lt;/math&gt; axis to the ray &lt;math&gt;OA&lt;/math&gt;. It will also be convenient to define &lt;math&gt;\angle XOB = \alpha_0&lt;/math&gt;. <br /> <br /> Firstly, since &lt;math&gt;B&lt;/math&gt; must lie in the minor arc &lt;math&gt;AD&lt;/math&gt;, we see that &lt;math&gt;\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)&lt;/math&gt;. However, since the midpoint of &lt;math&gt;AD&lt;/math&gt; must lie on &lt;math&gt;BC&lt;/math&gt;, and the highest possible &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;D&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt;, we see that the &lt;math&gt;y&lt;/math&gt;-coordinate cannot be lower than &lt;math&gt;3&lt;/math&gt;, that is, &lt;math&gt;\alpha \in \left[\sin^{-1}\frac{3}{5},\alpha_0\right)&lt;/math&gt;.<br /> <br /> Secondly, there is a theorem that says that, in a circle, if a chord is bisected by a radius, then they must be perpendicular. Therefore, suppose that &lt;math&gt;P&lt;/math&gt; is the intersection point of &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, so that by the theorem, &lt;math&gt;OP&lt;/math&gt; is perpendicular to &lt;math&gt;AD&lt;/math&gt;. So, if &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;, this means that &lt;math&gt;P&lt;/math&gt; is the only point on the chord &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;OP&lt;/math&gt; is perpendicular to &lt;math&gt;AD&lt;/math&gt;. Now suppose that &lt;math&gt;P=(p,4)&lt;/math&gt;, where &lt;math&gt;p \in (-3,3)&lt;/math&gt;. The fact that &lt;math&gt;OP&lt;/math&gt; must be perpendicular to &lt;math&gt;AD&lt;/math&gt; is equivalent to the following equation:<br /> <br /> &lt;cmath&gt; -1 = \left(\text{slope of } OP\right)\left(\text{slope of } AP\right)&lt;/cmath&gt;<br /> which becomes<br /> &lt;cmath&gt; -1 = \frac{4}{p} \cdot \frac{5\sin \alpha - 4}{5\cos \alpha - p}&lt;/cmath&gt;<br /> <br /> This rearranges to<br /> <br /> &lt;cmath&gt; p^2 - (5\cos \alpha)p + 16 - 20 \sin \alpha = 0&lt;/cmath&gt;<br /> <br /> Given that this equation must have only one real root &lt;math&gt;p\in (-3,3)&lt;/math&gt;, we study the following function:<br /> <br /> &lt;cmath&gt;f(x) = x^2 - (5\cos \alpha)x + 16 - 20 \sin \alpha&lt;/cmath&gt;<br /> <br /> First, by the fact that the equation &lt;math&gt;f(x)=0&lt;/math&gt; has real solutions, its discriminant &lt;math&gt;\Delta&lt;/math&gt; must be non-negative, so we calculate<br /> <br /> &lt;cmath&gt; \begin{split}\Delta &amp; = (5\cos \alpha)^2 - 4(16-20\sin \alpha) \\<br /> &amp; = 25 (1- \sin^2 \alpha) - 64 + 80 \sin \alpha \\<br /> &amp; = -25 \sin^2 \alpha + 80\sin \alpha - 39 \\<br /> &amp; = (13 - 5\sin \alpha)(5\sin \alpha - 3)\end{split}&lt;/cmath&gt;<br /> <br /> It is obvious that this is in fact non-negative. If it is actually zero, then &lt;math&gt;\sin \alpha = \frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\cos \alpha = \frac{4}{5}&lt;/math&gt;. In this case, &lt;math&gt;p = \frac{5\cos \alpha}{2} = 2 \in (-3,3)&lt;/math&gt;, so we have found a possible solution. We thus calculate &lt;math&gt;\sin(\text{central angle of minor arc } AB) = \sin (\alpha_0 - \alpha) = \frac{4}{5}\cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{5} = \frac{7}{25}&lt;/math&gt; by the subtraction formula for &lt;math&gt;\sin&lt;/math&gt;. This means that the answer is &lt;math&gt;7 \cdot 25 = 175&lt;/math&gt;.<br /> <br /> === Addendum to Solution 3 ===<br /> Since this is an AIME problem, we can assume that we are done since we have found one possible case. However, in a full-solution contest, we could not assume that the answer is unique, but would need to prove that this is the unique solution. This can be proven as follows.<br /> <br /> Suppose that &lt;math&gt;\Delta &gt; 0&lt;/math&gt;, which would mean that there could be two real roots of &lt;math&gt;f(x)&lt;/math&gt;, one lying in the interval &lt;math&gt;(-3,3)&lt;/math&gt;, and another outside of it. We also see, by [[Vieta's Formulas]], that the average of the two roots is &lt;math&gt;\frac{5\cos \alpha}{2}&lt;/math&gt;, which is non-negative, so the root outside of &lt;math&gt;(-3,3)&lt;/math&gt; must be no less than &lt;math&gt;3&lt;/math&gt;. By considering the graph of &lt;math&gt;y=f(x)&lt;/math&gt;, which is a &quot;U-shaped&quot; parabola, it is now evident that &lt;math&gt;f(-3) &gt; 0&lt;/math&gt; and &lt;math&gt;f(3)\leq 0&lt;/math&gt;. We can just use the second inequality:<br /> <br /> &lt;cmath&gt;0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt; 3\cos \alpha + 4 \sin \alpha \geq 5 &lt;/cmath&gt;<br /> <br /> The only way for this inequality to be satisfied is when &lt;math&gt;A=B&lt;/math&gt; (by applying the Cauchy-Schwarz inequality, or just plotting the line &lt;math&gt;3x+4y=5&lt;/math&gt; to see that point &lt;math&gt;A&lt;/math&gt; can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point &lt;math&gt;A&lt;/math&gt; lies in the half-plane above the line &lt;math&gt;3x+4y=5&lt;/math&gt;, inclusive, and the half-plane below the line &lt;math&gt;-3x+4y=5&lt;/math&gt;, exclusive. This can be seen to be impossible by drawing the lines and observing that the intersection of the two half-planes does not share any point with the circle.)<br /> ===Solution 4===<br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. Fix &lt;math&gt;B,C,&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt;. Then, as &lt;math&gt;D&lt;/math&gt; moves around the circle, the locus of the midpoints of &lt;math&gt;AD&lt;/math&gt; is clearly a circle. Since the problems gives that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; bisected by &lt;math&gt;BC&lt;/math&gt;, it follows that the circle with diameter &lt;math&gt;DO&lt;/math&gt; and &lt;math&gt;AO&lt;/math&gt; is tangent to &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> Now, let the intersection of &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt; and let the midpoint of &lt;math&gt;AO&lt;/math&gt; (the center of the circle tangent to &lt;math&gt;BC&lt;/math&gt; that we described beforehand) be &lt;math&gt;F&lt;/math&gt;. Drop the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call its intersection with &lt;math&gt;BC&lt;/math&gt; &lt;math&gt;K&lt;/math&gt;. Drop the perpendicular from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;KO&lt;/math&gt; and call its intersection with &lt;math&gt;KO&lt;/math&gt; &lt;math&gt;L&lt;/math&gt;. Clearly, &lt;math&gt;KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4&lt;/math&gt; and since &lt;math&gt;EF&lt;/math&gt; is radius, it equals &lt;math&gt;\frac{5}{2}&lt;/math&gt;. The same applies for &lt;math&gt;FO&lt;/math&gt;, which also equals &lt;math&gt;\frac{5}{2}&lt;/math&gt;. By the Pythagorean theorem, we deduce that &lt;math&gt;FL = 2&lt;/math&gt;, so &lt;math&gt;EK = 2&lt;/math&gt;. This is very important information! Now we know that &lt;math&gt;BE = 1&lt;/math&gt;, so by Power of a Point, &lt;math&gt;AE = ED = \sqrt{5}&lt;/math&gt;.<br /> <br /> We’re almost there! Since by the Pythagorean theorem, &lt;math&gt;ED^2 + EO^2 = 5&lt;/math&gt;, we deduce that &lt;math&gt;EO = 2\sqrt{5}&lt;/math&gt;. &lt;math&gt;EC=OC=5&lt;/math&gt;, so &lt;math&gt;\sin (CEO) = \frac{2\sqrt{5}}{5}&lt;/math&gt;. Furthermore, since &lt;math&gt;\sin (CEO) = \cos(DEC)&lt;/math&gt;, we know that &lt;math&gt;\cos (DEC) = \frac{2\sqrt{5}}{5}&lt;/math&gt;. By the law of cosines,<br /> &lt;cmath&gt;DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10&lt;/cmath&gt;Therefore, &lt;math&gt;DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}&lt;/math&gt;. Now, drop the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BA&lt;/math&gt; and call its intersection with &lt;math&gt;BA&lt;/math&gt; &lt;math&gt;Z&lt;/math&gt;. Then, by the Pythagorean theorem, &lt;math&gt;OZ = \frac{7\sqrt{2}}{2}&lt;/math&gt;. Thus, &lt;math&gt;\sin (BOZ) = \frac{\sqrt{2}}{10}&lt;/math&gt; and &lt;math&gt;\cos (BOZ) = \frac{7\sqrt{2}}{10}&lt;/math&gt;. As a result, &lt;math&gt;\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}&lt;/math&gt;. &lt;math&gt;7 \cdot 25 = \boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> [[Image:Dgram.png|thumb|none|800px]]<br /> <br /> Let I be the intersection of AD and BC.<br /> <br /> Lemma: &lt;math&gt;AI = ID&lt;/math&gt; if and only if &lt;math&gt;\angle AIO = 90&lt;/math&gt;.<br /> <br /> Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If &lt;math&gt;\angle AIO = 90&lt;/math&gt;, We can get &lt;math&gt;\triangle AIO \cong \triangle OID&lt;/math&gt;<br /> <br /> Let be this the circle with diameter AO.<br /> <br /> Thus, we get &lt;math&gt;\angle AIO = 90&lt;/math&gt;, implying I must lie on &lt;math&gt;\omega&lt;/math&gt;. I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC.<br /> <br /> Now, create a coordinate system such that the origin is O and the x-axis is perpendicular to BC. The positive x direction is from B to C.<br /> <br /> Let Z be (0,5).<br /> Let Y be (-5,0).<br /> Let X be the center of &lt;math&gt;\omega&lt;/math&gt;. Since &lt;math&gt;\omega&lt;/math&gt;'s radius is &lt;math&gt;\frac{5}{2}&lt;/math&gt;, the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so &lt;math&gt;sin(XOY) = sin(AOY) = \frac{3}{5}&lt;/math&gt;. &lt;math&gt;sin(BOZ) = \frac{3}{5}&lt;/math&gt;. If we let &lt;math&gt;sin(\theta) = \frac{3}{5}&lt;/math&gt;, we can find that what we are looking for is &lt;math&gt;sin(90 - 2\theta)&lt;/math&gt;, which we can evaluate and get &lt;math&gt;\frac{7}{25} \implies \boxed{175}&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_15&diff=132399 1983 AIME Problems/Problem 15 2020-08-23T18:37:58Z <p>Alexlikemath: Solution 5</p> <hr /> <div>== Problem ==<br /> The adjoining figure shows two intersecting chords in a circle, with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the sine of the central angle of minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this number is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> <br /> &lt;asy&gt;size(100);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=1;<br /> pair O1=(0,0);<br /> pair A=(-0.91,-0.41);<br /> pair B=(-0.99,0.13);<br /> pair C=(0.688,0.728);<br /> pair D=(-0.25,0.97);<br /> path C1=Circle(O1,1);<br /> draw(C1);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> draw(A--D);<br /> draw(B--C);<br /> pair F=intersectionpoint(A--D,B--C);<br /> add(pathticks(A--F,1,0.5,0,3.5));<br /> add(pathticks(F--D,1,0.5,0,3.5));<br /> &lt;/asy&gt;<br /> &lt;!-- [[Image:1983_AIME-15.png|200px]] --&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> -Credit to Adamz for diagram-<br /> &lt;asy&gt;<br /> size(10cm);<br /> import olympiad;<br /> pair O = (0,0);dot(O);label(&quot;$O$&quot;,O,SW);<br /> pair M = (4,0);dot(M);label(&quot;$M$&quot;,M,SE);<br /> pair N = (4,2);dot(N);label(&quot;$N$&quot;,N,NE);<br /> draw(circle(O,5));<br /> pair B = (4,3);dot(B);label(&quot;$B$&quot;,B,NE);<br /> pair C = (4,-3);dot(C);label(&quot;$C$&quot;,C,SE);<br /> draw(B--C);draw(O--M);<br /> pair P = (1.5,2);dot(P);label(&quot;$P$&quot;,P,W);<br /> draw(circle(P,2.5));<br /> pair A=(3,4);dot(A);label(&quot;$A$&quot;,A,NE);<br /> draw(O--A);<br /> draw(O--B);<br /> pair Q = (1.5,0); dot(Q); label(&quot;$Q$&quot;,Q,S);<br /> pair R = (3,0); dot(R); label(&quot;$R$&quot;,R,S);<br /> draw(P--Q,dotted); draw(A--R,dotted);<br /> pair D=(5,0); dot(D); label(&quot;$D$&quot;,D,E);<br /> draw(A--D);<br /> &lt;/asy&gt;Let &lt;math&gt;A&lt;/math&gt; be any fixed point on circle &lt;math&gt;O&lt;/math&gt;, and let &lt;math&gt;AD&lt;/math&gt; be a chord of circle &lt;math&gt;O&lt;/math&gt;. The [[locus]] of midpoints &lt;math&gt;N&lt;/math&gt; of the chord &lt;math&gt;AD&lt;/math&gt; is a circle &lt;math&gt;P&lt;/math&gt;, with diameter &lt;math&gt;AO&lt;/math&gt;. Generally, the circle &lt;math&gt;P&lt;/math&gt; can intersect the chord &lt;math&gt;BC&lt;/math&gt; at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle &lt;math&gt;P&lt;/math&gt; is tangent to &lt;math&gt;BC&lt;/math&gt; at point &lt;math&gt;N&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;M&lt;/math&gt; be the midpoint of the chord &lt;math&gt;BC&lt;/math&gt;. From right triangle &lt;math&gt;OMB&lt;/math&gt;, we have &lt;math&gt;OM = \sqrt{OB^2 - BM^2} =4&lt;/math&gt;. This gives &lt;math&gt;\tan \angle BOM = \frac{BM}{OM} = \frac 3 4&lt;/math&gt;.<br /> <br /> Notice that the distance &lt;math&gt;OM&lt;/math&gt; equals &lt;math&gt;PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the radius of circle &lt;math&gt;P&lt;/math&gt;. <br /> <br /> Hence &lt;cmath&gt;\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5&lt;/cmath&gt; (where &lt;math&gt;R&lt;/math&gt; represents the radius, &lt;math&gt;5&lt;/math&gt;, of the large circle given in the question). Therefore, since &lt;math&gt;\angle AOM&lt;/math&gt; is clearly acute, we see that &lt;cmath&gt;\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3&lt;/cmath&gt;<br /> <br /> Next, notice that &lt;math&gt;\angle AOB = \angle AOM - \angle BOM&lt;/math&gt;. We can therefore apply the subtraction formula for &lt;math&gt;\tan&lt;/math&gt; to obtain &lt;cmath&gt;\tan \angle AOB =\frac{\tan \angle AOM - \tan \angle BOM}{1 + \tan \angle AOM \cdot \tan \angle BOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}&lt;/cmath&gt; It follows that &lt;math&gt;\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}&lt;/math&gt;, such that the answer is &lt;math&gt;7 \cdot 25=\boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> This solution, while similar to Solution 1, is far more motivated and less contrived.<br /> <br /> Firstly, we note the statement in the problem that &quot;&lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; and bisected by &lt;math&gt;BC&lt;/math&gt;&quot; &amp;ndash; what is its significance? What is the criterion for this statement to be true?<br /> <br /> We consider the locus of midpoints of the chords from &lt;math&gt;A&lt;/math&gt;. It is well-known that this is the circle with diameter &lt;math&gt;AO&lt;/math&gt;, where &lt;math&gt;O&lt;/math&gt; is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor &lt;math&gt;\frac{1}{2}&lt;/math&gt; and center &lt;math&gt;A&lt;/math&gt;. Thus, the locus is the result of the dilation with scale factor &lt;math&gt;\frac{1}{2}&lt;/math&gt; and centre &lt;math&gt;A&lt;/math&gt; of circle &lt;math&gt;O&lt;/math&gt;. Let the center of this circle be &lt;math&gt;P&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt; if they cross at some point &lt;math&gt;N&lt;/math&gt; on the circle. Moreover, since &lt;math&gt;AD&lt;/math&gt; is the only chord, &lt;math&gt;BC&lt;/math&gt; must be tangent to the circle &lt;math&gt;P&lt;/math&gt;.<br /> <br /> The rest of this problem is straightforward.<br /> <br /> Our goal is to find &lt;math&gt;\sin \angle AOB = \sin{\left(\angle AOM - \angle BOM\right)}&lt;/math&gt;, where &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;. We have &lt;math&gt;BM=3&lt;/math&gt; and &lt;math&gt;OM=4&lt;/math&gt;.<br /> Let &lt;math&gt;R&lt;/math&gt; be the projection of &lt;math&gt;A&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;, and similarly let &lt;math&gt;Q&lt;/math&gt; be the projection of &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;. Then it remains to find &lt;math&gt;AR&lt;/math&gt; so that we can use the addition formula for &lt;math&gt;\sin&lt;/math&gt;.<br /> <br /> As &lt;math&gt;PN&lt;/math&gt; is a radius of circle &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;PN=2.5&lt;/math&gt;, and similarly, &lt;math&gt;PO=2.5&lt;/math&gt;. Since &lt;math&gt;OM=4&lt;/math&gt;, we have &lt;math&gt;OQ=OM-QM=OM-PN=4-2.5=1.5&lt;/math&gt;. Thus &lt;math&gt;PQ=\sqrt{2.5^2-1.5^2}=2&lt;/math&gt;.<br /> <br /> Further, we see that &lt;math&gt;\triangle OAR&lt;/math&gt; is a dilation of &lt;math&gt;\triangle OPQ&lt;/math&gt; about center &lt;math&gt;O&lt;/math&gt; with scale factor &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;AR=2PQ=4&lt;/math&gt;.<br /> <br /> Lastly, we apply the formula:<br /> &lt;cmath&gt; \sin{\left(\angle AOM - \angle BOM\right)} = \sin \angle AOM \cos \angle BOM - \sin \angle BOM \cos \angle AOM = \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)=\frac{7}{25}&lt;/cmath&gt;<br /> Thus the answer is &lt;math&gt;7\cdot25=\boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 3 (coordinate geometry) ===<br /> [[File:Aime1983p15s2.png|500px|link=]]<br /> <br /> Let the circle have equation &lt;math&gt;x^2 + y^2 = 25&lt;/math&gt;, with centre &lt;math&gt;O(0,0)&lt;/math&gt;. Since &lt;math&gt;BC=6&lt;/math&gt;, we can calculate (by the Pythagorean Theorem) that the distance from &lt;math&gt;O&lt;/math&gt; to the line &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. Therefore, we can let &lt;math&gt;B=(3,4)&lt;/math&gt; and &lt;math&gt;C=(-3,4)&lt;/math&gt;. Now, assume that &lt;math&gt;A&lt;/math&gt; is any point on the major arc BC, and &lt;math&gt;D&lt;/math&gt; any point on the minor arc BC. We can write &lt;math&gt;A=(5 \cos \alpha, 5 \sin \alpha)&lt;/math&gt;, where &lt;math&gt;\alpha&lt;/math&gt; is the angle measured from the positive &lt;math&gt;x&lt;/math&gt; axis to the ray &lt;math&gt;OA&lt;/math&gt;. It will also be convenient to define &lt;math&gt;\angle XOB = \alpha_0&lt;/math&gt;. <br /> <br /> Firstly, since &lt;math&gt;B&lt;/math&gt; must lie in the minor arc &lt;math&gt;AD&lt;/math&gt;, we see that &lt;math&gt;\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)&lt;/math&gt;. However, since the midpoint of &lt;math&gt;AD&lt;/math&gt; must lie on &lt;math&gt;BC&lt;/math&gt;, and the highest possible &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;D&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt;, we see that the &lt;math&gt;y&lt;/math&gt;-coordinate cannot be lower than &lt;math&gt;3&lt;/math&gt;, that is, &lt;math&gt;\alpha \in \left[\sin^{-1}\frac{3}{5},\alpha_0\right)&lt;/math&gt;.<br /> <br /> Secondly, there is a theorem that says that, in a circle, if a chord is bisected by a radius, then they must be perpendicular. Therefore, suppose that &lt;math&gt;P&lt;/math&gt; is the intersection point of &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, so that by the theorem, &lt;math&gt;OP&lt;/math&gt; is perpendicular to &lt;math&gt;AD&lt;/math&gt;. So, if &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;, this means that &lt;math&gt;P&lt;/math&gt; is the only point on the chord &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;OP&lt;/math&gt; is perpendicular to &lt;math&gt;AD&lt;/math&gt;. Now suppose that &lt;math&gt;P=(p,4)&lt;/math&gt;, where &lt;math&gt;p \in (-3,3)&lt;/math&gt;. The fact that &lt;math&gt;OP&lt;/math&gt; must be perpendicular to &lt;math&gt;AD&lt;/math&gt; is equivalent to the following equation:<br /> <br /> &lt;cmath&gt; -1 = \left(\text{slope of } OP\right)\left(\text{slope of } AP\right)&lt;/cmath&gt;<br /> which becomes<br /> &lt;cmath&gt; -1 = \frac{4}{p} \cdot \frac{5\sin \alpha - 4}{5\cos \alpha - p}&lt;/cmath&gt;<br /> <br /> This rearranges to<br /> <br /> &lt;cmath&gt; p^2 - (5\cos \alpha)p + 16 - 20 \sin \alpha = 0&lt;/cmath&gt;<br /> <br /> Given that this equation must have only one real root &lt;math&gt;p\in (-3,3)&lt;/math&gt;, we study the following function:<br /> <br /> &lt;cmath&gt;f(x) = x^2 - (5\cos \alpha)x + 16 - 20 \sin \alpha&lt;/cmath&gt;<br /> <br /> First, by the fact that the equation &lt;math&gt;f(x)=0&lt;/math&gt; has real solutions, its discriminant &lt;math&gt;\Delta&lt;/math&gt; must be non-negative, so we calculate<br /> <br /> &lt;cmath&gt; \begin{split}\Delta &amp; = (5\cos \alpha)^2 - 4(16-20\sin \alpha) \\<br /> &amp; = 25 (1- \sin^2 \alpha) - 64 + 80 \sin \alpha \\<br /> &amp; = -25 \sin^2 \alpha + 80\sin \alpha - 39 \\<br /> &amp; = (13 - 5\sin \alpha)(5\sin \alpha - 3)\end{split}&lt;/cmath&gt;<br /> <br /> It is obvious that this is in fact non-negative. If it is actually zero, then &lt;math&gt;\sin \alpha = \frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\cos \alpha = \frac{4}{5}&lt;/math&gt;. In this case, &lt;math&gt;p = \frac{5\cos \alpha}{2} = 2 \in (-3,3)&lt;/math&gt;, so we have found a possible solution. We thus calculate &lt;math&gt;\sin(\text{central angle of minor arc } AB) = \sin (\alpha_0 - \alpha) = \frac{4}{5}\cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{5} = \frac{7}{25}&lt;/math&gt; by the subtraction formula for &lt;math&gt;\sin&lt;/math&gt;. This means that the answer is &lt;math&gt;7 \cdot 25 = 175&lt;/math&gt;.<br /> <br /> === Addendum to Solution 3 ===<br /> Since this is an AIME problem, we can assume that we are done since we have found one possible case. However, in a full-solution contest, we could not assume that the answer is unique, but would need to prove that this is the unique solution. This can be proven as follows.<br /> <br /> Suppose that &lt;math&gt;\Delta &gt; 0&lt;/math&gt;, which would mean that there could be two real roots of &lt;math&gt;f(x)&lt;/math&gt;, one lying in the interval &lt;math&gt;(-3,3)&lt;/math&gt;, and another outside of it. We also see, by [[Vieta's Formulas]], that the average of the two roots is &lt;math&gt;\frac{5\cos \alpha}{2}&lt;/math&gt;, which is non-negative, so the root outside of &lt;math&gt;(-3,3)&lt;/math&gt; must be no less than &lt;math&gt;3&lt;/math&gt;. By considering the graph of &lt;math&gt;y=f(x)&lt;/math&gt;, which is a &quot;U-shaped&quot; parabola, it is now evident that &lt;math&gt;f(-3) &gt; 0&lt;/math&gt; and &lt;math&gt;f(3)\leq 0&lt;/math&gt;. We can just use the second inequality:<br /> <br /> &lt;cmath&gt;0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt; 3\cos \alpha + 4 \sin \alpha \geq 5 &lt;/cmath&gt;<br /> <br /> The only way for this inequality to be satisfied is when &lt;math&gt;A=B&lt;/math&gt; (by applying the Cauchy-Schwarz inequality, or just plotting the line &lt;math&gt;3x+4y=5&lt;/math&gt; to see that point &lt;math&gt;A&lt;/math&gt; can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point &lt;math&gt;A&lt;/math&gt; lies in the half-plane above the line &lt;math&gt;3x+4y=5&lt;/math&gt;, inclusive, and the half-plane below the line &lt;math&gt;-3x+4y=5&lt;/math&gt;, exclusive. This can be seen to be impossible by drawing the lines and observing that the intersection of the two half-planes does not share any point with the circle.)<br /> ===Solution 4===<br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. Fix &lt;math&gt;B,C,&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt;. Then, as &lt;math&gt;D&lt;/math&gt; moves around the circle, the locus of the midpoints of &lt;math&gt;AD&lt;/math&gt; is clearly a circle. Since the problems gives that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; bisected by &lt;math&gt;BC&lt;/math&gt;, it follows that the circle with diameter &lt;math&gt;DO&lt;/math&gt; and &lt;math&gt;AO&lt;/math&gt; is tangent to &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> Now, let the intersection of &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt; and let the midpoint of &lt;math&gt;AO&lt;/math&gt; (the center of the circle tangent to &lt;math&gt;BC&lt;/math&gt; that we described beforehand) be &lt;math&gt;F&lt;/math&gt;. Drop the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call its intersection with &lt;math&gt;BC&lt;/math&gt; &lt;math&gt;K&lt;/math&gt;. Drop the perpendicular from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;KO&lt;/math&gt; and call its intersection with &lt;math&gt;KO&lt;/math&gt; &lt;math&gt;L&lt;/math&gt;. Clearly, &lt;math&gt;KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4&lt;/math&gt; and since &lt;math&gt;EF&lt;/math&gt; is radius, it equals &lt;math&gt;\frac{5}{2}&lt;/math&gt;. The same applies for &lt;math&gt;FO&lt;/math&gt;, which also equals &lt;math&gt;\frac{5}{2}&lt;/math&gt;. By the Pythagorean theorem, we deduce that &lt;math&gt;FL = 2&lt;/math&gt;, so &lt;math&gt;EK = 2&lt;/math&gt;. This is very important information! Now we know that &lt;math&gt;BE = 1&lt;/math&gt;, so by Power of a Point, &lt;math&gt;AE = ED = \sqrt{5}&lt;/math&gt;.<br /> <br /> We’re almost there! Since by the Pythagorean theorem, &lt;math&gt;ED^2 + EO^2 = 5&lt;/math&gt;, we deduce that &lt;math&gt;EO = 2\sqrt{5}&lt;/math&gt;. &lt;math&gt;EC=OC=5&lt;/math&gt;, so &lt;math&gt;\sin (CEO) = \frac{2\sqrt{5}}{5}&lt;/math&gt;. Furthermore, since &lt;math&gt;\sin (CEO) = \cos(DEC)&lt;/math&gt;, we know that &lt;math&gt;\cos (DEC) = \frac{2\sqrt{5}}{5}&lt;/math&gt;. By the law of cosines,<br /> &lt;cmath&gt;DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10&lt;/cmath&gt;Therefore, &lt;math&gt;DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}&lt;/math&gt;. Now, drop the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BA&lt;/math&gt; and call its intersection with &lt;math&gt;BA&lt;/math&gt; &lt;math&gt;Z&lt;/math&gt;. Then, by the Pythagorean theorem, &lt;math&gt;OZ = \frac{7\sqrt{2}}{2}&lt;/math&gt;. Thus, &lt;math&gt;\sin (BOZ) = \frac{\sqrt{2}}{10}&lt;/math&gt; and &lt;math&gt;\cos (BOZ) = \frac{7\sqrt{2}}{10}&lt;/math&gt;. As a result, &lt;math&gt;\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}&lt;/math&gt;. &lt;math&gt;7 \cdot 25 = \boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> [[Image:Dgram.png|thumb|none|800px]]<br /> <br /> Let I be the intersection of AD and BC.<br /> <br /> Lemma: &lt;math&gt;AI = ID&lt;/math&gt; if and only if &lt;math&gt;\angle AIO = 90&lt;/math&gt;.<br /> <br /> Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If &lt;math&gt;\angle AIO = 90&lt;/math&gt;, We can get &lt;math&gt;\triangle AIO \cong \triangle OID&lt;/math&gt;<br /> <br /> Let be this the circle with diameter AO.<br /> <br /> Thus, we get &lt;math&gt;\angle AIO = 90&lt;/math&gt;, implying I must lie on &lt;math&gt;\omega&lt;/math&gt;. I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC.<br /> <br /> Let Y be a point such that &lt;math&gt;\angle DOY = 90&lt;/math&gt;, and minor arc DY contains A,B. <br /> Let X be the center of &lt;math&gt;\omega&lt;/math&gt;. Since &lt;math&gt;\omega&lt;/math&gt;'s radius is &lt;math&gt;\frac{5}{2}&lt;/math&gt;, the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so &lt;math&gt;sin(XOY) = sin(AOY) = \frac{3}{5}&lt;/math&gt;. &lt;math&gt;sin(BOD) = \frac{3}{5}&lt;/math&gt;. If we let &lt;math&gt;sin(\theta) = \frac{3}{5}&lt;/math&gt;, we can find that what we are looking for is &lt;math&gt;sin(90 - 2\theta)&lt;/math&gt;, which we can evaluate and get &lt;math&gt;\frac{7}{25} \implies \boxed{175}&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_15&diff=132397 1983 AIME Problems/Problem 15 2020-08-23T16:39:01Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> The adjoining figure shows two intersecting chords in a circle, with &lt;math&gt;B&lt;/math&gt; on minor arc &lt;math&gt;AD&lt;/math&gt;. Suppose that the radius of the circle is &lt;math&gt;5&lt;/math&gt;, that &lt;math&gt;BC=6&lt;/math&gt;, and that &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt;. Suppose further that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;. It follows that the sine of the central angle of minor arc &lt;math&gt;AB&lt;/math&gt; is a rational number. If this number is expressed as a fraction &lt;math&gt;\frac{m}{n}&lt;/math&gt; in lowest terms, what is the product &lt;math&gt;mn&lt;/math&gt;?<br /> <br /> &lt;asy&gt;size(100);<br /> defaultpen(linewidth(.8pt)+fontsize(11pt));<br /> dotfactor=1;<br /> pair O1=(0,0);<br /> pair A=(-0.91,-0.41);<br /> pair B=(-0.99,0.13);<br /> pair C=(0.688,0.728);<br /> pair D=(-0.25,0.97);<br /> path C1=Circle(O1,1);<br /> draw(C1);<br /> label(&quot;$A$&quot;,A,W);<br /> label(&quot;$B$&quot;,B,W);<br /> label(&quot;$C$&quot;,C,NE);<br /> label(&quot;$D$&quot;,D,N);<br /> draw(A--D);<br /> draw(B--C);<br /> pair F=intersectionpoint(A--D,B--C);<br /> add(pathticks(A--F,1,0.5,0,3.5));<br /> add(pathticks(F--D,1,0.5,0,3.5));<br /> &lt;/asy&gt;<br /> &lt;!-- [[Image:1983_AIME-15.png|200px]] --&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> -Credit to Adamz for diagram-<br /> &lt;asy&gt;<br /> size(10cm);<br /> import olympiad;<br /> pair O = (0,0);dot(O);label(&quot;$O$&quot;,O,SW);<br /> pair M = (4,0);dot(M);label(&quot;$M$&quot;,M,SE);<br /> pair N = (4,2);dot(N);label(&quot;$N$&quot;,N,NE);<br /> draw(circle(O,5));<br /> pair B = (4,3);dot(B);label(&quot;$B$&quot;,B,NE);<br /> pair C = (4,-3);dot(C);label(&quot;$C$&quot;,C,SE);<br /> draw(B--C);draw(O--M);<br /> pair P = (1.5,2);dot(P);label(&quot;$P$&quot;,P,W);<br /> draw(circle(P,2.5));<br /> pair A=(3,4);dot(A);label(&quot;$A$&quot;,A,NE);<br /> draw(O--A);<br /> draw(O--B);<br /> pair Q = (1.5,0); dot(Q); label(&quot;$Q$&quot;,Q,S);<br /> pair R = (3,0); dot(R); label(&quot;$R$&quot;,R,S);<br /> draw(P--Q,dotted); draw(A--R,dotted);<br /> pair D=(5,0); dot(D); label(&quot;$D$&quot;,D,E);<br /> draw(A--D);<br /> &lt;/asy&gt;Let &lt;math&gt;A&lt;/math&gt; be any fixed point on circle &lt;math&gt;O&lt;/math&gt;, and let &lt;math&gt;AD&lt;/math&gt; be a chord of circle &lt;math&gt;O&lt;/math&gt;. The [[locus]] of midpoints &lt;math&gt;N&lt;/math&gt; of the chord &lt;math&gt;AD&lt;/math&gt; is a circle &lt;math&gt;P&lt;/math&gt;, with diameter &lt;math&gt;AO&lt;/math&gt;. Generally, the circle &lt;math&gt;P&lt;/math&gt; can intersect the chord &lt;math&gt;BC&lt;/math&gt; at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle &lt;math&gt;P&lt;/math&gt; is tangent to &lt;math&gt;BC&lt;/math&gt; at point &lt;math&gt;N&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;M&lt;/math&gt; be the midpoint of the chord &lt;math&gt;BC&lt;/math&gt;. From right triangle &lt;math&gt;OMB&lt;/math&gt;, we have &lt;math&gt;OM = \sqrt{OB^2 - BM^2} =4&lt;/math&gt;. This gives &lt;math&gt;\tan \angle BOM = \frac{BM}{OM} = \frac 3 4&lt;/math&gt;.<br /> <br /> Notice that the distance &lt;math&gt;OM&lt;/math&gt; equals &lt;math&gt;PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)&lt;/math&gt;, where &lt;math&gt;r&lt;/math&gt; is the radius of circle &lt;math&gt;P&lt;/math&gt;. <br /> <br /> Hence &lt;cmath&gt;\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5&lt;/cmath&gt; (where &lt;math&gt;R&lt;/math&gt; represents the radius, &lt;math&gt;5&lt;/math&gt;, of the large circle given in the question). Therefore, since &lt;math&gt;\angle AOM&lt;/math&gt; is clearly acute, we see that &lt;cmath&gt;\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3&lt;/cmath&gt;<br /> <br /> Next, notice that &lt;math&gt;\angle AOB = \angle AOM - \angle BOM&lt;/math&gt;. We can therefore apply the subtraction formula for &lt;math&gt;\tan&lt;/math&gt; to obtain &lt;cmath&gt;\tan \angle AOB =\frac{\tan \angle AOM - \tan \angle BOM}{1 + \tan \angle AOM \cdot \tan \angle BOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}&lt;/cmath&gt; It follows that &lt;math&gt;\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}&lt;/math&gt;, such that the answer is &lt;math&gt;7 \cdot 25=\boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> This solution, while similar to Solution 1, is far more motivated and less contrived.<br /> <br /> Firstly, we note the statement in the problem that &quot;&lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; and bisected by &lt;math&gt;BC&lt;/math&gt;&quot; &amp;ndash; what is its significance? What is the criterion for this statement to be true?<br /> <br /> We consider the locus of midpoints of the chords from &lt;math&gt;A&lt;/math&gt;. It is well-known that this is the circle with diameter &lt;math&gt;AO&lt;/math&gt;, where &lt;math&gt;O&lt;/math&gt; is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor &lt;math&gt;\frac{1}{2}&lt;/math&gt; and center &lt;math&gt;A&lt;/math&gt;. Thus, the locus is the result of the dilation with scale factor &lt;math&gt;\frac{1}{2}&lt;/math&gt; and centre &lt;math&gt;A&lt;/math&gt; of circle &lt;math&gt;O&lt;/math&gt;. Let the center of this circle be &lt;math&gt;P&lt;/math&gt;.<br /> <br /> Now, &lt;math&gt;AD&lt;/math&gt; is bisected by &lt;math&gt;BC&lt;/math&gt; if they cross at some point &lt;math&gt;N&lt;/math&gt; on the circle. Moreover, since &lt;math&gt;AD&lt;/math&gt; is the only chord, &lt;math&gt;BC&lt;/math&gt; must be tangent to the circle &lt;math&gt;P&lt;/math&gt;.<br /> <br /> The rest of this problem is straightforward.<br /> <br /> Our goal is to find &lt;math&gt;\sin \angle AOB = \sin{\left(\angle AOM - \angle BOM\right)}&lt;/math&gt;, where &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;BC&lt;/math&gt;. We have &lt;math&gt;BM=3&lt;/math&gt; and &lt;math&gt;OM=4&lt;/math&gt;.<br /> Let &lt;math&gt;R&lt;/math&gt; be the projection of &lt;math&gt;A&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;, and similarly let &lt;math&gt;Q&lt;/math&gt; be the projection of &lt;math&gt;P&lt;/math&gt; onto &lt;math&gt;OM&lt;/math&gt;. Then it remains to find &lt;math&gt;AR&lt;/math&gt; so that we can use the addition formula for &lt;math&gt;\sin&lt;/math&gt;.<br /> <br /> As &lt;math&gt;PN&lt;/math&gt; is a radius of circle &lt;math&gt;P&lt;/math&gt;, &lt;math&gt;PN=2.5&lt;/math&gt;, and similarly, &lt;math&gt;PO=2.5&lt;/math&gt;. Since &lt;math&gt;OM=4&lt;/math&gt;, we have &lt;math&gt;OQ=OM-QM=OM-PN=4-2.5=1.5&lt;/math&gt;. Thus &lt;math&gt;PQ=\sqrt{2.5^2-1.5^2}=2&lt;/math&gt;.<br /> <br /> Further, we see that &lt;math&gt;\triangle OAR&lt;/math&gt; is a dilation of &lt;math&gt;\triangle OPQ&lt;/math&gt; about center &lt;math&gt;O&lt;/math&gt; with scale factor &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;AR=2PQ=4&lt;/math&gt;.<br /> <br /> Lastly, we apply the formula:<br /> &lt;cmath&gt; \sin{\left(\angle AOM - \angle BOM\right)} = \sin \angle AOM \cos \angle BOM - \sin \angle BOM \cos \angle AOM = \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)=\frac{7}{25}&lt;/cmath&gt;<br /> Thus the answer is &lt;math&gt;7\cdot25=\boxed{175}&lt;/math&gt;.<br /> <br /> === Solution 3 (coordinate geometry) ===<br /> [[File:Aime1983p15s2.png|500px|link=]]<br /> <br /> Let the circle have equation &lt;math&gt;x^2 + y^2 = 25&lt;/math&gt;, with centre &lt;math&gt;O(0,0)&lt;/math&gt;. Since &lt;math&gt;BC=6&lt;/math&gt;, we can calculate (by the Pythagorean Theorem) that the distance from &lt;math&gt;O&lt;/math&gt; to the line &lt;math&gt;BC&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;. Therefore, we can let &lt;math&gt;B=(3,4)&lt;/math&gt; and &lt;math&gt;C=(-3,4)&lt;/math&gt;. Now, assume that &lt;math&gt;A&lt;/math&gt; is any point on the major arc BC, and &lt;math&gt;D&lt;/math&gt; any point on the minor arc BC. We can write &lt;math&gt;A=(5 \cos \alpha, 5 \sin \alpha)&lt;/math&gt;, where &lt;math&gt;\alpha&lt;/math&gt; is the angle measured from the positive &lt;math&gt;x&lt;/math&gt; axis to the ray &lt;math&gt;OA&lt;/math&gt;. It will also be convenient to define &lt;math&gt;\angle XOB = \alpha_0&lt;/math&gt;. <br /> <br /> Firstly, since &lt;math&gt;B&lt;/math&gt; must lie in the minor arc &lt;math&gt;AD&lt;/math&gt;, we see that &lt;math&gt;\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)&lt;/math&gt;. However, since the midpoint of &lt;math&gt;AD&lt;/math&gt; must lie on &lt;math&gt;BC&lt;/math&gt;, and the highest possible &lt;math&gt;y&lt;/math&gt;-coordinate of &lt;math&gt;D&lt;/math&gt; is &lt;math&gt;5&lt;/math&gt;, we see that the &lt;math&gt;y&lt;/math&gt;-coordinate cannot be lower than &lt;math&gt;3&lt;/math&gt;, that is, &lt;math&gt;\alpha \in \left[\sin^{-1}\frac{3}{5},\alpha_0\right)&lt;/math&gt;.<br /> <br /> Secondly, there is a theorem that says that, in a circle, if a chord is bisected by a radius, then they must be perpendicular. Therefore, suppose that &lt;math&gt;P&lt;/math&gt; is the intersection point of &lt;math&gt;AD&lt;/math&gt; and &lt;math&gt;BC&lt;/math&gt;, so that by the theorem, &lt;math&gt;OP&lt;/math&gt; is perpendicular to &lt;math&gt;AD&lt;/math&gt;. So, if &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; which is bisected by &lt;math&gt;BC&lt;/math&gt;, this means that &lt;math&gt;P&lt;/math&gt; is the only point on the chord &lt;math&gt;BC&lt;/math&gt; such that &lt;math&gt;OP&lt;/math&gt; is perpendicular to &lt;math&gt;AD&lt;/math&gt;. Now suppose that &lt;math&gt;P=(p,4)&lt;/math&gt;, where &lt;math&gt;p \in (-3,3)&lt;/math&gt;. The fact that &lt;math&gt;OP&lt;/math&gt; must be perpendicular to &lt;math&gt;AD&lt;/math&gt; is equivalent to the following equation:<br /> <br /> &lt;cmath&gt; -1 = \left(\text{slope of } OP\right)\left(\text{slope of } AP\right)&lt;/cmath&gt;<br /> which becomes<br /> &lt;cmath&gt; -1 = \frac{4}{p} \cdot \frac{5\sin \alpha - 4}{5\cos \alpha - p}&lt;/cmath&gt;<br /> <br /> This rearranges to<br /> <br /> &lt;cmath&gt; p^2 - (5\cos \alpha)p + 16 - 20 \sin \alpha = 0&lt;/cmath&gt;<br /> <br /> Given that this equation must have only one real root &lt;math&gt;p\in (-3,3)&lt;/math&gt;, we study the following function:<br /> <br /> &lt;cmath&gt;f(x) = x^2 - (5\cos \alpha)x + 16 - 20 \sin \alpha&lt;/cmath&gt;<br /> <br /> First, by the fact that the equation &lt;math&gt;f(x)=0&lt;/math&gt; has real solutions, its discriminant &lt;math&gt;\Delta&lt;/math&gt; must be non-negative, so we calculate<br /> <br /> &lt;cmath&gt; \begin{split}\Delta &amp; = (5\cos \alpha)^2 - 4(16-20\sin \alpha) \\<br /> &amp; = 25 (1- \sin^2 \alpha) - 64 + 80 \sin \alpha \\<br /> &amp; = -25 \sin^2 \alpha + 80\sin \alpha - 39 \\<br /> &amp; = (13 - 5\sin \alpha)(5\sin \alpha - 3)\end{split}&lt;/cmath&gt;<br /> <br /> It is obvious that this is in fact non-negative. If it is actually zero, then &lt;math&gt;\sin \alpha = \frac{3}{5}&lt;/math&gt;, and &lt;math&gt;\cos \alpha = \frac{4}{5}&lt;/math&gt;. In this case, &lt;math&gt;p = \frac{5\cos \alpha}{2} = 2 \in (-3,3)&lt;/math&gt;, so we have found a possible solution. We thus calculate &lt;math&gt;\sin(\text{central angle of minor arc } AB) = \sin (\alpha_0 - \alpha) = \frac{4}{5}\cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{5} = \frac{7}{25}&lt;/math&gt; by the subtraction formula for &lt;math&gt;\sin&lt;/math&gt;. This means that the answer is &lt;math&gt;7 \cdot 25 = 175&lt;/math&gt;.<br /> <br /> === Addendum to Solution 3 ===<br /> Since this is an AIME problem, we can assume that we are done since we have found one possible case. However, in a full-solution contest, we could not assume that the answer is unique, but would need to prove that this is the unique solution. This can be proven as follows.<br /> <br /> Suppose that &lt;math&gt;\Delta &gt; 0&lt;/math&gt;, which would mean that there could be two real roots of &lt;math&gt;f(x)&lt;/math&gt;, one lying in the interval &lt;math&gt;(-3,3)&lt;/math&gt;, and another outside of it. We also see, by [[Vieta's Formulas]], that the average of the two roots is &lt;math&gt;\frac{5\cos \alpha}{2}&lt;/math&gt;, which is non-negative, so the root outside of &lt;math&gt;(-3,3)&lt;/math&gt; must be no less than &lt;math&gt;3&lt;/math&gt;. By considering the graph of &lt;math&gt;y=f(x)&lt;/math&gt;, which is a &quot;U-shaped&quot; parabola, it is now evident that &lt;math&gt;f(-3) &gt; 0&lt;/math&gt; and &lt;math&gt;f(3)\leq 0&lt;/math&gt;. We can just use the second inequality:<br /> <br /> &lt;cmath&gt;0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha&lt;/cmath&gt;<br /> so<br /> &lt;cmath&gt; 3\cos \alpha + 4 \sin \alpha \geq 5 &lt;/cmath&gt;<br /> <br /> The only way for this inequality to be satisfied is when &lt;math&gt;A=B&lt;/math&gt; (by applying the Cauchy-Schwarz inequality, or just plotting the line &lt;math&gt;3x+4y=5&lt;/math&gt; to see that point &lt;math&gt;A&lt;/math&gt; can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point &lt;math&gt;A&lt;/math&gt; lies in the half-plane above the line &lt;math&gt;3x+4y=5&lt;/math&gt;, inclusive, and the half-plane below the line &lt;math&gt;-3x+4y=5&lt;/math&gt;, exclusive. This can be seen to be impossible by drawing the lines and observing that the intersection of the two half-planes does not share any point with the circle.)<br /> ===Solution 4===<br /> Let the center of the circle be &lt;math&gt;O&lt;/math&gt;. Fix &lt;math&gt;B,C,&lt;/math&gt; and &lt;math&gt;A&lt;/math&gt;. Then, as &lt;math&gt;D&lt;/math&gt; moves around the circle, the locus of the midpoints of &lt;math&gt;AD&lt;/math&gt; is clearly a circle. Since the problems gives that &lt;math&gt;AD&lt;/math&gt; is the only chord starting at &lt;math&gt;A&lt;/math&gt; bisected by &lt;math&gt;BC&lt;/math&gt;, it follows that the circle with diameter &lt;math&gt;DO&lt;/math&gt; and &lt;math&gt;AO&lt;/math&gt; is tangent to &lt;math&gt;BC&lt;/math&gt;.<br /> <br /> Now, let the intersection of &lt;math&gt;BC&lt;/math&gt; and &lt;math&gt;AD&lt;/math&gt; be &lt;math&gt;E&lt;/math&gt; and let the midpoint of &lt;math&gt;AO&lt;/math&gt; (the center of the circle tangent to &lt;math&gt;BC&lt;/math&gt; that we described beforehand) be &lt;math&gt;F&lt;/math&gt;. Drop the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BC&lt;/math&gt; and call its intersection with &lt;math&gt;BC&lt;/math&gt; &lt;math&gt;K&lt;/math&gt;. Drop the perpendicular from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;KO&lt;/math&gt; and call its intersection with &lt;math&gt;KO&lt;/math&gt; &lt;math&gt;L&lt;/math&gt;. Clearly, &lt;math&gt;KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4&lt;/math&gt; and since &lt;math&gt;EF&lt;/math&gt; is radius, it equals &lt;math&gt;\frac{5}{2}&lt;/math&gt;. The same applies for &lt;math&gt;FO&lt;/math&gt;, which also equals &lt;math&gt;\frac{5}{2}&lt;/math&gt;. By the Pythagorean theorem, we deduce that &lt;math&gt;FL = 2&lt;/math&gt;, so &lt;math&gt;EK = 2&lt;/math&gt;. This is very important information! Now we know that &lt;math&gt;BE = 1&lt;/math&gt;, so by Power of a Point, &lt;math&gt;AE = ED = \sqrt{5}&lt;/math&gt;.<br /> <br /> We’re almost there! Since by the Pythagorean theorem, &lt;math&gt;ED^2 + EO^2 = 5&lt;/math&gt;, we deduce that &lt;math&gt;EO = 2\sqrt{5}&lt;/math&gt;. &lt;math&gt;EC=OC=5&lt;/math&gt;, so &lt;math&gt;\sin (CEO) = \frac{2\sqrt{5}}{5}&lt;/math&gt;. Furthermore, since &lt;math&gt;\sin (CEO) = \cos(DEC)&lt;/math&gt;, we know that &lt;math&gt;\cos (DEC) = \frac{2\sqrt{5}}{5}&lt;/math&gt;. By the law of cosines,<br /> &lt;cmath&gt;DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10&lt;/cmath&gt;Therefore, &lt;math&gt;DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}&lt;/math&gt;. Now, drop the altitude from &lt;math&gt;O&lt;/math&gt; to &lt;math&gt;BA&lt;/math&gt; and call its intersection with &lt;math&gt;BA&lt;/math&gt; &lt;math&gt;Z&lt;/math&gt;. Then, by the Pythagorean theorem, &lt;math&gt;OZ = \frac{7\sqrt{2}}{2}&lt;/math&gt;. Thus, &lt;math&gt;\sin (BOZ) = \frac{\sqrt{2}}{10}&lt;/math&gt; and &lt;math&gt;\cos (BOZ) = \frac{7\sqrt{2}}{10}&lt;/math&gt;. As a result, &lt;math&gt;\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}&lt;/math&gt;. &lt;math&gt;7 \cdot 25 = \boxed{175}&lt;/math&gt;.<br /> <br /> == Solution 5==<br /> <br /> [[Image:Dgram.png|thumb|none|800px]]<br /> <br /> Let I be the intersection of AD and BC.<br /> <br /> Lemma: &lt;math&gt;AI = ID&lt;/math&gt; if and only if &lt;math&gt;\angle AIO = 90&lt;/math&gt;.<br /> <br /> Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If &lt;math&gt;\angle AIO = 90&lt;/math&gt;, We can get &lt;math&gt;\triangle AIO \cong \triangle OID&lt;/math&gt;<br /> <br /> Let be this the circle with diameter AO.<br /> <br /> Thus, we get &lt;math&gt;\angle AIO = 90&lt;/math&gt;, implying I must lie on &lt;math&gt;\omega&lt;/math&gt;. I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC.<br /> <br /> Let Y be a point such that &lt;math&gt;\angle DOY = 90&lt;/math&gt;, and minor arc DY contains A,B. <br /> Let X be the center of &lt;math&gt;\omega&lt;/math&gt;. Since &lt;math&gt;\omega&lt;/math&gt;'s radius is &lt;math&gt;\frac{5}{2}&lt;/math&gt;, the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so &lt;math&gt;sin(XOY) = sin(AOY) = \frac{3}{5}&lt;/math&gt;. &lt;math&gt;sin(BOD) = \frac{3}{5}&lt;/math&gt;. If we let &lt;math&gt;sin(\theta) = \frac{3}{5}&lt;/math&gt;, we can find that what we are looking for is &lt;math&gt;sin(90 - 2\theta)&lt;/math&gt;, which we can evaluate and get &lt;math&gt;\frac{7}{25} \implies \boxed{175}&lt;/math&gt;<br /> == See Also ==<br /> {{AIME box|year=1983|num-b=14|after=Last question}}<br /> <br /> [[Category:Intermediate Geometry Problems]]</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=File:Dgram.png&diff=132386 File:Dgram.png 2020-08-23T15:44:15Z <p>Alexlikemath: </p> <hr /> <div></div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_5&diff=130511 2005 USAMO Problems/Problem 5 2020-08-04T04:05:37Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> (''Kiran Kedlaya'') Let &lt;math&gt;n&lt;/math&gt; be an integer greater than 1. Suppose &lt;math&gt;2n&lt;/math&gt; points are given in the plane, no three of which are collinear. Suppose &lt;math&gt;n&lt;/math&gt; of the given &lt;math&gt;2n&lt;/math&gt; points are colored blue and the other &lt;math&gt;n&lt;/math&gt; colored red. A line in the plane is called a ''balancing line'' if it passes through one blue and one red point and, for each side of the line, the number of blue points on that side is equal to the number of red points on the same side.<br /> <br /> Prove that there exist at least two balancing lines.<br /> <br /> == Solution ==<br /> Consider the convex hull of the the &lt;math&gt;2n&lt;/math&gt; points, or the points that would form the largest convex polygon. If the points in the convex hull contain both red and blue points, then there must be at least 2 edges of the graph of the convex hull such that the edge connects a blue and a red point. Drawing a line through those points would give a balancing line, as we have n-1 blue points and n-1 red points on one side, and 0 points on the other.<br /> <br /> Therefore it suffices to show that there exist at least 2 balancing lines when the convex hull is colored all the same color. <br /> <br /> Pick a random point on the convex hull, and without loss of generality we can say it is blue (if there are no red we can change all the colors, and end up with an equivalent setup). Consider a line going through it and not any other points. As we rotate the line clockwise, we encounter the red points in some order. Let the ith point encountered be &lt;math&gt;R_i&lt;/math&gt;. Let &lt;math&gt;b_i&lt;/math&gt; and &lt;math&gt;r_i&lt;/math&gt; be the number of points encountered before &lt;math&gt;R_i&lt;/math&gt;. Then &lt;math&gt;r_i=i-1&lt;/math&gt;.<br /> <br /> Define a sequence &lt;math&gt;f(i)=b_i-r_i&lt;/math&gt;. Then &lt;math&gt;f(1)=b_i-(1-1)=b_i\geq0&lt;/math&gt;<br /> &lt;math&gt;f(n)=b_n-(n-1)=b_n-n+1\leq0&lt;/math&gt;, because we can only encounter up to n-1 blue points.<br /> Thus, &lt;math&gt;f(i)&lt;/math&gt; goes from negative to positive as &lt;math&gt;i&lt;/math&gt; goes from 1 to &lt;math&gt;n&lt;/math&gt;. We can also see that &lt;math&gt;f(i)&lt;/math&gt; can only increase by one for each change in i, so we know &lt;math&gt;f(i)&lt;/math&gt; must be 0 for some value of &lt;math&gt;i&lt;/math&gt;. Take the first &lt;math&gt;f(i)&lt;/math&gt; where &lt;math&gt;f(i) = 0&lt;/math&gt;. Since &lt;math&gt;f(i-1)&lt;/math&gt; must have been positive, the ith point must have been red. Thus there is a balancing line for every point on the convex hull. Since a polygon must have at least 3 vertices, there must be at least &lt;math&gt;3&gt;2&lt;/math&gt; balancing lines for the set of points when the convex hull is all the same color, and the statement is true as we desired.<br /> <br /> {{alternate solutions}}<br /> <br /> == See Also==<br /> {{USAMO newbox|year=2005|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2005_USAMO_Problems/Problem_5&diff=130510 2005 USAMO Problems/Problem 5 2020-08-04T04:05:11Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> (''Kiran Kedlaya'') Let &lt;math&gt;n&lt;/math&gt; be an integer greater than 1. Suppose &lt;math&gt;2n&lt;/math&gt; points are given in the plane, no three of which are collinear. Suppose &lt;math&gt;n&lt;/math&gt; of the given &lt;math&gt;2n&lt;/math&gt; points are colored blue and the other &lt;math&gt;n&lt;/math&gt; colored red. A line in the plane is called a ''balancing line'' if it passes through one blue and one red point and, for each side of the line, the number of blue points on that side is equal to the number of red points on the same side.<br /> <br /> Prove that there exist at least two balancing lines.<br /> <br /> == Solution ==<br /> Consider the convex hull of the the &lt;math&gt;2n&lt;/math&gt; points, or the points that would form the largest convex polygon. If the points in the convex hull contain both red and blue points, then there must be at least 2 edges of the graph of the convex hull such that the edge connects a blue and a red point. Drawing a line through those points would give a balancing line, as we have n-1 blue points and n-1 red points on one side, and 0 points on the other.<br /> <br /> Therefore it suffices to show that there exist at least 2 balancing lines when the convex hull is colored all the same color. <br /> <br /> Pick a random point on the convex hull, and without loss of generality we can say it is blue (if there are no red we can change all the colors, and end up with an equivalent setup). Consider a line going through it and not any other points. As we rotate the line clockwise, we encounter the red points in some order. Let the ith point encountered be &lt;math&gt;R_i&lt;/math&gt;. Let &lt;math&gt;b_i&lt;/math&gt; and &lt;math&gt;r_i&lt;/math&gt; be the number of points encountered before &lt;math&gt;R_i&lt;/math&gt;. Then &lt;math&gt;r_i=i-1&lt;/math&gt;.<br /> <br /> Define a sequence &lt;math&gt;f(i)=b_i-r_i&lt;/math&gt;. Then &lt;math&gt;f(1)=b_i-(1-1)=b_i\geq0&lt;/math&gt;<br /> &lt;math&gt;f(n)=b_n-(n-1)=b_n-n+1\leq0&lt;/math&gt;, because we can only encounter up to n-1 blue points.<br /> Thus, &lt;math&gt;f(i)&lt;/math&gt; goes from negative to positive as &lt;math&gt;i&lt;/math&gt; goes from 1 to &lt;math&gt;n&lt;/math&gt;. We can also see that &lt;math&gt;f(i)&lt;/math&gt; can only increase by one for each change in i, so we know &lt;math&gt;f(i)&lt;/math&gt; must be 0 for some value of &lt;math&gt;i&lt;/math&gt;. Take the first f(i) where f(i) = 0. Since f(i-1) must have been positive, the ith point must have been red. Thus there is a balancing line for every point on the convex hull. Since a polygon must have at least 3 vertices, there must be at least &lt;math&gt;3&gt;2&lt;/math&gt; balancing lines for the set of points when the convex hull is all the same color, and the statement is true as we desired.<br /> <br /> {{alternate solutions}}<br /> <br /> == See Also==<br /> {{USAMO newbox|year=2005|num-b=4|num-a=6}}<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_5&diff=125883 2010 USAJMO Problems/Problem 5 2020-06-19T02:37:50Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Two permutations &lt;math&gt;a_1, a_2, \ldots, a_{2010}&lt;/math&gt; and<br /> &lt;math&gt;b_1, b_2, \ldots, b_{2010}&lt;/math&gt; of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt;<br /> are said to intersect if &lt;math&gt;a_k = b_k&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt; in the<br /> range &lt;math&gt;1 \le k\le 2010&lt;/math&gt;. Show that there exist &lt;math&gt;1006&lt;/math&gt; permutations<br /> of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt; such that any other such<br /> permutation is guaranteed to intersect at least one of these &lt;math&gt;1006&lt;/math&gt;<br /> permutations.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Let &lt;math&gt;m&lt;/math&gt; be the smallest positive integer with<br /> &lt;math&gt;m &gt; n - m&lt;/math&gt;. Since &lt;math&gt;n &gt; n - n = 0&lt;/math&gt;, &lt;math&gt;m \le n&lt;/math&gt;. Let &lt;math&gt;N = \{1, \ldots, n\}&lt;/math&gt;<br /> be the set of positive integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;M \subset N&lt;/math&gt;,<br /> be &lt;math&gt;M = \{1, \ldots, m\}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P_n&lt;/math&gt; be the set of of permutations of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C_m&lt;/math&gt; be the set of cyclic permutations of &lt;math&gt;M&lt;/math&gt;, there are &lt;math&gt;m&lt;/math&gt;<br /> cyclic permutations in all, and &lt;math&gt;C_m&lt;/math&gt; acts transitively on &lt;math&gt;M&lt;/math&gt;, i.e.<br /> for every pair of elements &lt;math&gt;a,b \in M&lt;/math&gt;, there is an element of &lt;math&gt;C_m&lt;/math&gt;<br /> that maps &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'_m \subset P_n&lt;/math&gt; be the permutations in &lt;math&gt;P_n&lt;/math&gt; that leave &lt;math&gt;N\setminus M&lt;/math&gt;<br /> fixed, and restricted to &lt;math&gt;M&lt;/math&gt; yield one of the permutations in &lt;math&gt;C_m&lt;/math&gt;.<br /> There is a natural one-to-one correspondence between &lt;math&gt;C'_m&lt;/math&gt; and &lt;math&gt;C_m&lt;/math&gt;.<br /> <br /> We claim that the &lt;math&gt;m&lt;/math&gt; permutations &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> Suppose, to the contrary, that there exists a permutation &lt;math&gt;p \in P_n&lt;/math&gt;<br /> that does not intersect any permutation in &lt;math&gt;C'_m&lt;/math&gt;. Since &lt;math&gt;C'_m&lt;/math&gt; acts<br /> transitively on &lt;math&gt;M \subset N&lt;/math&gt; the permutation &lt;math&gt;p&lt;/math&gt; cannot send any element of<br /> &lt;math&gt;M&lt;/math&gt; to any other element of &lt;math&gt;M&lt;/math&gt;, therefore it must send all the<br /> elements of &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;N\setminus M&lt;/math&gt;, but since &lt;math&gt;N\setminus M&lt;/math&gt; has &lt;math&gt;n<br /> - m&lt;/math&gt; elements and &lt;math&gt;m &gt; n - m&lt;/math&gt;, this is impossible<br /> by the pigeonhole principle. Therefore such a permutation cannot<br /> exist, and the permutations in &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n = 2010&lt;/math&gt; we get &lt;math&gt;m = 1006&lt;/math&gt;, which is the required special<br /> case of the general result above.<br /> <br /> ==Solution 2==<br /> I construct the following permutations by continuously rotating the first 1006 numbers):<br /> <br /> (1, 2, 3 ... 1005, 1006, 1007 ... 2009, 2010)<br /> <br /> (2, 3, 4 ... 1006, 1, 1007 ... 2009, 2010)<br /> <br /> ...<br /> <br /> ...<br /> <br /> ...<br /> <br /> (1006, 1, 2 ... 1005, 1007, ... 2009, 2010)<br /> <br /> <br /> <br /> I claim that these permutations above satisfy the property that any other permutation will intersect with at least one of them.<br /> <br /> '''Proof''':<br /> Assume for the sake of contradiction that there exists a permutation that won't intersect with these, say &lt;math&gt;(a_1, a_2, a_3 ... a_{2010})&lt;/math&gt;.<br /> <br /> '''Lemma''':<br /> If there exists a k such that &lt;math&gt;1 \leq k \leq 1006&lt;/math&gt; &lt;math&gt;a_k \leq 1006&lt;/math&gt;, we will get an intersection with the permutations. <br /> <br /> Note that for any 2 distinct permutations in our list, the numbers in kth index must be different, since each permutation is a rotation of an previous permutation. Also, the numbers can't exceed 1006, so, each number must occur exactly once in the kth index.<br /> <br /> '''End Lemma'''<br /> <br /> Using the lemma, in order to avoid intersections, we need &lt;math&gt;1007 \leq a_k \leq 2010&lt;/math&gt;, for all &lt;math&gt;1 \leq k \leq 1006&lt;/math&gt;.<br /> <br /> But, there are 1004 numbers such that &lt;math&gt;1007 \leq n \leq 2010&lt;/math&gt;, but we need to have 1004 numbers to fill in 1006 spots, and thus, by pigeonhole principle(the numbers are the holes, the spots are the pigeons), there must be 2 spots that have the same number! However, in a permutation, all numbers must be distinct, so we have a contradiction.<br /> <br /> So, we have shown such a set of permutations exists satisfying that any permutation is guaranteed to intersect one of them, and the proof is complete. &lt;math&gt;\blacksquare&lt;/math&gt; <br /> <br /> -Alexlikemath<br /> <br /> == See Also ==<br /> {{USAJMO newbox|year=2010|num-b=4|num-a=6}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_5&diff=125882 2010 USAJMO Problems/Problem 5 2020-06-19T02:36:43Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Two permutations &lt;math&gt;a_1, a_2, \ldots, a_{2010}&lt;/math&gt; and<br /> &lt;math&gt;b_1, b_2, \ldots, b_{2010}&lt;/math&gt; of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt;<br /> are said to intersect if &lt;math&gt;a_k = b_k&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt; in the<br /> range &lt;math&gt;1 \le k\le 2010&lt;/math&gt;. Show that there exist &lt;math&gt;1006&lt;/math&gt; permutations<br /> of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt; such that any other such<br /> permutation is guaranteed to intersect at least one of these &lt;math&gt;1006&lt;/math&gt;<br /> permutations.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Let &lt;math&gt;m&lt;/math&gt; be the smallest positive integer with<br /> &lt;math&gt;m &gt; n - m&lt;/math&gt;. Since &lt;math&gt;n &gt; n - n = 0&lt;/math&gt;, &lt;math&gt;m \le n&lt;/math&gt;. Let &lt;math&gt;N = \{1, \ldots, n\}&lt;/math&gt;<br /> be the set of positive integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;M \subset N&lt;/math&gt;,<br /> be &lt;math&gt;M = \{1, \ldots, m\}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P_n&lt;/math&gt; be the set of of permutations of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C_m&lt;/math&gt; be the set of cyclic permutations of &lt;math&gt;M&lt;/math&gt;, there are &lt;math&gt;m&lt;/math&gt;<br /> cyclic permutations in all, and &lt;math&gt;C_m&lt;/math&gt; acts transitively on &lt;math&gt;M&lt;/math&gt;, i.e.<br /> for every pair of elements &lt;math&gt;a,b \in M&lt;/math&gt;, there is an element of &lt;math&gt;C_m&lt;/math&gt;<br /> that maps &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'_m \subset P_n&lt;/math&gt; be the permutations in &lt;math&gt;P_n&lt;/math&gt; that leave &lt;math&gt;N\setminus M&lt;/math&gt;<br /> fixed, and restricted to &lt;math&gt;M&lt;/math&gt; yield one of the permutations in &lt;math&gt;C_m&lt;/math&gt;.<br /> There is a natural one-to-one correspondence between &lt;math&gt;C'_m&lt;/math&gt; and &lt;math&gt;C_m&lt;/math&gt;.<br /> <br /> We claim that the &lt;math&gt;m&lt;/math&gt; permutations &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> Suppose, to the contrary, that there exists a permutation &lt;math&gt;p \in P_n&lt;/math&gt;<br /> that does not intersect any permutation in &lt;math&gt;C'_m&lt;/math&gt;. Since &lt;math&gt;C'_m&lt;/math&gt; acts<br /> transitively on &lt;math&gt;M \subset N&lt;/math&gt; the permutation &lt;math&gt;p&lt;/math&gt; cannot send any element of<br /> &lt;math&gt;M&lt;/math&gt; to any other element of &lt;math&gt;M&lt;/math&gt;, therefore it must send all the<br /> elements of &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;N\setminus M&lt;/math&gt;, but since &lt;math&gt;N\setminus M&lt;/math&gt; has &lt;math&gt;n<br /> - m&lt;/math&gt; elements and &lt;math&gt;m &gt; n - m&lt;/math&gt;, this is impossible<br /> by the pigeonhole principle. Therefore such a permutation cannot<br /> exist, and the permutations in &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n = 2010&lt;/math&gt; we get &lt;math&gt;m = 1006&lt;/math&gt;, which is the required special<br /> case of the general result above.<br /> <br /> ==Solution 2==<br /> I construct the following permutations by continuously rotating the first 1006 numbers):<br /> <br /> (1, 2, 3 ... 1005, 1006, 1007 ... 2009, 2010)<br /> <br /> (2, 3, 4 ... 1006, 1, 1007 ... 2009, 2010)<br /> <br /> ...<br /> <br /> ...<br /> <br /> ...<br /> <br /> (1006, 1, 2 ... 1005, 1007, ... 2009, 2010)<br /> <br /> <br /> <br /> I claim that these permutations above satisfy the property that any other permutation will intersect with at least one of them.<br /> <br /> '''Proof''':<br /> Assume for the sake of contradiction that there exists a permutation that won't intersect with these, say &lt;math&gt;(a_1, a_2, a_3 ... a_{2010})&lt;/math&gt;.<br /> <br /> '''Lemma''':<br /> If there exists a k such that &lt;math&gt;1 \leq k \leq 1006&lt;/math&gt; &lt;math&gt;a_k \leq 1006&lt;/math&gt;, we will get an intersection with the permutations. <br /> <br /> Note that for any 2 distinct permutations in our list, the numbers in kth index must be different, since each permutation is a rotation of an previous permutation. Also, the numbers can't exceed 1006, so, each number must occur exactly once in the kth index.<br /> <br /> '''End Lemma'''<br /> <br /> Using the lemma, in order to avoid intersections, we need &lt;math&gt;1007 \leq a_k \leq 2010&lt;/math&gt;, for all &lt;math&gt;1 \leq k \leq 1006&lt;/math&gt;.<br /> <br /> But, there are 1004 numbers such that &lt;math&gt;1007 \leq n \leq 2010&lt;/math&gt;, but we need to have 1004 numbers to fill in 1006 spots, and thus, by pigeonhole principle(the numbers are the holes, the spots are the pigeons), there must be 2 spots that have the same number! However, in a permutation, all numbers must be distinct, so we have a contradiction.<br /> <br /> We have constructed a set of 1006 permutations. So, such a set of permutations exists satisfying that any permutation is guaranteed to intersect one of them, and the proof is complete. &lt;math&gt;\blacksquare&lt;/math&gt; <br /> <br /> -Alexlikemath<br /> <br /> == See Also ==<br /> {{USAJMO newbox|year=2010|num-b=4|num-a=6}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_5&diff=125881 2010 USAJMO Problems/Problem 5 2020-06-19T02:34:11Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Two permutations &lt;math&gt;a_1, a_2, \ldots, a_{2010}&lt;/math&gt; and<br /> &lt;math&gt;b_1, b_2, \ldots, b_{2010}&lt;/math&gt; of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt;<br /> are said to intersect if &lt;math&gt;a_k = b_k&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt; in the<br /> range &lt;math&gt;1 \le k\le 2010&lt;/math&gt;. Show that there exist &lt;math&gt;1006&lt;/math&gt; permutations<br /> of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt; such that any other such<br /> permutation is guaranteed to intersect at least one of these &lt;math&gt;1006&lt;/math&gt;<br /> permutations.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Let &lt;math&gt;m&lt;/math&gt; be the smallest positive integer with<br /> &lt;math&gt;m &gt; n - m&lt;/math&gt;. Since &lt;math&gt;n &gt; n - n = 0&lt;/math&gt;, &lt;math&gt;m \le n&lt;/math&gt;. Let &lt;math&gt;N = \{1, \ldots, n\}&lt;/math&gt;<br /> be the set of positive integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;M \subset N&lt;/math&gt;,<br /> be &lt;math&gt;M = \{1, \ldots, m\}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P_n&lt;/math&gt; be the set of of permutations of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C_m&lt;/math&gt; be the set of cyclic permutations of &lt;math&gt;M&lt;/math&gt;, there are &lt;math&gt;m&lt;/math&gt;<br /> cyclic permutations in all, and &lt;math&gt;C_m&lt;/math&gt; acts transitively on &lt;math&gt;M&lt;/math&gt;, i.e.<br /> for every pair of elements &lt;math&gt;a,b \in M&lt;/math&gt;, there is an element of &lt;math&gt;C_m&lt;/math&gt;<br /> that maps &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'_m \subset P_n&lt;/math&gt; be the permutations in &lt;math&gt;P_n&lt;/math&gt; that leave &lt;math&gt;N\setminus M&lt;/math&gt;<br /> fixed, and restricted to &lt;math&gt;M&lt;/math&gt; yield one of the permutations in &lt;math&gt;C_m&lt;/math&gt;.<br /> There is a natural one-to-one correspondence between &lt;math&gt;C'_m&lt;/math&gt; and &lt;math&gt;C_m&lt;/math&gt;.<br /> <br /> We claim that the &lt;math&gt;m&lt;/math&gt; permutations &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> Suppose, to the contrary, that there exists a permutation &lt;math&gt;p \in P_n&lt;/math&gt;<br /> that does not intersect any permutation in &lt;math&gt;C'_m&lt;/math&gt;. Since &lt;math&gt;C'_m&lt;/math&gt; acts<br /> transitively on &lt;math&gt;M \subset N&lt;/math&gt; the permutation &lt;math&gt;p&lt;/math&gt; cannot send any element of<br /> &lt;math&gt;M&lt;/math&gt; to any other element of &lt;math&gt;M&lt;/math&gt;, therefore it must send all the<br /> elements of &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;N\setminus M&lt;/math&gt;, but since &lt;math&gt;N\setminus M&lt;/math&gt; has &lt;math&gt;n<br /> - m&lt;/math&gt; elements and &lt;math&gt;m &gt; n - m&lt;/math&gt;, this is impossible<br /> by the pigeonhole principle. Therefore such a permutation cannot<br /> exist, and the permutations in &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n = 2010&lt;/math&gt; we get &lt;math&gt;m = 1006&lt;/math&gt;, which is the required special<br /> case of the general result above.<br /> <br /> ==Solution 2==<br /> I construct the following permutations by continuously rotating the first 1006 numbers):<br /> <br /> (1, 2, 3 ... 1005, 1006, 1007 ... 2009, 2010)<br /> <br /> (2, 3, 4 ... 1006, 1, 1007 ... 2009, 2010)<br /> <br /> ...<br /> <br /> ...<br /> <br /> ...<br /> <br /> (1006, 1, 2 ... 1005, 1007, ... 2009, 2010)<br /> <br /> <br /> <br /> I claim that these permutations above satisfy the property that any other permutation will intersect with at least one of them.<br /> <br /> '''Proof''':<br /> Assume for the sake of contradiction that there exists a permutation that won't intersect with these, say &lt;math&gt;(a_1, a_2, a_3 ... a_{2010})&lt;/math&gt;.<br /> <br /> '''Lemma''':<br /> If there exists a k such that &lt;math&gt;1 \leq k \leq 1006&lt;/math&gt; &lt;math&gt;a_k \leq 1006&lt;/math&gt;, we will get an intersection with the permutations. <br /> <br /> Note that for any 2 distinct permutations in our list, the numbers in kth index must be different, since each permutation is a rotation of an previous permutation. Also, the numbers can't exceed 1006, so, each number must occur exactly once in the kth index.<br /> <br /> '''End Lemma'''<br /> <br /> Using the lemma, in order to avoid intersections, we need &lt;math&gt;1007 \leq a_k \leq 2010&lt;/math&gt;, for all &lt;math&gt;1 \leq k \leq 1006&lt;/math&gt;.<br /> <br /> But, there are 1004 numbers such that &lt;math&gt;1007 \leq n \leq 2010&lt;/math&gt;, but we need to have 1004 numbers to fill in 1006 spots, and thus, by pigeonhole principle(the numbers are the holes, the spots are the pigeons), there must be 2 spots that have the same number! However, in a permutation, all numbers must be distinct, so we have a contradiction.<br /> <br /> We have constructed a set of 1006 permutations. So, such a permutation exists, and the proof is complete. &lt;math&gt;\blacksquare&lt;/math&gt; <br /> <br /> -Alexlikemath<br /> <br /> == See Also ==<br /> {{USAJMO newbox|year=2010|num-b=4|num-a=6}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_5&diff=125880 2010 USAJMO Problems/Problem 5 2020-06-19T02:30:04Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Two permutations &lt;math&gt;a_1, a_2, \ldots, a_{2010}&lt;/math&gt; and<br /> &lt;math&gt;b_1, b_2, \ldots, b_{2010}&lt;/math&gt; of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt;<br /> are said to intersect if &lt;math&gt;a_k = b_k&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt; in the<br /> range &lt;math&gt;1 \le k\le 2010&lt;/math&gt;. Show that there exist &lt;math&gt;1006&lt;/math&gt; permutations<br /> of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt; such that any other such<br /> permutation is guaranteed to intersect at least one of these &lt;math&gt;1006&lt;/math&gt;<br /> permutations.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Let &lt;math&gt;m&lt;/math&gt; be the smallest positive integer with<br /> &lt;math&gt;m &gt; n - m&lt;/math&gt;. Since &lt;math&gt;n &gt; n - n = 0&lt;/math&gt;, &lt;math&gt;m \le n&lt;/math&gt;. Let &lt;math&gt;N = \{1, \ldots, n\}&lt;/math&gt;<br /> be the set of positive integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;M \subset N&lt;/math&gt;,<br /> be &lt;math&gt;M = \{1, \ldots, m\}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P_n&lt;/math&gt; be the set of of permutations of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C_m&lt;/math&gt; be the set of cyclic permutations of &lt;math&gt;M&lt;/math&gt;, there are &lt;math&gt;m&lt;/math&gt;<br /> cyclic permutations in all, and &lt;math&gt;C_m&lt;/math&gt; acts transitively on &lt;math&gt;M&lt;/math&gt;, i.e.<br /> for every pair of elements &lt;math&gt;a,b \in M&lt;/math&gt;, there is an element of &lt;math&gt;C_m&lt;/math&gt;<br /> that maps &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'_m \subset P_n&lt;/math&gt; be the permutations in &lt;math&gt;P_n&lt;/math&gt; that leave &lt;math&gt;N\setminus M&lt;/math&gt;<br /> fixed, and restricted to &lt;math&gt;M&lt;/math&gt; yield one of the permutations in &lt;math&gt;C_m&lt;/math&gt;.<br /> There is a natural one-to-one correspondence between &lt;math&gt;C'_m&lt;/math&gt; and &lt;math&gt;C_m&lt;/math&gt;.<br /> <br /> We claim that the &lt;math&gt;m&lt;/math&gt; permutations &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> Suppose, to the contrary, that there exists a permutation &lt;math&gt;p \in P_n&lt;/math&gt;<br /> that does not intersect any permutation in &lt;math&gt;C'_m&lt;/math&gt;. Since &lt;math&gt;C'_m&lt;/math&gt; acts<br /> transitively on &lt;math&gt;M \subset N&lt;/math&gt; the permutation &lt;math&gt;p&lt;/math&gt; cannot send any element of<br /> &lt;math&gt;M&lt;/math&gt; to any other element of &lt;math&gt;M&lt;/math&gt;, therefore it must send all the<br /> elements of &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;N\setminus M&lt;/math&gt;, but since &lt;math&gt;N\setminus M&lt;/math&gt; has &lt;math&gt;n<br /> - m&lt;/math&gt; elements and &lt;math&gt;m &gt; n - m&lt;/math&gt;, this is impossible<br /> by the pigeonhole principle. Therefore such a permutation cannot<br /> exist, and the permutations in &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n = 2010&lt;/math&gt; we get &lt;math&gt;m = 1006&lt;/math&gt;, which is the required special<br /> case of the general result above.<br /> <br /> ==Solution 2==<br /> I construct the following permutations by continuously rotating the first 1006 numbers):<br /> <br /> (1, 2, 3 ... 1005, 1006, 1007 ... 2009, 2010)<br /> <br /> (2, 3, 4 ... 1006, 1, 1007 ... 2009, 2010)<br /> <br /> ...<br /> <br /> ...<br /> <br /> ...<br /> <br /> (1006, 1, 2 ... 1005, 1007, ... 2009, 2010)<br /> <br /> <br /> <br /> I claim that these permutations above satisfy the property that any other permutation will intersect with at least one of them.<br /> <br /> '''Proof''':<br /> Assume for the sake of contradiction that there exists a permutation that won't intersect with these, say &lt;math&gt;(a_1, a_2, a_3 ... a_{2010})&lt;/math&gt;.<br /> <br /> '''Lemma''':<br /> If there exists a k such that &lt;math&gt;1 \leq k \leq 1006&lt;/math&gt; &lt;math&gt;a_k \leq 1006&lt;/math&gt;, we will get an intersection with the permutations. <br /> <br /> Note that for any 2 distinct permutations in our list, the numbers in kth index must be different, since each permutation is a rotation of an previous permutation. Also, the numbers can't exceed 1006, so, each number must occur exactly once in the kth index.<br /> <br /> '''End Lemma'''<br /> <br /> Using the lemma, in order to avoid intersections, we need &lt;math&gt;1 \leq k \leq 1006&lt;/math&gt;, &lt;math&gt;1006 &lt; a_k \leq 2010&lt;/math&gt;.<br /> <br /> But, there are 1004 numbers such that &lt;math&gt;1006 &lt; n \leq 2010&lt;/math&gt;, but we need to have 1004 numbers to fill in 1006 spots, and thus, by pigeonhole principle(the numbers are the holes, the spots are the pigeons), there must be 2 spots that have the same number! However, in a permutation, all numbers must be distinct, so we have a contradiction.<br /> <br /> Thus, a set of 1006 permutations does exist satisfying the given conditions. Thus, the proof is complete. <br /> <br /> -Alexlikemath<br /> <br /> == See Also ==<br /> {{USAJMO newbox|year=2010|num-b=4|num-a=6}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_5&diff=125824 2010 USAJMO Problems/Problem 5 2020-06-18T18:01:53Z <p>Alexlikemath: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> Two permutations &lt;math&gt;a_1, a_2, \ldots, a_{2010}&lt;/math&gt; and<br /> &lt;math&gt;b_1, b_2, \ldots, b_{2010}&lt;/math&gt; of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt;<br /> are said to intersect if &lt;math&gt;a_k = b_k&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt; in the<br /> range &lt;math&gt;1 \le k\le 2010&lt;/math&gt;. Show that there exist &lt;math&gt;1006&lt;/math&gt; permutations<br /> of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt; such that any other such<br /> permutation is guaranteed to intersect at least one of these &lt;math&gt;1006&lt;/math&gt;<br /> permutations.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Let &lt;math&gt;m&lt;/math&gt; be the smallest positive integer with<br /> &lt;math&gt;m &gt; n - m&lt;/math&gt;. Since &lt;math&gt;n &gt; n - n = 0&lt;/math&gt;, &lt;math&gt;m \le n&lt;/math&gt;. Let &lt;math&gt;N = \{1, \ldots, n\}&lt;/math&gt;<br /> be the set of positive integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;M \subset N&lt;/math&gt;,<br /> be &lt;math&gt;M = \{1, \ldots, m\}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P_n&lt;/math&gt; be the set of of permutations of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C_m&lt;/math&gt; be the set of cyclic permutations of &lt;math&gt;M&lt;/math&gt;, there are &lt;math&gt;m&lt;/math&gt;<br /> cyclic permutations in all, and &lt;math&gt;C_m&lt;/math&gt; acts transitively on &lt;math&gt;M&lt;/math&gt;, i.e.<br /> for every pair of elements &lt;math&gt;a,b \in M&lt;/math&gt;, there is an element of &lt;math&gt;C_m&lt;/math&gt;<br /> that maps &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'_m \subset P_n&lt;/math&gt; be the permutations in &lt;math&gt;P_n&lt;/math&gt; that leave &lt;math&gt;N\setminus M&lt;/math&gt;<br /> fixed, and restricted to &lt;math&gt;M&lt;/math&gt; yield one of the permutations in &lt;math&gt;C_m&lt;/math&gt;.<br /> There is a natural one-to-one correspondence between &lt;math&gt;C'_m&lt;/math&gt; and &lt;math&gt;C_m&lt;/math&gt;.<br /> <br /> We claim that the &lt;math&gt;m&lt;/math&gt; permutations &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> Suppose, to the contrary, that there exists a permutation &lt;math&gt;p \in P_n&lt;/math&gt;<br /> that does not intersect any permutation in &lt;math&gt;C'_m&lt;/math&gt;. Since &lt;math&gt;C'_m&lt;/math&gt; acts<br /> transitively on &lt;math&gt;M \subset N&lt;/math&gt; the permutation &lt;math&gt;p&lt;/math&gt; cannot send any element of<br /> &lt;math&gt;M&lt;/math&gt; to any other element of &lt;math&gt;M&lt;/math&gt;, therefore it must send all the<br /> elements of &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;N\setminus M&lt;/math&gt;, but since &lt;math&gt;N\setminus M&lt;/math&gt; has &lt;math&gt;n<br /> - m&lt;/math&gt; elements and &lt;math&gt;m &gt; n - m&lt;/math&gt;, this is impossible<br /> by the pigeonhole principle. Therefore such a permutation cannot<br /> exist, and the permutations in &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n = 2010&lt;/math&gt; we get &lt;math&gt;m = 1006&lt;/math&gt;, which is the required special<br /> case of the general result above.<br /> <br /> ==Solution 2==<br /> I construct the following permutations by continuously rotating the first 1006 numbers):<br /> <br /> (1, 2, 3 ... 1005, 1006, 1007 ... 2009, 2010)<br /> <br /> (2, 3, 4 ... 1006, 1, 1007 ... 2009, 2010)<br /> <br /> ...<br /> <br /> ...<br /> <br /> ...<br /> <br /> (1006, 1, 2 ... 1005, 1007, ... 2009, 2010)<br /> <br /> <br /> <br /> I claim that these permutations satisfy the property that any other permutation will intersect with at least one of them.<br /> <br /> '''Proof''':<br /> Assume that there exists a permutation that won't intersect with these, say &lt;math&gt;(a_1, a_2, a_3 ... a_{2010})&lt;/math&gt;.<br /> <br /> '''Lemma''':<br /> If &lt;math&gt;a_k \leq 1006&lt;/math&gt;, &lt;math&gt;1 \leq k \leq 1006&lt;/math&gt;, we get an intersection. <br /> <br /> Note that for any 2 distinct permutations in our list, the numbers in kth index must be different, since each permutation is a rotation of an previous permutation. Also, the numbers can't exceed 1006, so, each number must occur exactly once in the kth index.<br /> <br /> Using the lemma, we get that for all &lt;math&gt;1 \geq k \geq 1006&lt;/math&gt;, &lt;math&gt;2010 \geq a_k &gt; 1006&lt;/math&gt;.<br /> <br /> But, there are 1004 numbers such that &lt;math&gt;2010 \geq n &gt; 1006&lt;/math&gt;, but we need to have 1004 numbers to fill in 1006 spots, and thus, by pigeonhole principle(the numbers are the holes, the spots are the pigeons), there must be 2 spots that have the same number! However, in a permutation, all numbers must be distinct, so we have a contradiction.<br /> <br /> Thus, such a set of 1006 permutations does exist satisfying the given conditions. Thus, the proof is complete. <br /> <br /> -Alexlikemath<br /> <br /> == See Also ==<br /> {{USAJMO newbox|year=2010|num-b=4|num-a=6}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_5&diff=125821 2010 USAJMO Problems/Problem 5 2020-06-18T17:30:11Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Two permutations &lt;math&gt;a_1, a_2, \ldots, a_{2010}&lt;/math&gt; and<br /> &lt;math&gt;b_1, b_2, \ldots, b_{2010}&lt;/math&gt; of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt;<br /> are said to intersect if &lt;math&gt;a_k = b_k&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt; in the<br /> range &lt;math&gt;1 \le k\le 2010&lt;/math&gt;. Show that there exist &lt;math&gt;1006&lt;/math&gt; permutations<br /> of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt; such that any other such<br /> permutation is guaranteed to intersect at least one of these &lt;math&gt;1006&lt;/math&gt;<br /> permutations.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Let &lt;math&gt;m&lt;/math&gt; be the smallest positive integer with<br /> &lt;math&gt;m &gt; n - m&lt;/math&gt;. Since &lt;math&gt;n &gt; n - n = 0&lt;/math&gt;, &lt;math&gt;m \le n&lt;/math&gt;. Let &lt;math&gt;N = \{1, \ldots, n\}&lt;/math&gt;<br /> be the set of positive integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;M \subset N&lt;/math&gt;,<br /> be &lt;math&gt;M = \{1, \ldots, m\}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P_n&lt;/math&gt; be the set of of permutations of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C_m&lt;/math&gt; be the set of cyclic permutations of &lt;math&gt;M&lt;/math&gt;, there are &lt;math&gt;m&lt;/math&gt;<br /> cyclic permutations in all, and &lt;math&gt;C_m&lt;/math&gt; acts transitively on &lt;math&gt;M&lt;/math&gt;, i.e.<br /> for every pair of elements &lt;math&gt;a,b \in M&lt;/math&gt;, there is an element of &lt;math&gt;C_m&lt;/math&gt;<br /> that maps &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'_m \subset P_n&lt;/math&gt; be the permutations in &lt;math&gt;P_n&lt;/math&gt; that leave &lt;math&gt;N\setminus M&lt;/math&gt;<br /> fixed, and restricted to &lt;math&gt;M&lt;/math&gt; yield one of the permutations in &lt;math&gt;C_m&lt;/math&gt;.<br /> There is a natural one-to-one correspondence between &lt;math&gt;C'_m&lt;/math&gt; and &lt;math&gt;C_m&lt;/math&gt;.<br /> <br /> We claim that the &lt;math&gt;m&lt;/math&gt; permutations &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> Suppose, to the contrary, that there exists a permutation &lt;math&gt;p \in P_n&lt;/math&gt;<br /> that does not intersect any permutation in &lt;math&gt;C'_m&lt;/math&gt;. Since &lt;math&gt;C'_m&lt;/math&gt; acts<br /> transitively on &lt;math&gt;M \subset N&lt;/math&gt; the permutation &lt;math&gt;p&lt;/math&gt; cannot send any element of<br /> &lt;math&gt;M&lt;/math&gt; to any other element of &lt;math&gt;M&lt;/math&gt;, therefore it must send all the<br /> elements of &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;N\setminus M&lt;/math&gt;, but since &lt;math&gt;N\setminus M&lt;/math&gt; has &lt;math&gt;n<br /> - m&lt;/math&gt; elements and &lt;math&gt;m &gt; n - m&lt;/math&gt;, this is impossible<br /> by the pigeonhole principle. Therefore such a permutation cannot<br /> exist, and the permutations in &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n = 2010&lt;/math&gt; we get &lt;math&gt;m = 1006&lt;/math&gt;, which is the required special<br /> case of the general result above.<br /> <br /> ==Solution 2==<br /> I claim that the permutations:<br /> <br /> [1, 2, 3 ... 1005, 1006, 1007 ... 2009, 2010]<br /> <br /> [2, 3, 4 ... 1006, 1, 1007 ... 2009, 2010] <br /> <br /> ...<br /> <br /> ...<br /> <br /> ...<br /> <br /> [1006, 1, 2 ... 1005, 1007, ... 2009, 2010]<br /> <br /> (constructed by continuously rotating the first 1006 numbers)<br /> <br /> satisfy the property that any other permutation will intersect with at least one of these permutations.<br /> <br /> '''Proof''':<br /> Assume that there exists a permutation that won't intersect with these, say &lt;math&gt;(a_1 ... a_{2010})&lt;/math&gt;. If &lt;math&gt;a_k \leq 1006&lt;/math&gt;, &lt;math&gt;1 \geq k \geq 1006&lt;/math&gt;, we get an intersection. So, for all &lt;math&gt;1 \geq k \geq 1006&lt;/math&gt;, &lt;math&gt;2010 \geq a_k &gt; 1006&lt;/math&gt;. There are 1004 numbers such that &lt;math&gt;2010 \geq a_k &gt; 1006&lt;/math&gt;, but we need to have 1006 numbers to fill in 1006 spots, and thus, by pigeonhole principle(the numbers are the holes, the spots are the pigeons), there must be 2 spots that have the same number! However, in a permutation, all numbers must be distinct, so we have a contradiction.<br /> <br /> Thus, such a set of 1006 permutations does exist satisfying that any other permutation does intersect at least 1 of the permutations. Thus, the proof is complete. <br /> <br /> -Alexlikemath<br /> == See Also ==<br /> {{USAJMO newbox|year=2010|num-b=4|num-a=6}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_5&diff=125820 2010 USAJMO Problems/Problem 5 2020-06-18T17:24:07Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Two permutations &lt;math&gt;a_1, a_2, \ldots, a_{2010}&lt;/math&gt; and<br /> &lt;math&gt;b_1, b_2, \ldots, b_{2010}&lt;/math&gt; of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt;<br /> are said to intersect if &lt;math&gt;a_k = b_k&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt; in the<br /> range &lt;math&gt;1 \le k\le 2010&lt;/math&gt;. Show that there exist &lt;math&gt;1006&lt;/math&gt; permutations<br /> of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt; such that any other such<br /> permutation is guaranteed to intersect at least one of these &lt;math&gt;1006&lt;/math&gt;<br /> permutations.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Let &lt;math&gt;m&lt;/math&gt; be the smallest positive integer with<br /> &lt;math&gt;m &gt; n - m&lt;/math&gt;. Since &lt;math&gt;n &gt; n - n = 0&lt;/math&gt;, &lt;math&gt;m \le n&lt;/math&gt;. Let &lt;math&gt;N = \{1, \ldots, n\}&lt;/math&gt;<br /> be the set of positive integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;M \subset N&lt;/math&gt;,<br /> be &lt;math&gt;M = \{1, \ldots, m\}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P_n&lt;/math&gt; be the set of of permutations of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C_m&lt;/math&gt; be the set of cyclic permutations of &lt;math&gt;M&lt;/math&gt;, there are &lt;math&gt;m&lt;/math&gt;<br /> cyclic permutations in all, and &lt;math&gt;C_m&lt;/math&gt; acts transitively on &lt;math&gt;M&lt;/math&gt;, i.e.<br /> for every pair of elements &lt;math&gt;a,b \in M&lt;/math&gt;, there is an element of &lt;math&gt;C_m&lt;/math&gt;<br /> that maps &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'_m \subset P_n&lt;/math&gt; be the permutations in &lt;math&gt;P_n&lt;/math&gt; that leave &lt;math&gt;N\setminus M&lt;/math&gt;<br /> fixed, and restricted to &lt;math&gt;M&lt;/math&gt; yield one of the permutations in &lt;math&gt;C_m&lt;/math&gt;.<br /> There is a natural one-to-one correspondence between &lt;math&gt;C'_m&lt;/math&gt; and &lt;math&gt;C_m&lt;/math&gt;.<br /> <br /> We claim that the &lt;math&gt;m&lt;/math&gt; permutations &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> Suppose, to the contrary, that there exists a permutation &lt;math&gt;p \in P_n&lt;/math&gt;<br /> that does not intersect any permutation in &lt;math&gt;C'_m&lt;/math&gt;. Since &lt;math&gt;C'_m&lt;/math&gt; acts<br /> transitively on &lt;math&gt;M \subset N&lt;/math&gt; the permutation &lt;math&gt;p&lt;/math&gt; cannot send any element of<br /> &lt;math&gt;M&lt;/math&gt; to any other element of &lt;math&gt;M&lt;/math&gt;, therefore it must send all the<br /> elements of &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;N\setminus M&lt;/math&gt;, but since &lt;math&gt;N\setminus M&lt;/math&gt; has &lt;math&gt;n<br /> - m&lt;/math&gt; elements and &lt;math&gt;m &gt; n - m&lt;/math&gt;, this is impossible<br /> by the pigeonhole principle. Therefore such a permutation cannot<br /> exist, and the permutations in &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n = 2010&lt;/math&gt; we get &lt;math&gt;m = 1006&lt;/math&gt;, which is the required special<br /> case of the general result above.<br /> <br /> ==Solution 2==<br /> I claim that the permutations:<br /> <br /> [1, 2, 3 ... 1005, 1006, 1007 ... 2009, 2010]<br /> <br /> [2, 3, 4 ... 1006, 1, 1007 ... 2009, 2010] <br /> <br /> ...<br /> <br /> ...<br /> <br /> ...<br /> <br /> [1006, 1, 2 ... 1005, 1007, ... 2009, 2010]<br /> <br /> (constructed by continuously rotating the first 1006 numbers)<br /> <br /> satisfy the property that any other permutation will intersect with at least one of these permutations.<br /> <br /> '''Proof''':<br /> Assume that there exists a permutation that won't intersect with these, say &lt;math&gt;(a_1 ... a_{2010})&lt;/math&gt;. If &lt;math&gt;a_k \leq 1006&lt;/math&gt;, &lt;math&gt;1 \geq k \geq 1006&lt;/math&gt;, we get an intersection. So, for all &lt;math&gt;1 \geq k \geq 1006&lt;/math&gt;, &lt;math&gt;2010 \leq a_k &gt; 1006&lt;/math&gt;. There are 1004 numbers such that &lt;math&gt;2010 \leq a_k &gt; 1006&lt;/math&gt;, but we need to have 1006 numbers to fill in 1006 spots, and thus, by pigeonhole principle(the numbers are the holes, the spots are the pigeons), there must be 2 spots that have the same number! However, in a permutation, all numbers must be distinct, so we have a contradiction.<br /> <br /> Thus, such a set of 1006 permutations does exist satisfying that any other permutation does intersect at least 1 of the permutations. Thus, the proof is complete. <br /> <br /> -Alexlikemath<br /> == See Also ==<br /> {{USAJMO newbox|year=2010|num-b=4|num-a=6}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_5&diff=125819 2010 USAJMO Problems/Problem 5 2020-06-18T17:23:13Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Two permutations &lt;math&gt;a_1, a_2, \ldots, a_{2010}&lt;/math&gt; and<br /> &lt;math&gt;b_1, b_2, \ldots, b_{2010}&lt;/math&gt; of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt;<br /> are said to intersect if &lt;math&gt;a_k = b_k&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt; in the<br /> range &lt;math&gt;1 \le k\le 2010&lt;/math&gt;. Show that there exist &lt;math&gt;1006&lt;/math&gt; permutations<br /> of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt; such that any other such<br /> permutation is guaranteed to intersect at least one of these &lt;math&gt;1006&lt;/math&gt;<br /> permutations.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Let &lt;math&gt;m&lt;/math&gt; be the smallest positive integer with<br /> &lt;math&gt;m &gt; n - m&lt;/math&gt;. Since &lt;math&gt;n &gt; n - n = 0&lt;/math&gt;, &lt;math&gt;m \le n&lt;/math&gt;. Let &lt;math&gt;N = \{1, \ldots, n\}&lt;/math&gt;<br /> be the set of positive integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;M \subset N&lt;/math&gt;,<br /> be &lt;math&gt;M = \{1, \ldots, m\}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P_n&lt;/math&gt; be the set of of permutations of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C_m&lt;/math&gt; be the set of cyclic permutations of &lt;math&gt;M&lt;/math&gt;, there are &lt;math&gt;m&lt;/math&gt;<br /> cyclic permutations in all, and &lt;math&gt;C_m&lt;/math&gt; acts transitively on &lt;math&gt;M&lt;/math&gt;, i.e.<br /> for every pair of elements &lt;math&gt;a,b \in M&lt;/math&gt;, there is an element of &lt;math&gt;C_m&lt;/math&gt;<br /> that maps &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'_m \subset P_n&lt;/math&gt; be the permutations in &lt;math&gt;P_n&lt;/math&gt; that leave &lt;math&gt;N\setminus M&lt;/math&gt;<br /> fixed, and restricted to &lt;math&gt;M&lt;/math&gt; yield one of the permutations in &lt;math&gt;C_m&lt;/math&gt;.<br /> There is a natural one-to-one correspondence between &lt;math&gt;C'_m&lt;/math&gt; and &lt;math&gt;C_m&lt;/math&gt;.<br /> <br /> We claim that the &lt;math&gt;m&lt;/math&gt; permutations &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> Suppose, to the contrary, that there exists a permutation &lt;math&gt;p \in P_n&lt;/math&gt;<br /> that does not intersect any permutation in &lt;math&gt;C'_m&lt;/math&gt;. Since &lt;math&gt;C'_m&lt;/math&gt; acts<br /> transitively on &lt;math&gt;M \subset N&lt;/math&gt; the permutation &lt;math&gt;p&lt;/math&gt; cannot send any element of<br /> &lt;math&gt;M&lt;/math&gt; to any other element of &lt;math&gt;M&lt;/math&gt;, therefore it must send all the<br /> elements of &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;N\setminus M&lt;/math&gt;, but since &lt;math&gt;N\setminus M&lt;/math&gt; has &lt;math&gt;n<br /> - m&lt;/math&gt; elements and &lt;math&gt;m &gt; n - m&lt;/math&gt;, this is impossible<br /> by the pigeonhole principle. Therefore such a permutation cannot<br /> exist, and the permutations in &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n = 2010&lt;/math&gt; we get &lt;math&gt;m = 1006&lt;/math&gt;, which is the required special<br /> case of the general result above.<br /> <br /> ==Solution 2==<br /> I claim that the permutations:<br /> <br /> [1, 2, 3 ... 1005, 1006, 1007 ... 2009, 2010]<br /> <br /> [2, 3, 4 ... 1006, 1, 1007 ... 2009, 2010] <br /> <br /> ...<br /> <br /> ...<br /> <br /> ...<br /> <br /> [1006, 1, 2 ... 1005, 1007, ... 2009, 2010]<br /> <br /> (constructed by continuously rotating the first 1006 numbers)<br /> <br /> satisfy the property that any other permutation will intersect with this.<br /> <br /> '''Proof''':<br /> Assume that there exists a permutation that won't intersect with these, say &lt;math&gt;(a_1 ... a_{2010})&lt;/math&gt;. If &lt;math&gt;a_k \leq 1006&lt;/math&gt;, &lt;math&gt;1 \geq k \geq 1006&lt;/math&gt;, we get an intersection. So, for all &lt;math&gt;1 \geq k \geq 1006&lt;/math&gt;, &lt;math&gt;2010 \leq a_k &gt; 1006&lt;/math&gt;. There are 1004 numbers such that &lt;math&gt;2010 \leq a_k &gt; 1006&lt;/math&gt;, but we need to have 1006 numbers to fill in 1006 spots, and thus, by pigeonhole principle(the numbers are the holes, the spots are the pigeons), there must be 2 spots that have the same number! However, in a permutation, all numbers must be distinct, so we have a contradiction.<br /> <br /> Thus, such a set of 1006 permutations does exist satisfying that any other permutation does intersect at least 1 of the permutations. Thus, the proof is complete. <br /> <br /> -Alexlikemath<br /> == See Also ==<br /> {{USAJMO newbox|year=2010|num-b=4|num-a=6}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_5&diff=125818 2010 USAJMO Problems/Problem 5 2020-06-18T17:21:09Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Two permutations &lt;math&gt;a_1, a_2, \ldots, a_{2010}&lt;/math&gt; and<br /> &lt;math&gt;b_1, b_2, \ldots, b_{2010}&lt;/math&gt; of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt;<br /> are said to intersect if &lt;math&gt;a_k = b_k&lt;/math&gt; for some value of &lt;math&gt;k&lt;/math&gt; in the<br /> range &lt;math&gt;1 \le k\le 2010&lt;/math&gt;. Show that there exist &lt;math&gt;1006&lt;/math&gt; permutations<br /> of the numbers &lt;math&gt;1, 2, \ldots, 2010&lt;/math&gt; such that any other such<br /> permutation is guaranteed to intersect at least one of these &lt;math&gt;1006&lt;/math&gt;<br /> permutations.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;n&lt;/math&gt; be a positive integer. Let &lt;math&gt;m&lt;/math&gt; be the smallest positive integer with<br /> &lt;math&gt;m &gt; n - m&lt;/math&gt;. Since &lt;math&gt;n &gt; n - n = 0&lt;/math&gt;, &lt;math&gt;m \le n&lt;/math&gt;. Let &lt;math&gt;N = \{1, \ldots, n\}&lt;/math&gt;<br /> be the set of positive integers from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;n&lt;/math&gt;. Let &lt;math&gt;M \subset N&lt;/math&gt;,<br /> be &lt;math&gt;M = \{1, \ldots, m\}&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;P_n&lt;/math&gt; be the set of of permutations of &lt;math&gt;N&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C_m&lt;/math&gt; be the set of cyclic permutations of &lt;math&gt;M&lt;/math&gt;, there are &lt;math&gt;m&lt;/math&gt;<br /> cyclic permutations in all, and &lt;math&gt;C_m&lt;/math&gt; acts transitively on &lt;math&gt;M&lt;/math&gt;, i.e.<br /> for every pair of elements &lt;math&gt;a,b \in M&lt;/math&gt;, there is an element of &lt;math&gt;C_m&lt;/math&gt;<br /> that maps &lt;math&gt;a&lt;/math&gt; to &lt;math&gt;b&lt;/math&gt;.<br /> <br /> Let &lt;math&gt;C'_m \subset P_n&lt;/math&gt; be the permutations in &lt;math&gt;P_n&lt;/math&gt; that leave &lt;math&gt;N\setminus M&lt;/math&gt;<br /> fixed, and restricted to &lt;math&gt;M&lt;/math&gt; yield one of the permutations in &lt;math&gt;C_m&lt;/math&gt;.<br /> There is a natural one-to-one correspondence between &lt;math&gt;C'_m&lt;/math&gt; and &lt;math&gt;C_m&lt;/math&gt;.<br /> <br /> We claim that the &lt;math&gt;m&lt;/math&gt; permutations &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> Suppose, to the contrary, that there exists a permutation &lt;math&gt;p \in P_n&lt;/math&gt;<br /> that does not intersect any permutation in &lt;math&gt;C'_m&lt;/math&gt;. Since &lt;math&gt;C'_m&lt;/math&gt; acts<br /> transitively on &lt;math&gt;M \subset N&lt;/math&gt; the permutation &lt;math&gt;p&lt;/math&gt; cannot send any element of<br /> &lt;math&gt;M&lt;/math&gt; to any other element of &lt;math&gt;M&lt;/math&gt;, therefore it must send all the<br /> elements of &lt;math&gt;M&lt;/math&gt; to &lt;math&gt;N\setminus M&lt;/math&gt;, but since &lt;math&gt;N\setminus M&lt;/math&gt; has &lt;math&gt;n<br /> - m&lt;/math&gt; elements and &lt;math&gt;m &gt; n - m&lt;/math&gt;, this is impossible<br /> by the pigeonhole principle. Therefore such a permutation cannot<br /> exist, and the permutations in &lt;math&gt;C'_m&lt;/math&gt; intersect every permutation in<br /> &lt;math&gt;P_n&lt;/math&gt;.<br /> <br /> For &lt;math&gt;n = 2010&lt;/math&gt; we get &lt;math&gt;m = 1006&lt;/math&gt;, which is the required special<br /> case of the general result above.<br /> <br /> ==Solution 2==<br /> I claim that the permutations:<br /> <br /> [1, 2, 3 ... 1005, 1006, 1007 ... 2009, 2010]<br /> [2, 3, 4 ... 1006, 1, 1007 ... 2009, 2010] <br /> .<br /> .<br /> .<br /> [1006, 1, 2 ... 1005, 1007, ... 2009, 2010]<br /> <br /> satisfy the property that any other permutation will intersect with this.<br /> <br /> '''Proof''':<br /> Assume that there exists a permutation that won't intersect with these, say &lt;math&gt;(a_1 ... a_{2010})&lt;/math&gt;. If &lt;math&gt;a_k \leq 1006&lt;/math&gt;, &lt;math&gt;1 \geq k \geq 1006&lt;/math&gt;, we get an intersection. So, for all &lt;math&gt;1 \geq k \geq 1006&lt;/math&gt;, &lt;math&gt;2010 \leq a_k &gt; 1006&lt;/math&gt;. There are 1004 numbers such that &lt;math&gt;2010 \leq a_k &gt; 1006&lt;/math&gt;, but we need to have 1006 numbers to fill in 1006 spots, and thus, by pigeonhole principle(the numbers are the holes, the spots are the pigeons), there must be 2 spots that have the same number! However, in a permutation, all numbers must be distinct, so we have a contradiction.<br /> <br /> Thus, such a set of 1006 permutations does exist satisfying that any other permutation does intersect at least 1 of the permutations. Thus, the proof is complete. <br /> == See Also ==<br /> {{USAJMO newbox|year=2010|num-b=4|num-a=6}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2013_USAJMO_Problems/Problem_1&diff=125511 2013 USAJMO Problems/Problem 1 2020-06-15T19:10:41Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> <br /> Are there integers &lt;math&gt; a &lt;/math&gt; and &lt;math&gt; b &lt;/math&gt; such that &lt;math&gt; a^5b+3 &lt;/math&gt; and &lt;math&gt; ab^5+3 &lt;/math&gt; are both perfect cubes of integers?<br /> <br /> ==Solution==<br /> <br /> No, such integers do not exist. This shall be proven by contradiction, by showing that if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube then &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> Remark that perfect cubes are always congruent to &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, or &lt;math&gt;-1&lt;/math&gt; modulo &lt;math&gt;9&lt;/math&gt;. Therefore, if &lt;math&gt;a^5b+3\equiv 0,1,\text{ or} -1\pmod{9}&lt;/math&gt;, then &lt;math&gt;a^5b\equiv 5,6,\text{ or }7\pmod{9}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a^5b\equiv 6\pmod 9&lt;/math&gt;, then note that &lt;math&gt;3|b&lt;/math&gt;. (This is because if &lt;math&gt;3|a&lt;/math&gt; then &lt;math&gt;a^5b\equiv 0\pmod 9&lt;/math&gt;.) Therefore &lt;math&gt;ab^5\equiv 0\pmod 9&lt;/math&gt; and &lt;math&gt;ab^5+3\equiv 3\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Otherwise, either &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt; or &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;. Note that since &lt;math&gt;a^6b^6&lt;/math&gt; is a perfect sixth power, and since neither &lt;math&gt;a&lt;/math&gt; nor &lt;math&gt;b&lt;/math&gt; contains a factor of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;a^6b^6\equiv 1\pmod 9&lt;/math&gt;. If &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 5ab^5\equiv 1\pmod 9\implies ab^5\equiv 2\pmod 9.&lt;/cmath&gt; Similarly, if &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 7ab^5\equiv 1\pmod 9\implies ab^5\equiv 4\pmod 9.&lt;/cmath&gt; Therefore &lt;math&gt;ab^5+3\equiv 5,7\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 2==<br /> We shall prove that such integers do not exist via contradiction.<br /> Suppose that &lt;math&gt;a^5b + 3 = x^3&lt;/math&gt; and &lt;math&gt;ab^5 + 3 = y^3&lt;/math&gt; for integers x and y. Rearranging terms gives &lt;math&gt;a^5b = x^3 - 3&lt;/math&gt; and &lt;math&gt;ab^5 = y^3 - 3&lt;/math&gt;. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = &lt;math&gt;(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}&lt;/math&gt; and b = &lt;math&gt;(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}&lt;/math&gt;. Consider a prime p in the prime factorization of &lt;math&gt;x^3 - 3&lt;/math&gt; and &lt;math&gt;y^3 - 3&lt;/math&gt;. If it has power &lt;math&gt;r_1&lt;/math&gt; in &lt;math&gt;x^3 - 3&lt;/math&gt; and power &lt;math&gt;r_2&lt;/math&gt; in &lt;math&gt;y^3 - 3&lt;/math&gt;, then &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; is a multiple of 24 and &lt;math&gt;5r_2&lt;/math&gt; - &lt;math&gt;r_1&lt;/math&gt; also is a multiple of 24. <br /> <br /> Adding and subtracting the divisions gives that &lt;math&gt;r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; divides 12. (actually, &lt;math&gt;r_1 - r_2&lt;/math&gt; is a multiple of 4, as you can verify if &lt;math&gt;\{^{5r_1 - r_2 = 24}_{5r_2 - r_1 = 48}&lt;/math&gt;. So the rest of the proof is invalid.) Because &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; also divides 12, &lt;math&gt;4r_1&lt;/math&gt; divides 12 and thus &lt;math&gt;r_1&lt;/math&gt; divides 3. Repeating this trick for all primes in &lt;math&gt;x^3 - 3&lt;/math&gt;, we see that &lt;math&gt;x^3 - 3&lt;/math&gt; is a perfect cube, say &lt;math&gt;q^3&lt;/math&gt;. Then &lt;math&gt;x^3 - q^3 = 3,&lt;/math&gt; and &lt;math&gt;(x-q)(x^2 + xq + q^2) = 3&lt;/math&gt;, so that &lt;math&gt;x - q = 1&lt;/math&gt; and &lt;math&gt;x^2 + xq + q^2 = 3&lt;/math&gt;. Clearly, this system of equations has no integer solutions for &lt;math&gt;x&lt;/math&gt; or &lt;math&gt;q&lt;/math&gt;, a contradiction, hence completing the proof.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;a^5b+3=x^3&lt;/math&gt; and &lt;math&gt;ab^5+3=y^3&lt;/math&gt;. Then, &lt;math&gt;a^5b=x^3-3&lt;/math&gt;, &lt;math&gt;ab^5=y^3-3&lt;/math&gt;, and &lt;cmath&gt;(ab)^6=(x^3-3)(y^3-3)&lt;/cmath&gt;<br /> Now take &lt;math&gt;\text{mod }9&lt;/math&gt; (recall that perfect cubes &lt;math&gt;\equiv -1,0,1\pmod{9}&lt;/math&gt; and perfect sixth powers &lt;math&gt;\equiv 0,1\pmod{9}&lt;/math&gt;) on both sides. There are &lt;math&gt;3\times 3=9&lt;/math&gt; cases to consider on what values &lt;math&gt;\text{mod }9&lt;/math&gt; that &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;y^3&lt;/math&gt; take. Checking these &lt;math&gt;9&lt;/math&gt; cases, we see that only &lt;math&gt;x^3\equiv y^3\equiv 0\pmod{9}&lt;/math&gt; or &lt;math&gt;x\equiv y\equiv 0\pmod{3}&lt;/math&gt; yield a valid residue &lt;math&gt;\text{mod }9&lt;/math&gt; (specifically, &lt;math&gt;(x^3-3)(y^3-3)\equiv 0\pmod{9}&lt;/math&gt;). But this means that &lt;math&gt;3\mid ab&lt;/math&gt;, so &lt;math&gt;729\mid (ab)^6&lt;/math&gt; so &lt;cmath&gt;729\mid (x^3-3)(y^3-3)\iff 729\mid (27x'^3-3)(27y'^3-3)\iff 81\mid (9x'^3-1)(9y'^3-1)&lt;/cmath&gt; contradiction.<br /> <br /> ==Solution 4==<br /> If &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube, then &lt;math&gt;a^5b&lt;/math&gt; can be one of &lt;math&gt;5,6,7 \pmod 9&lt;/math&gt;, so &lt;math&gt;(a^5b)^5 = a^{25} b^5&lt;/math&gt; can be one of &lt;math&gt;5^5 \equiv 2&lt;/math&gt;, &lt;math&gt;6^5 \equiv 0&lt;/math&gt;, or &lt;math&gt;7^5 \equiv 4 \pmod 9&lt;/math&gt;. If &lt;math&gt;a&lt;/math&gt; were divisible by &lt;math&gt;3&lt;/math&gt;, we'd have &lt;math&gt;a^5 b \equiv 0 \pmod 9&lt;/math&gt;, which we've ruled out. So &lt;math&gt;\gcd(a,9) = 1&lt;/math&gt;, which means &lt;math&gt;a^6 \equiv 1 \pmod 9&lt;/math&gt;, and therefore &lt;math&gt;a^{25} b^5 \equiv ab^5 \pmod 9&lt;/math&gt;.<br /> <br /> We've shown that &lt;math&gt;a b^5&lt;/math&gt; can be one of &lt;math&gt;0, 2, 4 \pmod 9&lt;/math&gt;, so &lt;math&gt;ab^5 + 3&lt;/math&gt; can be one of &lt;math&gt;3, 5, 7 \pmod 9&lt;/math&gt;. None of these are possibilities for a perfect cube, so if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube, &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> ==Solution 5==<br /> <br /> As in previous solutions, notice &lt;math&gt;ab^5,a^5b \equiv 5,6,7 \pmod 9&lt;/math&gt;. Now multiplying gives &lt;math&gt;a^6b^6&lt;/math&gt;, which is only &lt;math&gt;0,1 \pmod 9&lt;/math&gt;, so after testing all cases we find that &lt;math&gt;ab^5\equiv a^5b \equiv 6 \mod 9&lt;/math&gt;. Then since &lt;math&gt;\phi (9) = 6&lt;/math&gt;, &lt;math&gt;ab^5\equiv \frac{a}{b}\pmod 9&lt;/math&gt; and &lt;math&gt;a^5b \equiv \frac{b}{a}\pmod 9&lt;/math&gt; (Note that &lt;math&gt;a,b&lt;/math&gt; cannot be &lt;math&gt;0\pmod 9&lt;/math&gt;). Thus we find that the inverse of &lt;math&gt;6&lt;/math&gt; is itself under modulo &lt;math&gt;9&lt;/math&gt;, a contradiction.<br /> <br /> ==Solution 6==<br /> I claim there are no such a or b such that both expressions are cubes.<br /> <br /> Assume to the contrary &lt;math&gt;a^5b +3&lt;/math&gt; and &lt;math&gt;ab^5 + 3&lt;/math&gt; are cubes. <br /> <br /> '''Lemma 1''': If &lt;math&gt;a^5b +3&lt;/math&gt; and &lt;math&gt;ab^5 + 3&lt;/math&gt; are cubes, then &lt;math&gt;ab^5, a^5b \equiv 5,7 \pmod 9&lt;/math&gt;<br /> <br /> '''Proof''' Since cubes are congruent to any of &lt;math&gt;0, 1, -1 \pmod 9&lt;/math&gt;, &lt;math&gt;ab^5,a^5b \equiv 5,6,7 \pmod 9&lt;/math&gt;. But if &lt;math&gt;ab^5 \equiv 6 \pmod 9&lt;/math&gt;, &lt;math&gt;3|a&lt;/math&gt;, so &lt;math&gt;a^5b \equiv 0 \pmod 9&lt;/math&gt;, contradiction. A similar argument can be made for &lt;math&gt;ab^5 \neq 6 \pmod 9&lt;/math&gt;. <br /> <br /> <br /> '''Lemma 2''': If k is a perfect 6th power, then &lt;math&gt;k \equiv 0,1 \pmod 9&lt;/math&gt; <br /> <br /> '''Proof''': Since cubes are congruent to &lt;math&gt;0, 1, -1 \pmod 9&lt;/math&gt;, we can square, and get 6th powers are congruent to &lt;math&gt;0, 1 \pmod 9&lt;/math&gt;.<br /> <br /> <br /> Since &lt;math&gt;ab^5 \cdot a^5b = a^6 b^6 = (ab)^6&lt;/math&gt;, which is a perfect 6th power, by lemma 2, &lt;math&gt;ab^5 \cdot a^5b \equiv 0,1 \pmod 9&lt;/math&gt;.<br /> <br /> But, by lemma 1, &lt;math&gt;ab^5 \cdot a^5b \equiv 5 \cdot 5, 5 \cdot 7, 7 \cdot 7 \equiv 7, 8, 4 \pmod 9&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;ab^5 \cdot a^5b&lt;/math&gt;, which is an integer, can't go into any of the possible residue classes modulo 9, without breaking one of these lemmas. This is a contradiction, and the proof is complete. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -AlexLikeMath<br /> <br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2013_USAJMO_Problems/Problem_1&diff=125502 2013 USAJMO Problems/Problem 1 2020-06-15T18:18:35Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> <br /> Are there integers &lt;math&gt; a &lt;/math&gt; and &lt;math&gt; b &lt;/math&gt; such that &lt;math&gt; a^5b+3 &lt;/math&gt; and &lt;math&gt; ab^5+3 &lt;/math&gt; are both perfect cubes of integers?<br /> <br /> ==Solution==<br /> <br /> No, such integers do not exist. This shall be proven by contradiction, by showing that if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube then &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> Remark that perfect cubes are always congruent to &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, or &lt;math&gt;-1&lt;/math&gt; modulo &lt;math&gt;9&lt;/math&gt;. Therefore, if &lt;math&gt;a^5b+3\equiv 0,1,\text{ or} -1\pmod{9}&lt;/math&gt;, then &lt;math&gt;a^5b\equiv 5,6,\text{ or }7\pmod{9}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a^5b\equiv 6\pmod 9&lt;/math&gt;, then note that &lt;math&gt;3|b&lt;/math&gt;. (This is because if &lt;math&gt;3|a&lt;/math&gt; then &lt;math&gt;a^5b\equiv 0\pmod 9&lt;/math&gt;.) Therefore &lt;math&gt;ab^5\equiv 0\pmod 9&lt;/math&gt; and &lt;math&gt;ab^5+3\equiv 3\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Otherwise, either &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt; or &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;. Note that since &lt;math&gt;a^6b^6&lt;/math&gt; is a perfect sixth power, and since neither &lt;math&gt;a&lt;/math&gt; nor &lt;math&gt;b&lt;/math&gt; contains a factor of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;a^6b^6\equiv 1\pmod 9&lt;/math&gt;. If &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 5ab^5\equiv 1\pmod 9\implies ab^5\equiv 2\pmod 9.&lt;/cmath&gt; Similarly, if &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 7ab^5\equiv 1\pmod 9\implies ab^5\equiv 4\pmod 9.&lt;/cmath&gt; Therefore &lt;math&gt;ab^5+3\equiv 5,7\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 2==<br /> We shall prove that such integers do not exist via contradiction.<br /> Suppose that &lt;math&gt;a^5b + 3 = x^3&lt;/math&gt; and &lt;math&gt;ab^5 + 3 = y^3&lt;/math&gt; for integers x and y. Rearranging terms gives &lt;math&gt;a^5b = x^3 - 3&lt;/math&gt; and &lt;math&gt;ab^5 = y^3 - 3&lt;/math&gt;. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = &lt;math&gt;(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}&lt;/math&gt; and b = &lt;math&gt;(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}&lt;/math&gt;. Consider a prime p in the prime factorization of &lt;math&gt;x^3 - 3&lt;/math&gt; and &lt;math&gt;y^3 - 3&lt;/math&gt;. If it has power &lt;math&gt;r_1&lt;/math&gt; in &lt;math&gt;x^3 - 3&lt;/math&gt; and power &lt;math&gt;r_2&lt;/math&gt; in &lt;math&gt;y^3 - 3&lt;/math&gt;, then &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; is a multiple of 24 and &lt;math&gt;5r_2&lt;/math&gt; - &lt;math&gt;r_1&lt;/math&gt; also is a multiple of 24. <br /> <br /> Adding and subtracting the divisions gives that &lt;math&gt;r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; divides 12. (actually, &lt;math&gt;r_1 - r_2&lt;/math&gt; is a multiple of 4, as you can verify if &lt;math&gt;\{^{5r_1 - r_2 = 24}_{5r_2 - r_1 = 48}&lt;/math&gt;. So the rest of the proof is invalid.) Because &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; also divides 12, &lt;math&gt;4r_1&lt;/math&gt; divides 12 and thus &lt;math&gt;r_1&lt;/math&gt; divides 3. Repeating this trick for all primes in &lt;math&gt;x^3 - 3&lt;/math&gt;, we see that &lt;math&gt;x^3 - 3&lt;/math&gt; is a perfect cube, say &lt;math&gt;q^3&lt;/math&gt;. Then &lt;math&gt;x^3 - q^3 = 3,&lt;/math&gt; and &lt;math&gt;(x-q)(x^2 + xq + q^2) = 3&lt;/math&gt;, so that &lt;math&gt;x - q = 1&lt;/math&gt; and &lt;math&gt;x^2 + xq + q^2 = 3&lt;/math&gt;. Clearly, this system of equations has no integer solutions for &lt;math&gt;x&lt;/math&gt; or &lt;math&gt;q&lt;/math&gt;, a contradiction, hence completing the proof.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;a^5b+3=x^3&lt;/math&gt; and &lt;math&gt;ab^5+3=y^3&lt;/math&gt;. Then, &lt;math&gt;a^5b=x^3-3&lt;/math&gt;, &lt;math&gt;ab^5=y^3-3&lt;/math&gt;, and &lt;cmath&gt;(ab)^6=(x^3-3)(y^3-3)&lt;/cmath&gt;<br /> Now take &lt;math&gt;\text{mod }9&lt;/math&gt; (recall that perfect cubes &lt;math&gt;\equiv -1,0,1\pmod{9}&lt;/math&gt; and perfect sixth powers &lt;math&gt;\equiv 0,1\pmod{9}&lt;/math&gt;) on both sides. There are &lt;math&gt;3\times 3=9&lt;/math&gt; cases to consider on what values &lt;math&gt;\text{mod }9&lt;/math&gt; that &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;y^3&lt;/math&gt; take. Checking these &lt;math&gt;9&lt;/math&gt; cases, we see that only &lt;math&gt;x^3\equiv y^3\equiv 0\pmod{9}&lt;/math&gt; or &lt;math&gt;x\equiv y\equiv 0\pmod{3}&lt;/math&gt; yield a valid residue &lt;math&gt;\text{mod }9&lt;/math&gt; (specifically, &lt;math&gt;(x^3-3)(y^3-3)\equiv 0\pmod{9}&lt;/math&gt;). But this means that &lt;math&gt;3\mid ab&lt;/math&gt;, so &lt;math&gt;729\mid (ab)^6&lt;/math&gt; so &lt;cmath&gt;729\mid (x^3-3)(y^3-3)\iff 729\mid (27x'^3-3)(27y'^3-3)\iff 81\mid (9x'^3-1)(9y'^3-1)&lt;/cmath&gt; contradiction.<br /> <br /> ==Solution 4==<br /> If &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube, then &lt;math&gt;a^5b&lt;/math&gt; can be one of &lt;math&gt;5,6,7 \pmod 9&lt;/math&gt;, so &lt;math&gt;(a^5b)^5 = a^{25} b^5&lt;/math&gt; can be one of &lt;math&gt;5^5 \equiv 2&lt;/math&gt;, &lt;math&gt;6^5 \equiv 0&lt;/math&gt;, or &lt;math&gt;7^5 \equiv 4 \pmod 9&lt;/math&gt;. If &lt;math&gt;a&lt;/math&gt; were divisible by &lt;math&gt;3&lt;/math&gt;, we'd have &lt;math&gt;a^5 b \equiv 0 \pmod 9&lt;/math&gt;, which we've ruled out. So &lt;math&gt;\gcd(a,9) = 1&lt;/math&gt;, which means &lt;math&gt;a^6 \equiv 1 \pmod 9&lt;/math&gt;, and therefore &lt;math&gt;a^{25} b^5 \equiv ab^5 \pmod 9&lt;/math&gt;.<br /> <br /> We've shown that &lt;math&gt;a b^5&lt;/math&gt; can be one of &lt;math&gt;0, 2, 4 \pmod 9&lt;/math&gt;, so &lt;math&gt;ab^5 + 3&lt;/math&gt; can be one of &lt;math&gt;3, 5, 7 \pmod 9&lt;/math&gt;. None of these are possibilities for a perfect cube, so if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube, &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> ==Solution 5==<br /> <br /> As in previous solutions, notice &lt;math&gt;ab^5,a^5b \equiv 5,6,7 \pmod 9&lt;/math&gt;. Now multiplying gives &lt;math&gt;a^6b^6&lt;/math&gt;, which is only &lt;math&gt;0,1 \pmod 9&lt;/math&gt;, so after testing all cases we find that &lt;math&gt;ab^5\equiv a^5b \equiv 6 \mod 9&lt;/math&gt;. Then since &lt;math&gt;\phi (9) = 6&lt;/math&gt;, &lt;math&gt;ab^5\equiv \frac{a}{b}\pmod 9&lt;/math&gt; and &lt;math&gt;a^5b \equiv \frac{b}{a}\pmod 9&lt;/math&gt; (Note that &lt;math&gt;a,b&lt;/math&gt; cannot be &lt;math&gt;0\pmod 9&lt;/math&gt;). Thus we find that the inverse of &lt;math&gt;6&lt;/math&gt; is itself under modulo &lt;math&gt;9&lt;/math&gt;, a contradiction.<br /> <br /> ==Solution 6==<br /> I claim there are no such a or b such that both expressions are cubes.<br /> <br /> Assume to the contrary &lt;math&gt;a^5b +3&lt;/math&gt; and &lt;math&gt;ab^5 + 3&lt;/math&gt; are cubes. <br /> <br /> '''Lemma 1''': If &lt;math&gt;a^5b +3&lt;/math&gt; and &lt;math&gt;ab^5 + 3&lt;/math&gt; are cubes &lt;math&gt;ab^5,a^5b \equiv 5,7 \pmod 9&lt;/math&gt;<br /> <br /> '''Proof''' Since cubes are congruent to any of &lt;math&gt;0, 1, -1 \pmod 9&lt;/math&gt;, &lt;math&gt;ab^5,a^5b \equiv 5,6,7 \pmod 9&lt;/math&gt;. But if &lt;math&gt;ab^5 \equiv 6 \pmod 9&lt;/math&gt;, &lt;math&gt;3|a&lt;/math&gt;, so &lt;math&gt;a^5b \equiv 0 \pmod 9&lt;/math&gt;, contradiction. A similar argument can be made for &lt;math&gt;ab^5 \neq 6 \pmod 9&lt;/math&gt;. <br /> <br /> <br /> '''Lemma 2''': If k is a perfect 6th power, &lt;math&gt;k \equiv 0,1 \pmod 9&lt;/math&gt; <br /> <br /> '''Proof''': since cubes are congruent to &lt;math&gt;0, 1, -1 \pmod 9&lt;/math&gt;, 6th powers are congruent to &lt;math&gt;0, 1 \pmod 9&lt;/math&gt;.<br /> <br /> <br /> Since &lt;math&gt;ab^5 \cdot a^5b = a^6 b^6 = (ab)^6&lt;/math&gt;, which is a perfect 6th power, by lemma 2, &lt;math&gt;ab^5 \cdot a^5b \equiv 0,1 \pmod 9&lt;/math&gt;.<br /> <br /> But, by lemma 1, &lt;math&gt;ab^5 \cdot a^5b \equiv 5 \cdot 5, 5 \cdot 7, 7 \cdot 7 \equiv 7, 8, 4 \pmod 9&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;ab^5 \cdot a^5b&lt;/math&gt;, which is an integer, can't go into any of the possible residue classes modulo 9, without breaking one of these lemmas. This is a contradiction, and the proof is complete. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -AlexLikeMath<br /> <br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2013_USAJMO_Problems/Problem_1&diff=125497 2013 USAJMO Problems/Problem 1 2020-06-15T18:14:38Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> <br /> Are there integers &lt;math&gt; a &lt;/math&gt; and &lt;math&gt; b &lt;/math&gt; such that &lt;math&gt; a^5b+3 &lt;/math&gt; and &lt;math&gt; ab^5+3 &lt;/math&gt; are both perfect cubes of integers?<br /> <br /> ==Solution==<br /> <br /> No, such integers do not exist. This shall be proven by contradiction, by showing that if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube then &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> Remark that perfect cubes are always congruent to &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, or &lt;math&gt;-1&lt;/math&gt; modulo &lt;math&gt;9&lt;/math&gt;. Therefore, if &lt;math&gt;a^5b+3\equiv 0,1,\text{ or} -1\pmod{9}&lt;/math&gt;, then &lt;math&gt;a^5b\equiv 5,6,\text{ or }7\pmod{9}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a^5b\equiv 6\pmod 9&lt;/math&gt;, then note that &lt;math&gt;3|b&lt;/math&gt;. (This is because if &lt;math&gt;3|a&lt;/math&gt; then &lt;math&gt;a^5b\equiv 0\pmod 9&lt;/math&gt;.) Therefore &lt;math&gt;ab^5\equiv 0\pmod 9&lt;/math&gt; and &lt;math&gt;ab^5+3\equiv 3\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Otherwise, either &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt; or &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;. Note that since &lt;math&gt;a^6b^6&lt;/math&gt; is a perfect sixth power, and since neither &lt;math&gt;a&lt;/math&gt; nor &lt;math&gt;b&lt;/math&gt; contains a factor of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;a^6b^6\equiv 1\pmod 9&lt;/math&gt;. If &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 5ab^5\equiv 1\pmod 9\implies ab^5\equiv 2\pmod 9.&lt;/cmath&gt; Similarly, if &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 7ab^5\equiv 1\pmod 9\implies ab^5\equiv 4\pmod 9.&lt;/cmath&gt; Therefore &lt;math&gt;ab^5+3\equiv 5,7\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 2==<br /> We shall prove that such integers do not exist via contradiction.<br /> Suppose that &lt;math&gt;a^5b + 3 = x^3&lt;/math&gt; and &lt;math&gt;ab^5 + 3 = y^3&lt;/math&gt; for integers x and y. Rearranging terms gives &lt;math&gt;a^5b = x^3 - 3&lt;/math&gt; and &lt;math&gt;ab^5 = y^3 - 3&lt;/math&gt;. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = &lt;math&gt;(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}&lt;/math&gt; and b = &lt;math&gt;(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}&lt;/math&gt;. Consider a prime p in the prime factorization of &lt;math&gt;x^3 - 3&lt;/math&gt; and &lt;math&gt;y^3 - 3&lt;/math&gt;. If it has power &lt;math&gt;r_1&lt;/math&gt; in &lt;math&gt;x^3 - 3&lt;/math&gt; and power &lt;math&gt;r_2&lt;/math&gt; in &lt;math&gt;y^3 - 3&lt;/math&gt;, then &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; is a multiple of 24 and &lt;math&gt;5r_2&lt;/math&gt; - &lt;math&gt;r_1&lt;/math&gt; also is a multiple of 24. <br /> <br /> Adding and subtracting the divisions gives that &lt;math&gt;r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; divides 12. (actually, &lt;math&gt;r_1 - r_2&lt;/math&gt; is a multiple of 4, as you can verify if &lt;math&gt;\{^{5r_1 - r_2 = 24}_{5r_2 - r_1 = 48}&lt;/math&gt;. So the rest of the proof is invalid.) Because &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; also divides 12, &lt;math&gt;4r_1&lt;/math&gt; divides 12 and thus &lt;math&gt;r_1&lt;/math&gt; divides 3. Repeating this trick for all primes in &lt;math&gt;x^3 - 3&lt;/math&gt;, we see that &lt;math&gt;x^3 - 3&lt;/math&gt; is a perfect cube, say &lt;math&gt;q^3&lt;/math&gt;. Then &lt;math&gt;x^3 - q^3 = 3,&lt;/math&gt; and &lt;math&gt;(x-q)(x^2 + xq + q^2) = 3&lt;/math&gt;, so that &lt;math&gt;x - q = 1&lt;/math&gt; and &lt;math&gt;x^2 + xq + q^2 = 3&lt;/math&gt;. Clearly, this system of equations has no integer solutions for &lt;math&gt;x&lt;/math&gt; or &lt;math&gt;q&lt;/math&gt;, a contradiction, hence completing the proof.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;a^5b+3=x^3&lt;/math&gt; and &lt;math&gt;ab^5+3=y^3&lt;/math&gt;. Then, &lt;math&gt;a^5b=x^3-3&lt;/math&gt;, &lt;math&gt;ab^5=y^3-3&lt;/math&gt;, and &lt;cmath&gt;(ab)^6=(x^3-3)(y^3-3)&lt;/cmath&gt;<br /> Now take &lt;math&gt;\text{mod }9&lt;/math&gt; (recall that perfect cubes &lt;math&gt;\equiv -1,0,1\pmod{9}&lt;/math&gt; and perfect sixth powers &lt;math&gt;\equiv 0,1\pmod{9}&lt;/math&gt;) on both sides. There are &lt;math&gt;3\times 3=9&lt;/math&gt; cases to consider on what values &lt;math&gt;\text{mod }9&lt;/math&gt; that &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;y^3&lt;/math&gt; take. Checking these &lt;math&gt;9&lt;/math&gt; cases, we see that only &lt;math&gt;x^3\equiv y^3\equiv 0\pmod{9}&lt;/math&gt; or &lt;math&gt;x\equiv y\equiv 0\pmod{3}&lt;/math&gt; yield a valid residue &lt;math&gt;\text{mod }9&lt;/math&gt; (specifically, &lt;math&gt;(x^3-3)(y^3-3)\equiv 0\pmod{9}&lt;/math&gt;). But this means that &lt;math&gt;3\mid ab&lt;/math&gt;, so &lt;math&gt;729\mid (ab)^6&lt;/math&gt; so &lt;cmath&gt;729\mid (x^3-3)(y^3-3)\iff 729\mid (27x'^3-3)(27y'^3-3)\iff 81\mid (9x'^3-1)(9y'^3-1)&lt;/cmath&gt; contradiction.<br /> <br /> ==Solution 4==<br /> If &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube, then &lt;math&gt;a^5b&lt;/math&gt; can be one of &lt;math&gt;5,6,7 \pmod 9&lt;/math&gt;, so &lt;math&gt;(a^5b)^5 = a^{25} b^5&lt;/math&gt; can be one of &lt;math&gt;5^5 \equiv 2&lt;/math&gt;, &lt;math&gt;6^5 \equiv 0&lt;/math&gt;, or &lt;math&gt;7^5 \equiv 4 \pmod 9&lt;/math&gt;. If &lt;math&gt;a&lt;/math&gt; were divisible by &lt;math&gt;3&lt;/math&gt;, we'd have &lt;math&gt;a^5 b \equiv 0 \pmod 9&lt;/math&gt;, which we've ruled out. So &lt;math&gt;\gcd(a,9) = 1&lt;/math&gt;, which means &lt;math&gt;a^6 \equiv 1 \pmod 9&lt;/math&gt;, and therefore &lt;math&gt;a^{25} b^5 \equiv ab^5 \pmod 9&lt;/math&gt;.<br /> <br /> We've shown that &lt;math&gt;a b^5&lt;/math&gt; can be one of &lt;math&gt;0, 2, 4 \pmod 9&lt;/math&gt;, so &lt;math&gt;ab^5 + 3&lt;/math&gt; can be one of &lt;math&gt;3, 5, 7 \pmod 9&lt;/math&gt;. None of these are possibilities for a perfect cube, so if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube, &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> ==Solution 5==<br /> <br /> As in previous solutions, notice &lt;math&gt;ab^5,a^5b \equiv 5,6,7 \pmod 9&lt;/math&gt;. Now multiplying gives &lt;math&gt;a^6b^6&lt;/math&gt;, which is only &lt;math&gt;0,1 \pmod 9&lt;/math&gt;, so after testing all cases we find that &lt;math&gt;ab^5\equiv a^5b \equiv 6 \mod 9&lt;/math&gt;. Then since &lt;math&gt;\phi (9) = 6&lt;/math&gt;, &lt;math&gt;ab^5\equiv \frac{a}{b}\pmod 9&lt;/math&gt; and &lt;math&gt;a^5b \equiv \frac{b}{a}\pmod 9&lt;/math&gt; (Note that &lt;math&gt;a,b&lt;/math&gt; cannot be &lt;math&gt;0\pmod 9&lt;/math&gt;). Thus we find that the inverse of &lt;math&gt;6&lt;/math&gt; is itself under modulo &lt;math&gt;9&lt;/math&gt;, a contradiction.<br /> <br /> ==Solution 6==<br /> I claim there are no such a or b such that both expressions are cubes.<br /> <br /> Assume to the contrary &lt;math&gt;a^5b +3&lt;/math&gt; and &lt;math&gt;ab^5 + 3&lt;/math&gt; are cubes. <br /> <br /> '''Lemma 1''': If &lt;math&gt;a^5b +3&lt;/math&gt; and &lt;math&gt;ab^5 + 3&lt;/math&gt; are cubes &lt;math&gt;ab^5,a^5b \equiv 5,7 \pmod 9&lt;/math&gt;<br /> <br /> '''Proof''' Since cubes are congruent to any of &lt;math&gt;0, 1, -1 \pmod 9&lt;/math&gt;, &lt;math&gt;ab^5,a^5b \equiv 5,6,7 \pmod 9&lt;/math&gt;. But if &lt;math&gt;ab^5 \equiv 6 \pmod 9&lt;/math&gt;, &lt;math&gt;3|a&lt;/math&gt;, so &lt;math&gt;a^5b \equiv 0 \pmod 9&lt;/math&gt;, contradiction. A similar argument can be made for &lt;math&gt;ab^5 \neq 6 \pmod 9&lt;/math&gt;. <br /> <br /> <br /> '''Lemma 2''': If k is a perfect 6th power, &lt;math&gt;k \equiv 0,1 \pmod 9&lt;/math&gt; <br /> <br /> '''Proof''': since cubes are congruent to &lt;math&gt;0, 1, -1 \pmod 9&lt;/math&gt;, 6th powers are congruent to &lt;math&gt;0, 1 \pmod 9&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;ab^5 \cdot a^5b = a^6 b^6 = (ab)^6&lt;/math&gt;, which is a perfect 6th power, by lemma 2, &lt;math&gt;ab^5 \cdot a^5b \equiv 0,1 \pmod 9&lt;/math&gt;.<br /> <br /> But, by lemma 1, &lt;math&gt;ab^5 \cdot a^5b \equiv 5 \cdot 5, 5 \cdot 7, 7 \cdot 7 \equiv 7, 8, 4 \pmod 9&lt;/math&gt;. <br /> <br /> So, &lt;math&gt;ab^5 \cdot a^5b&lt;/math&gt;, which is an integer, can't go into any of the possible residue classes modulo 9, without breaking one of these lemmas. This is a contradiction, and the proof is complete. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -AlexLikeMath<br /> <br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2013_USAJMO_Problems/Problem_1&diff=125496 2013 USAJMO Problems/Problem 1 2020-06-15T17:53:17Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> <br /> Are there integers &lt;math&gt; a &lt;/math&gt; and &lt;math&gt; b &lt;/math&gt; such that &lt;math&gt; a^5b+3 &lt;/math&gt; and &lt;math&gt; ab^5+3 &lt;/math&gt; are both perfect cubes of integers?<br /> <br /> ==Solution==<br /> <br /> No, such integers do not exist. This shall be proven by contradiction, by showing that if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube then &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> Remark that perfect cubes are always congruent to &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, or &lt;math&gt;-1&lt;/math&gt; modulo &lt;math&gt;9&lt;/math&gt;. Therefore, if &lt;math&gt;a^5b+3\equiv 0,1,\text{ or} -1\pmod{9}&lt;/math&gt;, then &lt;math&gt;a^5b\equiv 5,6,\text{ or }7\pmod{9}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a^5b\equiv 6\pmod 9&lt;/math&gt;, then note that &lt;math&gt;3|b&lt;/math&gt;. (This is because if &lt;math&gt;3|a&lt;/math&gt; then &lt;math&gt;a^5b\equiv 0\pmod 9&lt;/math&gt;.) Therefore &lt;math&gt;ab^5\equiv 0\pmod 9&lt;/math&gt; and &lt;math&gt;ab^5+3\equiv 3\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Otherwise, either &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt; or &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;. Note that since &lt;math&gt;a^6b^6&lt;/math&gt; is a perfect sixth power, and since neither &lt;math&gt;a&lt;/math&gt; nor &lt;math&gt;b&lt;/math&gt; contains a factor of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;a^6b^6\equiv 1\pmod 9&lt;/math&gt;. If &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 5ab^5\equiv 1\pmod 9\implies ab^5\equiv 2\pmod 9.&lt;/cmath&gt; Similarly, if &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 7ab^5\equiv 1\pmod 9\implies ab^5\equiv 4\pmod 9.&lt;/cmath&gt; Therefore &lt;math&gt;ab^5+3\equiv 5,7\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 2==<br /> We shall prove that such integers do not exist via contradiction.<br /> Suppose that &lt;math&gt;a^5b + 3 = x^3&lt;/math&gt; and &lt;math&gt;ab^5 + 3 = y^3&lt;/math&gt; for integers x and y. Rearranging terms gives &lt;math&gt;a^5b = x^3 - 3&lt;/math&gt; and &lt;math&gt;ab^5 = y^3 - 3&lt;/math&gt;. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = &lt;math&gt;(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}&lt;/math&gt; and b = &lt;math&gt;(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}&lt;/math&gt;. Consider a prime p in the prime factorization of &lt;math&gt;x^3 - 3&lt;/math&gt; and &lt;math&gt;y^3 - 3&lt;/math&gt;. If it has power &lt;math&gt;r_1&lt;/math&gt; in &lt;math&gt;x^3 - 3&lt;/math&gt; and power &lt;math&gt;r_2&lt;/math&gt; in &lt;math&gt;y^3 - 3&lt;/math&gt;, then &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; is a multiple of 24 and &lt;math&gt;5r_2&lt;/math&gt; - &lt;math&gt;r_1&lt;/math&gt; also is a multiple of 24. <br /> <br /> Adding and subtracting the divisions gives that &lt;math&gt;r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; divides 12. (actually, &lt;math&gt;r_1 - r_2&lt;/math&gt; is a multiple of 4, as you can verify if &lt;math&gt;\{^{5r_1 - r_2 = 24}_{5r_2 - r_1 = 48}&lt;/math&gt;. So the rest of the proof is invalid.) Because &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; also divides 12, &lt;math&gt;4r_1&lt;/math&gt; divides 12 and thus &lt;math&gt;r_1&lt;/math&gt; divides 3. Repeating this trick for all primes in &lt;math&gt;x^3 - 3&lt;/math&gt;, we see that &lt;math&gt;x^3 - 3&lt;/math&gt; is a perfect cube, say &lt;math&gt;q^3&lt;/math&gt;. Then &lt;math&gt;x^3 - q^3 = 3,&lt;/math&gt; and &lt;math&gt;(x-q)(x^2 + xq + q^2) = 3&lt;/math&gt;, so that &lt;math&gt;x - q = 1&lt;/math&gt; and &lt;math&gt;x^2 + xq + q^2 = 3&lt;/math&gt;. Clearly, this system of equations has no integer solutions for &lt;math&gt;x&lt;/math&gt; or &lt;math&gt;q&lt;/math&gt;, a contradiction, hence completing the proof.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;a^5b+3=x^3&lt;/math&gt; and &lt;math&gt;ab^5+3=y^3&lt;/math&gt;. Then, &lt;math&gt;a^5b=x^3-3&lt;/math&gt;, &lt;math&gt;ab^5=y^3-3&lt;/math&gt;, and &lt;cmath&gt;(ab)^6=(x^3-3)(y^3-3)&lt;/cmath&gt;<br /> Now take &lt;math&gt;\text{mod }9&lt;/math&gt; (recall that perfect cubes &lt;math&gt;\equiv -1,0,1\pmod{9}&lt;/math&gt; and perfect sixth powers &lt;math&gt;\equiv 0,1\pmod{9}&lt;/math&gt;) on both sides. There are &lt;math&gt;3\times 3=9&lt;/math&gt; cases to consider on what values &lt;math&gt;\text{mod }9&lt;/math&gt; that &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;y^3&lt;/math&gt; take. Checking these &lt;math&gt;9&lt;/math&gt; cases, we see that only &lt;math&gt;x^3\equiv y^3\equiv 0\pmod{9}&lt;/math&gt; or &lt;math&gt;x\equiv y\equiv 0\pmod{3}&lt;/math&gt; yield a valid residue &lt;math&gt;\text{mod }9&lt;/math&gt; (specifically, &lt;math&gt;(x^3-3)(y^3-3)\equiv 0\pmod{9}&lt;/math&gt;). But this means that &lt;math&gt;3\mid ab&lt;/math&gt;, so &lt;math&gt;729\mid (ab)^6&lt;/math&gt; so &lt;cmath&gt;729\mid (x^3-3)(y^3-3)\iff 729\mid (27x'^3-3)(27y'^3-3)\iff 81\mid (9x'^3-1)(9y'^3-1)&lt;/cmath&gt; contradiction.<br /> <br /> ==Solution 4==<br /> If &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube, then &lt;math&gt;a^5b&lt;/math&gt; can be one of &lt;math&gt;5,6,7 \pmod 9&lt;/math&gt;, so &lt;math&gt;(a^5b)^5 = a^{25} b^5&lt;/math&gt; can be one of &lt;math&gt;5^5 \equiv 2&lt;/math&gt;, &lt;math&gt;6^5 \equiv 0&lt;/math&gt;, or &lt;math&gt;7^5 \equiv 4 \pmod 9&lt;/math&gt;. If &lt;math&gt;a&lt;/math&gt; were divisible by &lt;math&gt;3&lt;/math&gt;, we'd have &lt;math&gt;a^5 b \equiv 0 \pmod 9&lt;/math&gt;, which we've ruled out. So &lt;math&gt;\gcd(a,9) = 1&lt;/math&gt;, which means &lt;math&gt;a^6 \equiv 1 \pmod 9&lt;/math&gt;, and therefore &lt;math&gt;a^{25} b^5 \equiv ab^5 \pmod 9&lt;/math&gt;.<br /> <br /> We've shown that &lt;math&gt;a b^5&lt;/math&gt; can be one of &lt;math&gt;0, 2, 4 \pmod 9&lt;/math&gt;, so &lt;math&gt;ab^5 + 3&lt;/math&gt; can be one of &lt;math&gt;3, 5, 7 \pmod 9&lt;/math&gt;. None of these are possibilities for a perfect cube, so if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube, &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> ==Solution 5==<br /> <br /> As in previous solutions, notice &lt;math&gt;ab^5,a^5b \equiv 5,6,7 \pmod 9&lt;/math&gt;. Now multiplying gives &lt;math&gt;a^6b^6&lt;/math&gt;, which is only &lt;math&gt;0,1 \pmod 9&lt;/math&gt;, so after testing all cases we find that &lt;math&gt;ab^5\equiv a^5b \equiv 6 \mod 9&lt;/math&gt;. Then since &lt;math&gt;\phi (9) = 6&lt;/math&gt;, &lt;math&gt;ab^5\equiv \frac{a}{b}\pmod 9&lt;/math&gt; and &lt;math&gt;a^5b \equiv \frac{b}{a}\pmod 9&lt;/math&gt; (Note that &lt;math&gt;a,b&lt;/math&gt; cannot be &lt;math&gt;0\pmod 9&lt;/math&gt;). Thus we find that the inverse of &lt;math&gt;6&lt;/math&gt; is itself under modulo &lt;math&gt;9&lt;/math&gt;, a contradiction.<br /> <br /> ==Solution 6==<br /> I claim there are no such a or b such that both expressions are cubes.<br /> <br /> Assume to the contrary &lt;math&gt;a^5b +3&lt;/math&gt; and &lt;math&gt;ab^5 + 3&lt;/math&gt; are cubes. <br /> <br /> Since cubes are congruent to any of &lt;math&gt;0, 1, -1 \pmod 9&lt;/math&gt;, &lt;math&gt;ab^5,a^5b \equiv 5,6,7 \pmod 9&lt;/math&gt;. But if &lt;math&gt;ab^5 \equiv 6 \pmod 9&lt;/math&gt;, &lt;math&gt;3|a&lt;/math&gt;, so &lt;math&gt;a^5b \equiv 0 \pmod 9&lt;/math&gt;, contradiction. <br /> <br /> A similar argument can be made for &lt;math&gt;ab^5 \neq 6 \pmod 9&lt;/math&gt;. Thus, we get that:<br /> <br /> &lt;cmath&gt;ab^5,a^5b \equiv 5,7 \pmod 9&lt;/cmath&gt;<br /> <br /> A critical idea is that since cubes are congruent to &lt;math&gt;0, 1, -1 \pmod 9&lt;/math&gt;, 6th powers are congruent to &lt;math&gt;0, 1 \pmod 9&lt;/math&gt;. And, &lt;math&gt;ab^5 \cdot a^5b = a^6 b^6 = (ab)^6&lt;/math&gt;, making the the product of &lt;math&gt;ab^5, a^5b&lt;/math&gt; a perfect 6th power. <br /> <br /> But, we can substitiute our possible values of &lt;math&gt;ab^5, a^5b&lt;/math&gt;. &lt;math&gt;5 \cdot 5 \equiv 7 \pmod 9&lt;/math&gt;, &lt;math&gt;5 \cdot 7 \equiv 8 \pmod 9&lt;/math&gt;, and &lt;math&gt;7 \cdot 7 \equiv 4 \pmod 9&lt;/math&gt;. None of these are congruent to &lt;math&gt;0, 1 \pmod 9&lt;/math&gt;, but perfect 6th powers have to be congruent to &lt;math&gt;0, 1 \pmod 9&lt;/math&gt;. We have a contradiction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -AlexLikeMath<br /> <br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2013_USAJMO_Problems/Problem_1&diff=125495 2013 USAJMO Problems/Problem 1 2020-06-15T17:46:06Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> <br /> Are there integers &lt;math&gt; a &lt;/math&gt; and &lt;math&gt; b &lt;/math&gt; such that &lt;math&gt; a^5b+3 &lt;/math&gt; and &lt;math&gt; ab^5+3 &lt;/math&gt; are both perfect cubes of integers?<br /> <br /> ==Solution==<br /> <br /> No, such integers do not exist. This shall be proven by contradiction, by showing that if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube then &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> Remark that perfect cubes are always congruent to &lt;math&gt;0&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, or &lt;math&gt;-1&lt;/math&gt; modulo &lt;math&gt;9&lt;/math&gt;. Therefore, if &lt;math&gt;a^5b+3\equiv 0,1,\text{ or} -1\pmod{9}&lt;/math&gt;, then &lt;math&gt;a^5b\equiv 5,6,\text{ or }7\pmod{9}&lt;/math&gt;.<br /> <br /> If &lt;math&gt;a^5b\equiv 6\pmod 9&lt;/math&gt;, then note that &lt;math&gt;3|b&lt;/math&gt;. (This is because if &lt;math&gt;3|a&lt;/math&gt; then &lt;math&gt;a^5b\equiv 0\pmod 9&lt;/math&gt;.) Therefore &lt;math&gt;ab^5\equiv 0\pmod 9&lt;/math&gt; and &lt;math&gt;ab^5+3\equiv 3\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Otherwise, either &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt; or &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;. Note that since &lt;math&gt;a^6b^6&lt;/math&gt; is a perfect sixth power, and since neither &lt;math&gt;a&lt;/math&gt; nor &lt;math&gt;b&lt;/math&gt; contains a factor of &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;a^6b^6\equiv 1\pmod 9&lt;/math&gt;. If &lt;math&gt;a^5b\equiv 5\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 5ab^5\equiv 1\pmod 9\implies ab^5\equiv 2\pmod 9.&lt;/cmath&gt; Similarly, if &lt;math&gt;a^5b\equiv 7\pmod 9&lt;/math&gt;, then &lt;cmath&gt;a^6b^6\equiv (a^5b)(ab^5)\equiv 7ab^5\equiv 1\pmod 9\implies ab^5\equiv 4\pmod 9.&lt;/cmath&gt; Therefore &lt;math&gt;ab^5+3\equiv 5,7\pmod 9&lt;/math&gt;, contradiction.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 2==<br /> We shall prove that such integers do not exist via contradiction.<br /> Suppose that &lt;math&gt;a^5b + 3 = x^3&lt;/math&gt; and &lt;math&gt;ab^5 + 3 = y^3&lt;/math&gt; for integers x and y. Rearranging terms gives &lt;math&gt;a^5b = x^3 - 3&lt;/math&gt; and &lt;math&gt;ab^5 = y^3 - 3&lt;/math&gt;. Solving for a and b (by first multiplying the equations together and taking the sixth root) gives a = &lt;math&gt;(x^3 - 3)^\frac{5}{24} (y^3 - 3)^\frac{-1}{24}&lt;/math&gt; and b = &lt;math&gt;(y^3 - 3)^\frac{5}{24} (x^3 - 3)^\frac{-1}{24}&lt;/math&gt;. Consider a prime p in the prime factorization of &lt;math&gt;x^3 - 3&lt;/math&gt; and &lt;math&gt;y^3 - 3&lt;/math&gt;. If it has power &lt;math&gt;r_1&lt;/math&gt; in &lt;math&gt;x^3 - 3&lt;/math&gt; and power &lt;math&gt;r_2&lt;/math&gt; in &lt;math&gt;y^3 - 3&lt;/math&gt;, then &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; is a multiple of 24 and &lt;math&gt;5r_2&lt;/math&gt; - &lt;math&gt;r_1&lt;/math&gt; also is a multiple of 24. <br /> <br /> Adding and subtracting the divisions gives that &lt;math&gt;r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; divides 12. (actually, &lt;math&gt;r_1 - r_2&lt;/math&gt; is a multiple of 4, as you can verify if &lt;math&gt;\{^{5r_1 - r_2 = 24}_{5r_2 - r_1 = 48}&lt;/math&gt;. So the rest of the proof is invalid.) Because &lt;math&gt;5r_1&lt;/math&gt; - &lt;math&gt;r_2&lt;/math&gt; also divides 12, &lt;math&gt;4r_1&lt;/math&gt; divides 12 and thus &lt;math&gt;r_1&lt;/math&gt; divides 3. Repeating this trick for all primes in &lt;math&gt;x^3 - 3&lt;/math&gt;, we see that &lt;math&gt;x^3 - 3&lt;/math&gt; is a perfect cube, say &lt;math&gt;q^3&lt;/math&gt;. Then &lt;math&gt;x^3 - q^3 = 3,&lt;/math&gt; and &lt;math&gt;(x-q)(x^2 + xq + q^2) = 3&lt;/math&gt;, so that &lt;math&gt;x - q = 1&lt;/math&gt; and &lt;math&gt;x^2 + xq + q^2 = 3&lt;/math&gt;. Clearly, this system of equations has no integer solutions for &lt;math&gt;x&lt;/math&gt; or &lt;math&gt;q&lt;/math&gt;, a contradiction, hence completing the proof.<br /> <br /> Therefore no such integers exist.<br /> <br /> ==Solution 3==<br /> Let &lt;math&gt;a^5b+3=x^3&lt;/math&gt; and &lt;math&gt;ab^5+3=y^3&lt;/math&gt;. Then, &lt;math&gt;a^5b=x^3-3&lt;/math&gt;, &lt;math&gt;ab^5=y^3-3&lt;/math&gt;, and &lt;cmath&gt;(ab)^6=(x^3-3)(y^3-3)&lt;/cmath&gt;<br /> Now take &lt;math&gt;\text{mod }9&lt;/math&gt; (recall that perfect cubes &lt;math&gt;\equiv -1,0,1\pmod{9}&lt;/math&gt; and perfect sixth powers &lt;math&gt;\equiv 0,1\pmod{9}&lt;/math&gt;) on both sides. There are &lt;math&gt;3\times 3=9&lt;/math&gt; cases to consider on what values &lt;math&gt;\text{mod }9&lt;/math&gt; that &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;y^3&lt;/math&gt; take. Checking these &lt;math&gt;9&lt;/math&gt; cases, we see that only &lt;math&gt;x^3\equiv y^3\equiv 0\pmod{9}&lt;/math&gt; or &lt;math&gt;x\equiv y\equiv 0\pmod{3}&lt;/math&gt; yield a valid residue &lt;math&gt;\text{mod }9&lt;/math&gt; (specifically, &lt;math&gt;(x^3-3)(y^3-3)\equiv 0\pmod{9}&lt;/math&gt;). But this means that &lt;math&gt;3\mid ab&lt;/math&gt;, so &lt;math&gt;729\mid (ab)^6&lt;/math&gt; so &lt;cmath&gt;729\mid (x^3-3)(y^3-3)\iff 729\mid (27x'^3-3)(27y'^3-3)\iff 81\mid (9x'^3-1)(9y'^3-1)&lt;/cmath&gt; contradiction.<br /> <br /> ==Solution 4==<br /> If &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube, then &lt;math&gt;a^5b&lt;/math&gt; can be one of &lt;math&gt;5,6,7 \pmod 9&lt;/math&gt;, so &lt;math&gt;(a^5b)^5 = a^{25} b^5&lt;/math&gt; can be one of &lt;math&gt;5^5 \equiv 2&lt;/math&gt;, &lt;math&gt;6^5 \equiv 0&lt;/math&gt;, or &lt;math&gt;7^5 \equiv 4 \pmod 9&lt;/math&gt;. If &lt;math&gt;a&lt;/math&gt; were divisible by &lt;math&gt;3&lt;/math&gt;, we'd have &lt;math&gt;a^5 b \equiv 0 \pmod 9&lt;/math&gt;, which we've ruled out. So &lt;math&gt;\gcd(a,9) = 1&lt;/math&gt;, which means &lt;math&gt;a^6 \equiv 1 \pmod 9&lt;/math&gt;, and therefore &lt;math&gt;a^{25} b^5 \equiv ab^5 \pmod 9&lt;/math&gt;.<br /> <br /> We've shown that &lt;math&gt;a b^5&lt;/math&gt; can be one of &lt;math&gt;0, 2, 4 \pmod 9&lt;/math&gt;, so &lt;math&gt;ab^5 + 3&lt;/math&gt; can be one of &lt;math&gt;3, 5, 7 \pmod 9&lt;/math&gt;. None of these are possibilities for a perfect cube, so if &lt;math&gt;a^5b+3&lt;/math&gt; is a perfect cube, &lt;math&gt;ab^5+3&lt;/math&gt; cannot be.<br /> <br /> ==Solution 5==<br /> <br /> As in previous solutions, notice &lt;math&gt;ab^5,a^5b \equiv 5,6,7 \pmod 9&lt;/math&gt;. Now multiplying gives &lt;math&gt;a^6b^6&lt;/math&gt;, which is only &lt;math&gt;0,1 \pmod 9&lt;/math&gt;, so after testing all cases we find that &lt;math&gt;ab^5\equiv a^5b \equiv 6 \mod 9&lt;/math&gt;. Then since &lt;math&gt;\phi (9) = 6&lt;/math&gt;, &lt;math&gt;ab^5\equiv \frac{a}{b}\pmod 9&lt;/math&gt; and &lt;math&gt;a^5b \equiv \frac{b}{a}\pmod 9&lt;/math&gt; (Note that &lt;math&gt;a,b&lt;/math&gt; cannot be &lt;math&gt;0\pmod 9&lt;/math&gt;). Thus we find that the inverse of &lt;math&gt;6&lt;/math&gt; is itself under modulo &lt;math&gt;9&lt;/math&gt;, a contradiction.<br /> <br /> ==Solution 6==<br /> <br /> Assume to the contrary &lt;math&gt;a^5b +3&lt;/math&gt; and &lt;math&gt;ab^5 + 3&lt;/math&gt; are cubes. Since cubes are congruent to any of &lt;math&gt;0, 1, -1 \pmod 9&lt;/math&gt;, &lt;math&gt;ab^5,a^5b \equiv 5,6,7 \pmod 9&lt;/math&gt;. But if &lt;math&gt;ab^5 \equiv 6 \pmod 9&lt;/math&gt;, &lt;math&gt;3|a&lt;/math&gt;, so &lt;math&gt;a^5b \equiv 0 \pmod 9&lt;/math&gt;, contradiction. A similar argument can be made for &lt;math&gt;ab^5 \neq 6 \pmod 9&lt;/math&gt;. Thus, we get that:<br /> <br /> &lt;cmath&gt;ab^5,a^5b \equiv 5,7 \pmod 9&lt;/cmath&gt;<br /> <br /> This will be important later on.<br /> <br /> A critical idea is that since cubes are congruent to &lt;math&gt;0, 1, -1 \pmod 9&lt;/math&gt;, 6th powers are congruent to &lt;math&gt;0, 1 \pmod 9&lt;/math&gt;. And, &lt;math&gt;ab^5 \cdot a^5b = a^6 b^6 = (ab)^6&lt;/math&gt;, making the the product of &lt;math&gt;ab^5, a^5b&lt;/math&gt; a perfect 6th power. <br /> <br /> But, we can substitiute our possible values of &lt;math&gt;ab^5, a^5b&lt;/math&gt;. &lt;math&gt;5 \cdot 5 \equiv 7 \pmod 9&lt;/math&gt;, &lt;math&gt;5 \cdot 7 \equiv 8 \pmod 9&lt;/math&gt;, and &lt;math&gt;7 \cdot 7 \equiv 4 \pmod 9&lt;/math&gt;. None of these are congruent to &lt;math&gt;0, 1 \pmod 9&lt;/math&gt;, but perfect 6th powers have to be congruent to &lt;math&gt;0, 1 \pmod 9&lt;/math&gt;. We have a contradiction. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> <br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_6&diff=125160 2017 USAJMO Problems/Problem 6 2020-06-12T03:54:53Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is colored either red or blue, with exactly &lt;math&gt;n&lt;/math&gt; of them red and exactly &lt;math&gt;n&lt;/math&gt; of them blue. Let &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; be any ordering of the red points. Let &lt;math&gt;B_1&lt;/math&gt; be the nearest blue point to &lt;math&gt;R_1&lt;/math&gt; traveling counterclockwise around the circle starting from &lt;math&gt;R_1&lt;/math&gt;. Then let &lt;math&gt;B_2&lt;/math&gt; be the nearest of the remaining blue points to &lt;math&gt;R_2&lt;/math&gt; traveling counterclockwise around the circle from &lt;math&gt;R_2&lt;/math&gt;, and so on, until we have labeled all the blue points &lt;math&gt;B_1, \ldots, B_n&lt;/math&gt;. Show that the number of counterclockwise arcs of the form &lt;math&gt;R_i \rightarrow B_i&lt;/math&gt; that contain the point &lt;math&gt;(1,0)&lt;/math&gt; is independent of the way we chose the ordering &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; of the red points. <br /> <br /> ==Solution==<br /> I define a sequence to be, starting at &lt;math&gt;(1,0)&lt;/math&gt; and tracing the circle counterclockwise, and writing the color of the points in that order - either R or B. For example, possible sequences include &lt;math&gt;RB&lt;/math&gt;, &lt;math&gt;RBBR&lt;/math&gt;, &lt;math&gt;BBRRRB&lt;/math&gt;, &lt;math&gt;BRBRRBBR&lt;/math&gt;, etc.<br /> Note that choosing an &lt;math&gt;R_1&lt;/math&gt; is equivalent to choosing an &lt;math&gt;R&lt;/math&gt; in a sequence, and &lt;math&gt;B_1&lt;/math&gt; is defined as the &lt;math&gt;B&lt;/math&gt; closest to &lt;math&gt;R_1&lt;/math&gt; when moving rightwards. If no &lt;math&gt;B&lt;/math&gt;s exist to the right of &lt;math&gt;R_1&lt;/math&gt;, start from the far left. For example, if I have the above example &lt;math&gt;RBBR&lt;/math&gt;, and I define the 2nd &lt;math&gt;R&lt;/math&gt; to be &lt;math&gt;R_1&lt;/math&gt;, then the first &lt;math&gt;B&lt;/math&gt; will be &lt;math&gt;B_1&lt;/math&gt;. Because no &lt;math&gt;R&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt; can be named twice, I can simply remove &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;B_1&lt;/math&gt; from my sequence when I choose them. I define this to be a move. Hence, a possible move sequence of &lt;math&gt;BBRRRB&lt;/math&gt; is:<br /> &lt;math&gt;BBR_1RRB_1\implies B_2BRR_2\implies B_3R_3&lt;/math&gt;<br /> ---------<br /> Note that, if, in a move, &lt;math&gt;B_n&lt;/math&gt; appears to the left of &lt;math&gt;R_n&lt;/math&gt;, then &lt;math&gt;\stackrel{\frown}{R_nB_n}&lt;/math&gt; intersects &lt;math&gt;(1,0)&lt;/math&gt;<br /> <br /> Now, I define a commencing &lt;math&gt;B&lt;/math&gt; to be a &lt;math&gt;B&lt;/math&gt; which appears to the left of all &lt;math&gt;R&lt;/math&gt;s, and a terminating &lt;math&gt;R&lt;/math&gt; to be a &lt;math&gt;R&lt;/math&gt; which appears to the right of all &lt;math&gt;B&lt;/math&gt;s. Let the amount of commencing &lt;math&gt;B&lt;/math&gt;s be &lt;math&gt;j&lt;/math&gt;, and the amount of terminating &lt;math&gt;R&lt;/math&gt;s be &lt;math&gt;k&lt;/math&gt;, I claim that the number of arcs which cross &lt;math&gt;(1,0)&lt;/math&gt; is constant, and it is equal to &lt;math&gt;\text{max}(j,k)&lt;/math&gt;. I will show this with induction.<br /> <br /> Base case is when &lt;math&gt;n=1&lt;/math&gt;. In this case, there are only two possible sequences - &lt;math&gt;RB&lt;/math&gt; and &lt;math&gt;BR&lt;/math&gt;. In the first case, &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; does not cross &lt;math&gt;(1, 0)&lt;/math&gt;, but both &lt;math&gt;j&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; are &lt;math&gt;0&lt;/math&gt;, so &lt;math&gt;\text{max}(j,k)=0&lt;/math&gt;. In the second example, &lt;math&gt;j=1&lt;/math&gt;, &lt;math&gt;k=1&lt;/math&gt;, so &lt;math&gt;\text{max}(j,k)=1&lt;/math&gt;. &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; crosses &lt;math&gt;(1,0)&lt;/math&gt; since &lt;math&gt;B_1&lt;/math&gt; appears to the left of &lt;math&gt;R_1&lt;/math&gt;, so there is one arc which intersects. Hence, the base case is proved.<br /> <br /> For the inductive step, suppose that for a positive number &lt;math&gt;n&lt;/math&gt;, the number of arcs which cross &lt;math&gt;(1,0)&lt;/math&gt; is constant, and given by &lt;math&gt;\text{max}(j, k)&lt;/math&gt; for any configuration. Now, I will show it for &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> Suppose I first choose &lt;math&gt;R_1&lt;/math&gt; such that &lt;math&gt;B_1&lt;/math&gt; is to the right of &lt;math&gt;R_1&lt;/math&gt; in the sequence. This implies that &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; does not cross &lt;math&gt;(1,0)&lt;/math&gt;. But, neither &lt;math&gt;R_1&lt;/math&gt; nor &lt;math&gt;B_1&lt;/math&gt; is a commencing &lt;math&gt;B&lt;/math&gt; or terminating &lt;math&gt;R&lt;/math&gt;. These numbers remain constant, and now after this move we have a sequence of length &lt;math&gt;2n&lt;/math&gt;. Hence, by assumption, the total amount of arcs is &lt;math&gt;0+\text{max}(j,k)=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> -------------------------<br /> *Here is a counter-case. &lt;math&gt;BRR_{1}B_{1}RR&lt;/math&gt; : j = 1, k = 2 =&gt; &lt;math&gt;BRRR&lt;/math&gt; : j = 1, k = 3. These numbers may not remain constant.<br /> Thus, this solution probably doesn't work.<br /> -------------------------<br /> <br /> Now suppose that &lt;math&gt;R_1&lt;/math&gt; appears to the right of &lt;math&gt;B_1&lt;/math&gt;, but &lt;math&gt;B_1&lt;/math&gt; is not a commencing &lt;math&gt;B&lt;/math&gt;. This implies that there are no commencing &lt;math&gt;B&lt;/math&gt;s in the series, because there are no &lt;math&gt;B&lt;/math&gt;s to the left of &lt;math&gt;B_1&lt;/math&gt;, so &lt;math&gt;j=0&lt;/math&gt;. Note that this arc does intersect &lt;math&gt;(1,0)&lt;/math&gt;, and &lt;math&gt;R_1&lt;/math&gt; must be a terminating &lt;math&gt;R&lt;/math&gt;. &lt;math&gt;R_1&lt;/math&gt; must be a terminating &lt;math&gt;R&lt;/math&gt; because there are no &lt;math&gt;B&lt;/math&gt;s to the right of &lt;math&gt;R_1&lt;/math&gt;, or else that &lt;math&gt;B&lt;/math&gt; would be &lt;math&gt;B_1&lt;/math&gt;. The &lt;math&gt;2n&lt;/math&gt; length sequence that remains has &lt;math&gt;0&lt;/math&gt; commencing &lt;math&gt;B&lt;/math&gt;s and &lt;math&gt;k-1&lt;/math&gt; terminating &lt;math&gt;R&lt;/math&gt;s. Hence, by assumption, the total amount of arcs is &lt;math&gt;1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> Finally, suppose that &lt;math&gt;R_1&lt;/math&gt; appears to the right of &lt;math&gt;B_1&lt;/math&gt;, and &lt;math&gt;B_1&lt;/math&gt; is a commencing &lt;math&gt;B&lt;/math&gt;. We know that this arc will cross &lt;math&gt;(1,0)&lt;/math&gt;. Analogous to the previous case, &lt;math&gt;R_1&lt;/math&gt; is a terminating &lt;math&gt;R&lt;/math&gt;, so the &lt;math&gt;2n&lt;/math&gt; length sequence which remains has &lt;math&gt;j-1&lt;/math&gt; commencing &lt;math&gt;B&lt;/math&gt;s and &lt;math&gt;k-1&lt;/math&gt; terminating &lt;math&gt;R&lt;/math&gt;s. Hence, by assumption, the total amount of arcs is &lt;math&gt;1+\text{max}(j-1,k-1)=1+\text{max}(j,k)-1=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> There are no more possible cases, hence the induction is complete, and the number of arcs which intersect &lt;math&gt;(1,0)&lt;/math&gt; is indeed a constant which is given by &lt;math&gt;\text{max}(j,k)&lt;/math&gt;.<br /> <br /> -william122<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=5|aftertext=|after=Last Problem}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_6&diff=125159 2017 USAJMO Problems/Problem 6 2020-06-12T03:54:06Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is colored either red or blue, with exactly &lt;math&gt;n&lt;/math&gt; of them red and exactly &lt;math&gt;n&lt;/math&gt; of them blue. Let &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; be any ordering of the red points. Let &lt;math&gt;B_1&lt;/math&gt; be the nearest blue point to &lt;math&gt;R_1&lt;/math&gt; traveling counterclockwise around the circle starting from &lt;math&gt;R_1&lt;/math&gt;. Then let &lt;math&gt;B_2&lt;/math&gt; be the nearest of the remaining blue points to &lt;math&gt;R_2&lt;/math&gt; traveling counterclockwise around the circle from &lt;math&gt;R_2&lt;/math&gt;, and so on, until we have labeled all the blue points &lt;math&gt;B_1, \ldots, B_n&lt;/math&gt;. Show that the number of counterclockwise arcs of the form &lt;math&gt;R_i \rightarrow B_i&lt;/math&gt; that contain the point &lt;math&gt;(1,0)&lt;/math&gt; is independent of the way we chose the ordering &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; of the red points. <br /> <br /> ==Solution==<br /> I define a sequence to be, starting at &lt;math&gt;(1,0)&lt;/math&gt; and tracing the circle counterclockwise, and writing the color of the points in that order - either R or B. For example, possible sequences include &lt;math&gt;RB&lt;/math&gt;, &lt;math&gt;RBBR&lt;/math&gt;, &lt;math&gt;BBRRRB&lt;/math&gt;, &lt;math&gt;BRBRRBBR&lt;/math&gt;, etc.<br /> Note that choosing an &lt;math&gt;R_1&lt;/math&gt; is equivalent to choosing an &lt;math&gt;R&lt;/math&gt; in a sequence, and &lt;math&gt;B_1&lt;/math&gt; is defined as the &lt;math&gt;B&lt;/math&gt; closest to &lt;math&gt;R_1&lt;/math&gt; when moving rightwards. If no &lt;math&gt;B&lt;/math&gt;s exist to the right of &lt;math&gt;R_1&lt;/math&gt;, start from the far left. For example, if I have the above example &lt;math&gt;RBBR&lt;/math&gt;, and I define the 2nd &lt;math&gt;R&lt;/math&gt; to be &lt;math&gt;R_1&lt;/math&gt;, then the first &lt;math&gt;B&lt;/math&gt; will be &lt;math&gt;B_1&lt;/math&gt;. Because no &lt;math&gt;R&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt; can be named twice, I can simply remove &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;B_1&lt;/math&gt; from my sequence when I choose them. I define this to be a move. Hence, a possible move sequence of &lt;math&gt;BBRRRB&lt;/math&gt; is:<br /> &lt;math&gt;BBR_1RRB_1\implies B_2BRR_2\implies B_3R_3&lt;/math&gt;<br /> ---------<br /> Note that, if, in a move, &lt;math&gt;B_n&lt;/math&gt; appears to the left of &lt;math&gt;R_n&lt;/math&gt;, then &lt;math&gt;\stackrel{\frown}{R_nB_n}&lt;/math&gt; intersects &lt;math&gt;(1,0)&lt;/math&gt;<br /> <br /> Now, I define a commencing &lt;math&gt;B&lt;/math&gt; to be a &lt;math&gt;B&lt;/math&gt; which appears to the left of all &lt;math&gt;R&lt;/math&gt;s, and a terminating &lt;math&gt;R&lt;/math&gt; to be a &lt;math&gt;R&lt;/math&gt; which appears to the right of all &lt;math&gt;B&lt;/math&gt;s. Let the amount of commencing &lt;math&gt;B&lt;/math&gt;s be &lt;math&gt;j&lt;/math&gt;, and the amount of terminating &lt;math&gt;R&lt;/math&gt;s be &lt;math&gt;k&lt;/math&gt;, I claim that the number of arcs which cross &lt;math&gt;(1,0)&lt;/math&gt; is constant, and it is equal to &lt;math&gt;\text{max}(j,k)&lt;/math&gt;. I will show this with induction.<br /> <br /> Base case is when &lt;math&gt;n=1&lt;/math&gt;. In this case, there are only two possible sequences - &lt;math&gt;RB&lt;/math&gt; and &lt;math&gt;BR&lt;/math&gt;. In the first case, &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; does not cross &lt;math&gt;(1, 0)&lt;/math&gt;, but both &lt;math&gt;j&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; are &lt;math&gt;0&lt;/math&gt;, so &lt;math&gt;\text{max}(j,k)=0&lt;/math&gt;. In the second example, &lt;math&gt;j=1&lt;/math&gt;, &lt;math&gt;k=1&lt;/math&gt;, so &lt;math&gt;\text{max}(j,k)=1&lt;/math&gt;. &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; crosses &lt;math&gt;(1,0)&lt;/math&gt; since &lt;math&gt;B_1&lt;/math&gt; appears to the left of &lt;math&gt;R_1&lt;/math&gt;, so there is one arc which intersects. Hence, the base case is proved.<br /> <br /> For the inductive step, suppose that for a positive number &lt;math&gt;n&lt;/math&gt;, the number of arcs which cross &lt;math&gt;(1,0)&lt;/math&gt; is constant, and given by &lt;math&gt;\text{max}(j, k)&lt;/math&gt; for any configuration. Now, I will show it for &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> Suppose I first choose &lt;math&gt;R_1&lt;/math&gt; such that &lt;math&gt;B_1&lt;/math&gt; is to the right of &lt;math&gt;R_1&lt;/math&gt; in the sequence. This implies that &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; does not cross &lt;math&gt;(1,0)&lt;/math&gt;. But, neither &lt;math&gt;R_1&lt;/math&gt; nor &lt;math&gt;B_1&lt;/math&gt; is a commencing &lt;math&gt;B&lt;/math&gt; or terminating &lt;math&gt;R&lt;/math&gt;. These numbers remain constant, and now after this move we have a sequence of length &lt;math&gt;2n&lt;/math&gt;. Hence, by assumption, the total amount of arcs is &lt;math&gt;0+\text{max}(j,k)=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> *Here is a counter-case. &lt;math&gt;BRR_{1}B_{1}RR&lt;/math&gt; : j = 1, k = 2 =&gt; &lt;math&gt;BRRR&lt;/math&gt; : j = 1, k = 3. These numbers may not remain constant.<br /> Thus, this solution probably doesn't work.<br /> <br /> Now suppose that &lt;math&gt;R_1&lt;/math&gt; appears to the right of &lt;math&gt;B_1&lt;/math&gt;, but &lt;math&gt;B_1&lt;/math&gt; is not a commencing &lt;math&gt;B&lt;/math&gt;. This implies that there are no commencing &lt;math&gt;B&lt;/math&gt;s in the series, because there are no &lt;math&gt;B&lt;/math&gt;s to the left of &lt;math&gt;B_1&lt;/math&gt;, so &lt;math&gt;j=0&lt;/math&gt;. Note that this arc does intersect &lt;math&gt;(1,0)&lt;/math&gt;, and &lt;math&gt;R_1&lt;/math&gt; must be a terminating &lt;math&gt;R&lt;/math&gt;. &lt;math&gt;R_1&lt;/math&gt; must be a terminating &lt;math&gt;R&lt;/math&gt; because there are no &lt;math&gt;B&lt;/math&gt;s to the right of &lt;math&gt;R_1&lt;/math&gt;, or else that &lt;math&gt;B&lt;/math&gt; would be &lt;math&gt;B_1&lt;/math&gt;. The &lt;math&gt;2n&lt;/math&gt; length sequence that remains has &lt;math&gt;0&lt;/math&gt; commencing &lt;math&gt;B&lt;/math&gt;s and &lt;math&gt;k-1&lt;/math&gt; terminating &lt;math&gt;R&lt;/math&gt;s. Hence, by assumption, the total amount of arcs is &lt;math&gt;1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> Finally, suppose that &lt;math&gt;R_1&lt;/math&gt; appears to the right of &lt;math&gt;B_1&lt;/math&gt;, and &lt;math&gt;B_1&lt;/math&gt; is a commencing &lt;math&gt;B&lt;/math&gt;. We know that this arc will cross &lt;math&gt;(1,0)&lt;/math&gt;. Analogous to the previous case, &lt;math&gt;R_1&lt;/math&gt; is a terminating &lt;math&gt;R&lt;/math&gt;, so the &lt;math&gt;2n&lt;/math&gt; length sequence which remains has &lt;math&gt;j-1&lt;/math&gt; commencing &lt;math&gt;B&lt;/math&gt;s and &lt;math&gt;k-1&lt;/math&gt; terminating &lt;math&gt;R&lt;/math&gt;s. Hence, by assumption, the total amount of arcs is &lt;math&gt;1+\text{max}(j-1,k-1)=1+\text{max}(j,k)-1=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> There are no more possible cases, hence the induction is complete, and the number of arcs which intersect &lt;math&gt;(1,0)&lt;/math&gt; is indeed a constant which is given by &lt;math&gt;\text{max}(j,k)&lt;/math&gt;.<br /> <br /> -william122<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=5|aftertext=|after=Last Problem}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_6&diff=125158 2017 USAJMO Problems/Problem 6 2020-06-12T03:52:36Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is colored either red or blue, with exactly &lt;math&gt;n&lt;/math&gt; of them red and exactly &lt;math&gt;n&lt;/math&gt; of them blue. Let &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; be any ordering of the red points. Let &lt;math&gt;B_1&lt;/math&gt; be the nearest blue point to &lt;math&gt;R_1&lt;/math&gt; traveling counterclockwise around the circle starting from &lt;math&gt;R_1&lt;/math&gt;. Then let &lt;math&gt;B_2&lt;/math&gt; be the nearest of the remaining blue points to &lt;math&gt;R_2&lt;/math&gt; traveling counterclockwise around the circle from &lt;math&gt;R_2&lt;/math&gt;, and so on, until we have labeled all the blue points &lt;math&gt;B_1, \ldots, B_n&lt;/math&gt;. Show that the number of counterclockwise arcs of the form &lt;math&gt;R_i \rightarrow B_i&lt;/math&gt; that contain the point &lt;math&gt;(1,0)&lt;/math&gt; is independent of the way we chose the ordering &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; of the red points. <br /> <br /> ==Solution==<br /> I define a sequence to be, starting at &lt;math&gt;(1,0)&lt;/math&gt; and tracing the circle counterclockwise, and writing the color of the points in that order - either R or B. For example, possible sequences include &lt;math&gt;RB&lt;/math&gt;, &lt;math&gt;RBBR&lt;/math&gt;, &lt;math&gt;BBRRRB&lt;/math&gt;, &lt;math&gt;BRBRRBBR&lt;/math&gt;, etc.<br /> Note that choosing an &lt;math&gt;R_1&lt;/math&gt; is equivalent to choosing an &lt;math&gt;R&lt;/math&gt; in a sequence, and &lt;math&gt;B_1&lt;/math&gt; is defined as the &lt;math&gt;B&lt;/math&gt; closest to &lt;math&gt;R_1&lt;/math&gt; when moving rightwards. If no &lt;math&gt;B&lt;/math&gt;s exist to the right of &lt;math&gt;R_1&lt;/math&gt;, start from the far left. For example, if I have the above example &lt;math&gt;RBBR&lt;/math&gt;, and I define the 2nd &lt;math&gt;R&lt;/math&gt; to be &lt;math&gt;R_1&lt;/math&gt;, then the first &lt;math&gt;B&lt;/math&gt; will be &lt;math&gt;B_1&lt;/math&gt;. Because no &lt;math&gt;R&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt; can be named twice, I can simply remove &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;B_1&lt;/math&gt; from my sequence when I choose them. I define this to be a move. Hence, a possible move sequence of &lt;math&gt;BBRRRB&lt;/math&gt; is:<br /> &lt;math&gt;BBR_1RRB_1\implies B_2BRR_2\implies B_3R_3&lt;/math&gt;<br /> ---------<br /> Note that, if, in a move, &lt;math&gt;B_n&lt;/math&gt; appears to the left of &lt;math&gt;R_n&lt;/math&gt;, then &lt;math&gt;\stackrel{\frown}{R_nB_n}&lt;/math&gt; intersects &lt;math&gt;(1,0)&lt;/math&gt;<br /> <br /> Now, I define a commencing &lt;math&gt;B&lt;/math&gt; to be a &lt;math&gt;B&lt;/math&gt; which appears to the left of all &lt;math&gt;R&lt;/math&gt;s, and a terminating &lt;math&gt;R&lt;/math&gt; to be a &lt;math&gt;R&lt;/math&gt; which appears to the right of all &lt;math&gt;B&lt;/math&gt;s. Let the amount of commencing &lt;math&gt;B&lt;/math&gt;s be &lt;math&gt;j&lt;/math&gt;, and the amount of terminating &lt;math&gt;R&lt;/math&gt;s be &lt;math&gt;k&lt;/math&gt;, I claim that the number of arcs which cross &lt;math&gt;(1,0)&lt;/math&gt; is constant, and it is equal to &lt;math&gt;\text{max}(j,k)&lt;/math&gt;. I will show this with induction.<br /> <br /> Base case is when &lt;math&gt;n=1&lt;/math&gt;. In this case, there are only two possible sequences - &lt;math&gt;RB&lt;/math&gt; and &lt;math&gt;BR&lt;/math&gt;. In the first case, &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; does not cross &lt;math&gt;(1, 0)&lt;/math&gt;, but both &lt;math&gt;j&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; are &lt;math&gt;0&lt;/math&gt;, so &lt;math&gt;\text{max}(j,k)=0&lt;/math&gt;. In the second example, &lt;math&gt;j=1&lt;/math&gt;, &lt;math&gt;k=1&lt;/math&gt;, so &lt;math&gt;\text{max}(j,k)=1&lt;/math&gt;. &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; crosses &lt;math&gt;(1,0)&lt;/math&gt; since &lt;math&gt;B_1&lt;/math&gt; appears to the left of &lt;math&gt;R_1&lt;/math&gt;, so there is one arc which intersects. Hence, the base case is proved.<br /> <br /> For the inductive step, suppose that for a positive number &lt;math&gt;n&lt;/math&gt;, the number of arcs which cross &lt;math&gt;(1,0)&lt;/math&gt; is constant, and given by &lt;math&gt;\text{max}(j, k)&lt;/math&gt; for any configuration. Now, I will show it for &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> Suppose I first choose &lt;math&gt;R_1&lt;/math&gt; such that &lt;math&gt;B_1&lt;/math&gt; is to the right of &lt;math&gt;R_1&lt;/math&gt; in the sequence. This implies that &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; does not cross &lt;math&gt;(1,0)&lt;/math&gt;. But, neither &lt;math&gt;R_1&lt;/math&gt; nor &lt;math&gt;B_1&lt;/math&gt; is a commencing &lt;math&gt;B&lt;/math&gt; or terminating &lt;math&gt;R&lt;/math&gt;. These numbers remain constant, and now after this move we have a sequence of length &lt;math&gt;2n&lt;/math&gt;. Hence, by assumption, the total amount of arcs is &lt;math&gt;0+\text{max}(j,k)=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> *Here is a counter-case. &lt;math&gt;BRR_{1}B_{1}RR&lt;/math&gt; : j = 1, k = 2 =&gt; &lt;math&gt;BRRR&lt;/math&gt; : j = 1, k = 3. These numbers may not remain constant.<br /> <br /> Now suppose that &lt;math&gt;R_1&lt;/math&gt; appears to the right of &lt;math&gt;B_1&lt;/math&gt;, but &lt;math&gt;B_1&lt;/math&gt; is not a commencing &lt;math&gt;B&lt;/math&gt;. This implies that there are no commencing &lt;math&gt;B&lt;/math&gt;s in the series, because there are no &lt;math&gt;B&lt;/math&gt;s to the left of &lt;math&gt;B_1&lt;/math&gt;, so &lt;math&gt;j=0&lt;/math&gt;. Note that this arc does intersect &lt;math&gt;(1,0)&lt;/math&gt;, and &lt;math&gt;R_1&lt;/math&gt; must be a terminating &lt;math&gt;R&lt;/math&gt;. &lt;math&gt;R_1&lt;/math&gt; must be a terminating &lt;math&gt;R&lt;/math&gt; because there are no &lt;math&gt;B&lt;/math&gt;s to the right of &lt;math&gt;R_1&lt;/math&gt;, or else that &lt;math&gt;B&lt;/math&gt; would be &lt;math&gt;B_1&lt;/math&gt;. The &lt;math&gt;2n&lt;/math&gt; length sequence that remains has &lt;math&gt;0&lt;/math&gt; commencing &lt;math&gt;B&lt;/math&gt;s and &lt;math&gt;k-1&lt;/math&gt; terminating &lt;math&gt;R&lt;/math&gt;s. Hence, by assumption, the total amount of arcs is &lt;math&gt;1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> Finally, suppose that &lt;math&gt;R_1&lt;/math&gt; appears to the right of &lt;math&gt;B_1&lt;/math&gt;, and &lt;math&gt;B_1&lt;/math&gt; is a commencing &lt;math&gt;B&lt;/math&gt;. We know that this arc will cross &lt;math&gt;(1,0)&lt;/math&gt;. Analogous to the previous case, &lt;math&gt;R_1&lt;/math&gt; is a terminating &lt;math&gt;R&lt;/math&gt;, so the &lt;math&gt;2n&lt;/math&gt; length sequence which remains has &lt;math&gt;j-1&lt;/math&gt; commencing &lt;math&gt;B&lt;/math&gt;s and &lt;math&gt;k-1&lt;/math&gt; terminating &lt;math&gt;R&lt;/math&gt;s. Hence, by assumption, the total amount of arcs is &lt;math&gt;1+\text{max}(j-1,k-1)=1+\text{max}(j,k)-1=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> There are no more possible cases, hence the induction is complete, and the number of arcs which intersect &lt;math&gt;(1,0)&lt;/math&gt; is indeed a constant which is given by &lt;math&gt;\text{max}(j,k)&lt;/math&gt;.<br /> <br /> -william122<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=5|aftertext=|after=Last Problem}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_6&diff=125157 2017 USAJMO Problems/Problem 6 2020-06-12T03:42:37Z <p>Alexlikemath: </p> <hr /> <div>==Problem==<br /> Let &lt;math&gt;P_1, \ldots, P_{2n}&lt;/math&gt; be &lt;math&gt;2n&lt;/math&gt; distinct points on the unit circle &lt;math&gt;x^2 + y^2 = 1&lt;/math&gt; other than &lt;math&gt;(1,0)&lt;/math&gt;. Each point is colored either red or blue, with exactly &lt;math&gt;n&lt;/math&gt; of them red and exactly &lt;math&gt;n&lt;/math&gt; of them blue. Let &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; be any ordering of the red points. Let &lt;math&gt;B_1&lt;/math&gt; be the nearest blue point to &lt;math&gt;R_1&lt;/math&gt; traveling counterclockwise around the circle starting from &lt;math&gt;R_1&lt;/math&gt;. Then let &lt;math&gt;B_2&lt;/math&gt; be the nearest of the remaining blue points to &lt;math&gt;R_2&lt;/math&gt; traveling counterclockwise around the circle from &lt;math&gt;R_2&lt;/math&gt;, and so on, until we have labeled all the blue points &lt;math&gt;B_1, \ldots, B_n&lt;/math&gt;. Show that the number of counterclockwise arcs of the form &lt;math&gt;R_i \rightarrow B_i&lt;/math&gt; that contain the point &lt;math&gt;(1,0)&lt;/math&gt; is independent of the way we chose the ordering &lt;math&gt;R_1, \ldots, R_n&lt;/math&gt; of the red points. <br /> <br /> ==Solution==<br /> I define a sequence to be, starting at &lt;math&gt;(1,0)&lt;/math&gt; and tracing the circle counterclockwise, and writing the color of the points in that order - either R or B. For example, possible sequences include &lt;math&gt;RB&lt;/math&gt;, &lt;math&gt;RBBR&lt;/math&gt;, &lt;math&gt;BBRRRB&lt;/math&gt;, &lt;math&gt;BRBRRBBR&lt;/math&gt;, etc.<br /> Note that choosing an &lt;math&gt;R_1&lt;/math&gt; is equivalent to choosing an &lt;math&gt;R&lt;/math&gt; in a sequence, and &lt;math&gt;B_1&lt;/math&gt; is defined as the &lt;math&gt;B&lt;/math&gt; closest to &lt;math&gt;R_1&lt;/math&gt; when moving rightwards. If no &lt;math&gt;B&lt;/math&gt;s exist to the right of &lt;math&gt;R_1&lt;/math&gt;, start from the far left. For example, if I have the above example &lt;math&gt;RBBR&lt;/math&gt;, and I define the 2nd &lt;math&gt;R&lt;/math&gt; to be &lt;math&gt;R_1&lt;/math&gt;, then the first &lt;math&gt;B&lt;/math&gt; will be &lt;math&gt;B_1&lt;/math&gt;. Because no &lt;math&gt;R&lt;/math&gt; or &lt;math&gt;B&lt;/math&gt; can be named twice, I can simply remove &lt;math&gt;R_1&lt;/math&gt; and &lt;math&gt;B_1&lt;/math&gt; from my sequence when I choose them. I define this to be a move. Hence, a possible move sequence of &lt;math&gt;BBRRRB&lt;/math&gt; is:<br /> &lt;math&gt;BBR_1RRB_1\implies B_2BRR_2\implies B_3R_3&lt;/math&gt;<br /> ---------<br /> Note that, if, in a move, &lt;math&gt;B_n&lt;/math&gt; appears to the left of &lt;math&gt;R_n&lt;/math&gt;, then &lt;math&gt;\stackrel{\frown}{R_nB_n}&lt;/math&gt; intersects &lt;math&gt;(1,0)&lt;/math&gt;<br /> <br /> Now, I define a commencing &lt;math&gt;B&lt;/math&gt; to be a &lt;math&gt;B&lt;/math&gt; which appears to the left of all &lt;math&gt;R&lt;/math&gt;s, and a terminating &lt;math&gt;R&lt;/math&gt; to be a &lt;math&gt;R&lt;/math&gt; which appears to the right of all &lt;math&gt;B&lt;/math&gt;s. Let the amount of commencing &lt;math&gt;B&lt;/math&gt;s be &lt;math&gt;j&lt;/math&gt;, and the amount of terminating &lt;math&gt;R&lt;/math&gt;s be &lt;math&gt;k&lt;/math&gt;, I claim that the number of arcs which cross &lt;math&gt;(1,0)&lt;/math&gt; is constant, and it is equal to &lt;math&gt;\text{max}(j,k)&lt;/math&gt;. I will show this with induction.<br /> <br /> Base case is when &lt;math&gt;n=1&lt;/math&gt;. In this case, there are only two possible sequences - &lt;math&gt;RB&lt;/math&gt; and &lt;math&gt;BR&lt;/math&gt;. In the first case, &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; does not cross &lt;math&gt;(1, 0)&lt;/math&gt;, but both &lt;math&gt;j&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; are &lt;math&gt;0&lt;/math&gt;, so &lt;math&gt;\text{max}(j,k)=0&lt;/math&gt;. In the second example, &lt;math&gt;j=1&lt;/math&gt;, &lt;math&gt;k=1&lt;/math&gt;, so &lt;math&gt;\text{max}(j,k)=1&lt;/math&gt;. &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; crosses &lt;math&gt;(1,0)&lt;/math&gt; since &lt;math&gt;B_1&lt;/math&gt; appears to the left of &lt;math&gt;R_1&lt;/math&gt;, so there is one arc which intersects. Hence, the base case is proved.<br /> <br /> For the inductive step, suppose that for a positive number &lt;math&gt;n&lt;/math&gt;, the number of arcs which cross &lt;math&gt;(1,0)&lt;/math&gt; is constant, and given by &lt;math&gt;\text{max}(j, k)&lt;/math&gt; for any configuration. Now, I will show it for &lt;math&gt;n+1&lt;/math&gt;.<br /> <br /> Suppose I first choose &lt;math&gt;R_1&lt;/math&gt; such that &lt;math&gt;B_1&lt;/math&gt; is to the right of &lt;math&gt;R_1&lt;/math&gt; in the sequence. This implies that &lt;math&gt;\stackrel{\frown}{R_1B_1}&lt;/math&gt; does not cross &lt;math&gt;(1,0)&lt;/math&gt;. But, neither &lt;math&gt;R_1&lt;/math&gt; nor &lt;math&gt;B_1&lt;/math&gt; is a commencing &lt;math&gt;B&lt;/math&gt; or terminating &lt;math&gt;R&lt;/math&gt;. These numbers remain constant, and now after this move we have a sequence of length &lt;math&gt;2n&lt;/math&gt;. Hence, by assumption, the total amount of arcs is &lt;math&gt;0+\text{max}(j,k)=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> *Here is a counter-case. &lt;math&gt;BRR_1B_1RR&lt;/math&gt; j = 1, k = 2 =&gt; &lt;math&gt;BRRR&lt;/math&gt; j = 1, k = 3. These numbers may not remain constant.<br /> <br /> Now suppose that &lt;math&gt;R_1&lt;/math&gt; appears to the right of &lt;math&gt;B_1&lt;/math&gt;, but &lt;math&gt;B_1&lt;/math&gt; is not a commencing &lt;math&gt;B&lt;/math&gt;. This implies that there are no commencing &lt;math&gt;B&lt;/math&gt;s in the series, because there are no &lt;math&gt;B&lt;/math&gt;s to the left of &lt;math&gt;B_1&lt;/math&gt;, so &lt;math&gt;j=0&lt;/math&gt;. Note that this arc does intersect &lt;math&gt;(1,0)&lt;/math&gt;, and &lt;math&gt;R_1&lt;/math&gt; must be a terminating &lt;math&gt;R&lt;/math&gt;. &lt;math&gt;R_1&lt;/math&gt; must be a terminating &lt;math&gt;R&lt;/math&gt; because there are no &lt;math&gt;B&lt;/math&gt;s to the right of &lt;math&gt;R_1&lt;/math&gt;, or else that &lt;math&gt;B&lt;/math&gt; would be &lt;math&gt;B_1&lt;/math&gt;. The &lt;math&gt;2n&lt;/math&gt; length sequence that remains has &lt;math&gt;0&lt;/math&gt; commencing &lt;math&gt;B&lt;/math&gt;s and &lt;math&gt;k-1&lt;/math&gt; terminating &lt;math&gt;R&lt;/math&gt;s. Hence, by assumption, the total amount of arcs is &lt;math&gt;1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> Finally, suppose that &lt;math&gt;R_1&lt;/math&gt; appears to the right of &lt;math&gt;B_1&lt;/math&gt;, and &lt;math&gt;B_1&lt;/math&gt; is a commencing &lt;math&gt;B&lt;/math&gt;. We know that this arc will cross &lt;math&gt;(1,0)&lt;/math&gt;. Analogous to the previous case, &lt;math&gt;R_1&lt;/math&gt; is a terminating &lt;math&gt;R&lt;/math&gt;, so the &lt;math&gt;2n&lt;/math&gt; length sequence which remains has &lt;math&gt;j-1&lt;/math&gt; commencing &lt;math&gt;B&lt;/math&gt;s and &lt;math&gt;k-1&lt;/math&gt; terminating &lt;math&gt;R&lt;/math&gt;s. Hence, by assumption, the total amount of arcs is &lt;math&gt;1+\text{max}(j-1,k-1)=1+\text{max}(j,k)-1=\text{max}(j,k)&lt;/math&gt;.<br /> <br /> There are no more possible cases, hence the induction is complete, and the number of arcs which intersect &lt;math&gt;(1,0)&lt;/math&gt; is indeed a constant which is given by &lt;math&gt;\text{max}(j,k)&lt;/math&gt;.<br /> <br /> -william122<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=5|aftertext=|after=Last Problem}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_2&diff=124985 2017 USAJMO Problems/Problem 2 2020-06-11T17:24:13Z <p>Alexlikemath: Simplified Evan Chen's Solution</p> <hr /> <div>==Problem:==<br /> Consider the equation <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = (x-y)^7.&lt;/cmath&gt;<br /> <br /> (a) Prove that there are infinitely many pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation. <br /> <br /> (b) Describe all pairs &lt;math&gt;(x,y)&lt;/math&gt; of positive integers satisfying the equation.<br /> <br /> ==Solution 1==<br /> We have &lt;math&gt;(3x^3+xy^2)(yx^2+3y^3)=(x-y)^7&lt;/math&gt;, which can be expressed as &lt;math&gt;xy(3x^2+y^2)(x^2+3y^2)=(x-y)^7&lt;/math&gt;. At this point, we <br /> <br /> think of substitution. A substitution of form &lt;math&gt;a=x+y, b=x-y&lt;/math&gt; is slightly derailed by the leftover x and y terms, so <br /> <br /> instead, seeing the xy in front, we substitute &lt;math&gt;x=a+b, y=a-b&lt;/math&gt;. This leaves us with &lt;math&gt;(a^2-b^2)(4a^2+4ab+4b^2)(4a^2-<br /> 4ab+4b^2)=128b^7&lt;/math&gt;, so &lt;math&gt;(a^2-b^2)(a^2+ab+b^2)(a^2-ab+b^2)=8b^7&lt;/math&gt;. Expanding yields &lt;math&gt;a^6-b^6=8b^7&lt;/math&gt;. Rearranging, we have <br /> <br /> &lt;math&gt;b^6(8b+1)=a^6&lt;/math&gt;. To satisfy this equation in integers, &lt;math&gt;8b+1&lt;/math&gt; must obviously be a &lt;math&gt;6th&lt;/math&gt; power, and further inspection shows <br /> <br /> that it must also be odd. Also, since it is a square and all odd squares are 1 mod 8, every odd sixth power gives a <br /> <br /> solution. Since the problem asks for positive integers, the pair &lt;math&gt;(a,b)=(0,0)&lt;/math&gt; does not work. We go to the next highest odd <br /> <br /> &lt;math&gt;6th&lt;/math&gt; power, &lt;math&gt;3^6&lt;/math&gt; or &lt;math&gt;729&lt;/math&gt;. In this case, &lt;math&gt;b=91&lt;/math&gt;, so the LHS is &lt;math&gt;91^6\cdot3^6=273^6&lt;/math&gt;, so &lt;math&gt;a=273&lt;/math&gt;. Using the original <br /> <br /> substitution yields &lt;math&gt;(x,y)=(364,182)&lt;/math&gt; as the first solution. We have shown part a by showing that there are an infinite <br /> <br /> number of positive integer solutions for &lt;math&gt;(a,b)&lt;/math&gt;, which can then be manipulated into solutions for &lt;math&gt;(x,y)&lt;/math&gt;. To solve part <br /> <br /> b, we look back at the original method of generating solutions. Define &lt;math&gt;a_n&lt;/math&gt; and &lt;math&gt;b_n&lt;/math&gt; to be the pair representing the nth <br /> <br /> solution. Since the nth odd number is &lt;math&gt;2n+1&lt;/math&gt;, &lt;math&gt;b_n=\frac{(2n+1)^6-1}{8}&lt;/math&gt;. It follows that &lt;math&gt;a_n=(2n+1)b_n=\frac{(2n+1)^7-<br /> (2n+1)}{8}&lt;/math&gt;. From our original substitution, &lt;math&gt;(x,y)=\left(\frac{(2n+1)^7+(2n+1)^6-2n-2}{8}, \frac{(2n+1)^7-(2n+1)^6-2n}<br /> {8}\right)&lt;/math&gt;.<br /> <br /> ==Solution 2 (and motivation)==<br /> First, we shall prove a lemma:<br /> <br /> LEMMA:<br /> <br /> &lt;cmath&gt;\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right) = \frac{(x+y)^6-(x-y)^6}{4}&lt;/cmath&gt;<br /> PROOF: Expanding and simplifying the right side, we find that &lt;cmath&gt;\frac{(x+y)^6-(x-y)^6}{4}=\frac{12x^5y+40x^3y^3+12xy^5}{4}&lt;/cmath&gt;&lt;cmath&gt;=3x^5y+10x^3y^3+3xy^5&lt;/cmath&gt;&lt;cmath&gt;=\left(3x^3 + xy^2 \right) \left(x^2y + 3y^3 \right)&lt;/cmath&gt;<br /> which proves our lemma.<br /> <br /> <br /> Now, we have that &lt;cmath&gt;\frac{(x+y)^6-(x-y)^6}{4}=(x-y)^7&lt;/cmath&gt;<br /> Rearranging and getting rid of the denominator, we have that &lt;cmath&gt;(x+y)^6=4(x-y)^7+(x-y)^6&lt;/cmath&gt;<br /> Factoring, we have &lt;cmath&gt;(x+y)^6=(x-y)^6(4(x-y)+1)&lt;/cmath&gt;Dividing both sides, we have &lt;cmath&gt;\left(\frac{x+y}{x-y}\right)^6=4x-4y+1&lt;/cmath&gt;<br /> Now, since the LHS is the 6th power of a rational number, and the RHS is an integer, the RHS must be a perfect 6th power. Define &lt;math&gt;a=\frac{x+y}{x-y}&lt;/math&gt;. By inspection, &lt;math&gt;a&lt;/math&gt; must be a positive odd integer satistisfying &lt;math&gt;a \geq 3&lt;/math&gt;. We also have &lt;cmath&gt;a^6=4x-4y+1&lt;/cmath&gt; Now, we can solve for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; in terms of &lt;math&gt;a&lt;/math&gt;:<br /> &lt;math&gt;x-y=\frac{a^6-1}{4}&lt;/math&gt; and &lt;math&gt;x+y=a(x-y)=\frac{a(a^6-1)}{4}&lt;/math&gt;.<br /> Now we have: &lt;cmath&gt;(x,y)=\left(\frac{(a+1)(a^6-1)}{8},\frac{(a-1)(a^6-1)}{8}\right)&lt;/cmath&gt;<br /> and it is trivial to check that this parameterization works for all such &lt;math&gt;a&lt;/math&gt; (to keep &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; integral), which implies part (a).<br /> <br /> <br /> MOTIVATION FOR LEMMA:<br /> I expanded the LHS, noticed the coefficients were &lt;math&gt;(3,10,3)&lt;/math&gt;, and immediately thought of binomial coefficients. Looking at Pascal's triangle, it was then easy to find and prove the lemma.<br /> <br /> -sunfishho<br /> <br /> ==Solution &lt;math&gt;(1.5, x)&lt;/math&gt; where &lt;math&gt;|x|&gt;0&lt;/math&gt;== <br /> So named because it is a mix of solutions 1 and 2 but differs in other aspects.<br /> After fruitless searching, let &lt;math&gt;x+y=a&lt;/math&gt;, &lt;math&gt;x-y=b&lt;/math&gt;. Clearly &lt;math&gt;a, b &gt; 0&lt;/math&gt;. Then &lt;math&gt;x=\frac{a+b}{2}, y=\frac{a-b}{2}, x^2+y^2=\frac{a^2+b^2}{2}, x^2=\frac{a^2+2ab+b^2}{4}, y^2=\frac{a^2-2ab+b^2}{4}, xy=\frac{a^2-b^2}{4}&lt;/math&gt;.<br /> <br /> Change the LHS to &lt;math&gt;xy(3x^2+y^2)(x^2+3y^2)=(a-b)(a+b)(a^2+ab+b^2)(a^2-ab+b^2)*\frac{1}{4}=(a^3-b^3)(a^3+b^3)*\frac{1}{4}=(a^6-b^6)*\frac{1}{4}&lt;/math&gt;. Change the RHS to &lt;math&gt;b^7&lt;/math&gt;. Therefore &lt;math&gt;(\frac{a}{b})^6=4b+1&lt;/math&gt;. Let &lt;math&gt;n=\frac{a}{b}&lt;/math&gt;, and note that &lt;math&gt;n&lt;/math&gt; is an integer. Therefore &lt;math&gt;a=n(\frac{n^6-1}{4}), b=\frac{n^6-1}{4}&lt;/math&gt;. Because &lt;math&gt;n^6 \equiv 1 (\mod{4})&lt;/math&gt;, &lt;math&gt;n&lt;/math&gt; is odd and &lt;math&gt;&gt;1&lt;/math&gt; because &lt;math&gt;b&gt;0&lt;/math&gt;. Therefore, substituting for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; we get: &lt;math&gt;x=\frac{(n^4+n^2+1)(n-1)(n+1)^2}{8}, y=\frac{(n+1)(n-1)^2(n^4+n^2+1)}{8}&lt;/math&gt;.<br /> <br /> ==Simplified Evan Chen's Solution==<br /> <br /> Let &lt;math&gt;x = da&lt;/math&gt;, &lt;math&gt;y = db&lt;/math&gt;, such that &lt;math&gt;gcd(a,b) = 1&lt;/math&gt;. It's obvious &lt;math&gt;x &gt; y&lt;/math&gt;, so &lt;math&gt;a &gt; b&lt;/math&gt; <br /> <br /> By plugging it in to the original equation, and simplifying, we get<br /> <br /> &lt;cmath&gt;d = \frac{ab(a^2 + 3b^2)(3a^2 + b^2)}{(a-b)^7}&lt;/cmath&gt;<br /> <br /> Since d is an integer, we got:<br /> <br /> &lt;cmath&gt;(a-b)^7 \mid ab(a^2 + 3b^2)(3a^2 + b^2) \: \: \: \: \: (*)&lt;/cmath&gt;<br /> <br /> 1. Since &lt;math&gt;gcd(a,b) = 1&lt;/math&gt;, a and b can't both be even<br /> <br /> 2. If both a and b are odd. <br /> &lt;math&gt;a-b&lt;/math&gt; will be even. The left hand side (*) has at least 7 factors of 2. <br /> Evaluating the (*) RHS term by term, we get:<br /> &lt;cmath&gt;a \equiv 1 \pmod{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;b \equiv 1 \pmod{2}&lt;/cmath&gt;<br /> &lt;cmath&gt;a^2 + 3b^2 \equiv 4 \pmod{8}&lt;/cmath&gt;<br /> &lt;cmath&gt;3a^2 + b^2 \equiv 4 \pmod{8}&lt;/cmath&gt;<br /> <br /> the (*) RHS has only 4 factors of 2. Contradiction.<br /> <br /> 3. Thus, &lt;math&gt;a-b&lt;/math&gt; has to be odd. We want to prove &lt;math&gt;a-b=1&lt;/math&gt;<br /> <br /> We know: <br /> &lt;cmath&gt;a \equiv b \pmod{a-b}&lt;/cmath&gt;<br /> So <br /> &lt;cmath&gt;ab(a^2 + 3b^2)(3a^2 + b^2) \equiv 16a^6 \pmod{a-b}&lt;/cmath&gt;<br /> We can assume if &lt;math&gt;a-b \neq 1&lt;/math&gt;<br /> <br /> Since &lt;math&gt;a-b&lt;/math&gt; is odd, &lt;math&gt;gcd(16, a-b) = 1&lt;/math&gt;<br /> <br /> Since &lt;math&gt;gcd(a, a-b) = 1&lt;/math&gt;, &lt;math&gt;gcd(a^6, a-b) = 1&lt;/math&gt; <br /> <br /> &lt;cmath&gt;ab(a^2 + 3b^2)(3a^2 + b^2) \equiv 16a^6 \neq 0 \pmod{a-b}&lt;/cmath&gt;<br /> contradiction. So, &lt;math&gt;a-b = 1&lt;/math&gt;. <br /> <br /> If &lt;math&gt;a-b=1&lt;/math&gt;, obviously (*) will work, d will be an integer. <br /> <br /> So (*) &lt;math&gt;\Leftrightarrow&lt;/math&gt; &lt;math&gt;a-b=1&lt;/math&gt;. The proof is complete.<br /> <br /> -Alexlikemath<br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|num-b=1|num-a=3}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=124920 2017 USAJMO Problems/Problem 1 2020-06-11T01:40:43Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;a = 2n-1&lt;/math&gt; and &lt;math&gt;b = 2n+1&lt;/math&gt;. We see that &lt;math&gt;(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}&lt;/math&gt;. Therefore, we have &lt;math&gt;(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}&lt;/math&gt;, as desired. <br /> <br /> (Credits to mathmaster2012)<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;x&lt;/math&gt; be odd where &lt;math&gt;x&gt;1&lt;/math&gt;. We have &lt;math&gt;x^2-1=(x-1)(x+1),&lt;/math&gt; so &lt;math&gt;x^2-1 \equiv 0 \pmod{2x+2}.&lt;/math&gt; This means that &lt;math&gt;x^{x+2}-x^x \equiv 0 \pmod{2x+2},&lt;/math&gt; and since x is odd, &lt;math&gt;x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; or &lt;math&gt;x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; as desired.<br /> <br /> ==Solution 3==<br /> Because problems such as this usually are related to expressions along the lines of &lt;math&gt;x\pm1&lt;/math&gt;, it's tempting to try these. After a few cases, we see that &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; is convenient due to the repeated occurrence of &lt;math&gt;4x&lt;/math&gt; when squared and added. We rewrite the given expressions as: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.&lt;/cmath&gt; After repeatedly factoring the initial equation,we can get: &lt;cmath&gt;\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).&lt;/cmath&gt; Expanding each of the squares, we can compute each product independently then sum them: &lt;cmath&gt;\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},&lt;/cmath&gt; &lt;cmath&gt;\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.&lt;/cmath&gt; Now we place the values back into the expression: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.&lt;/cmath&gt; Plugging any positive integer value for &lt;math&gt;x&lt;/math&gt; into &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs &lt;math&gt;\left(a,b\right)&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> -fatant<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;a = 2x + 1&lt;/math&gt; and &lt;math&gt;b = 2x-1&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;4&lt;/math&gt;.We seek to show that &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt; because that will show that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> Claim 1: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4&lt;/math&gt;.<br /> We have that the remainder when &lt;math&gt;2x+1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt; and the remainder when &lt;math&gt;2x-1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always &lt;math&gt;1&lt;/math&gt;. Therefore, the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 2: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x&lt;/math&gt;<br /> We know that &lt;math&gt;(2x+1) \mod x \equiv 1&lt;/math&gt; and &lt;math&gt;(2x-1) \mod x \equiv 3&lt;/math&gt;, so the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 3: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt;<br /> Trivial given claim &lt;math&gt;1,2&lt;/math&gt;. &lt;math&gt;\boxed{}&lt;/math&gt;<br /> <br /> ~AopsUser101<br /> <br /> ==Solution 5==<br /> I claim &lt;math&gt;(a,b) = (2n-1,2n+1)&lt;/math&gt;, &lt;math&gt;n (\in \mathbb{N}) \geq 2&lt;/math&gt; always satisfies above conditions.<br /> <br /> Note: We could have also substituted 2n with 2^n or 4n, 8n, ... any sequence of numbers such that they are all even. The proof will work the same.<br /> <br /> Proof:<br /> <br /> Since there are infinitely many integers larger than or equal to 2, there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt;. <br /> <br /> We only need to prove:<br /> <br /> &lt;math&gt;a^b+b^a \equiv 0 \pmod{a+b}&lt;/math&gt;<br /> <br /> We can expand &lt;math&gt;a^b + b^a = (2n-1)^{2n+1} + (2n+1)^{2n-1}&lt;/math&gt; using binomial theorem. However, since &lt;math&gt;a + b = 2n-1 + 2n+1 = 4n&lt;/math&gt;, all the &lt;math&gt;2n&lt;/math&gt; terms (with more than 2 powers of) when evaluated modulo &lt;math&gt;4n&lt;/math&gt; equal to 0, and thus can be omitted. We are left with the terms: &lt;math&gt;(2n+1)(2n)^1-1+(2n-1)(2n)^1+1 = 4n \cdot 2n&lt;/math&gt;, which is divisible by &lt;math&gt;4n&lt;/math&gt;.<br /> <br /> &lt;math&gt;(2n-1)^{2n+1} + (2n+1)^{2n-1} \equiv (2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n \equiv 0 \pmod{4n}&lt;/math&gt;<br /> <br /> The proof is complete. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -AlexLikeMath<br /> <br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=124795 2017 USAJMO Problems/Problem 1 2020-06-10T04:25:59Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;a = 2n-1&lt;/math&gt; and &lt;math&gt;b = 2n+1&lt;/math&gt;. We see that &lt;math&gt;(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}&lt;/math&gt;. Therefore, we have &lt;math&gt;(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}&lt;/math&gt;, as desired. <br /> <br /> (Credits to mathmaster2012)<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;x&lt;/math&gt; be odd where &lt;math&gt;x&gt;1&lt;/math&gt;. We have &lt;math&gt;x^2-1=(x-1)(x+1),&lt;/math&gt; so &lt;math&gt;x^2-1 \equiv 0 \pmod{2x+2}.&lt;/math&gt; This means that &lt;math&gt;x^{x+2}-x^x \equiv 0 \pmod{2x+2},&lt;/math&gt; and since x is odd, &lt;math&gt;x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; or &lt;math&gt;x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; as desired.<br /> <br /> ==Solution 3==<br /> Because problems such as this usually are related to expressions along the lines of &lt;math&gt;x\pm1&lt;/math&gt;, it's tempting to try these. After a few cases, we see that &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; is convenient due to the repeated occurrence of &lt;math&gt;4x&lt;/math&gt; when squared and added. We rewrite the given expressions as: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.&lt;/cmath&gt; After repeatedly factoring the initial equation,we can get: &lt;cmath&gt;\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).&lt;/cmath&gt; Expanding each of the squares, we can compute each product independently then sum them: &lt;cmath&gt;\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},&lt;/cmath&gt; &lt;cmath&gt;\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.&lt;/cmath&gt; Now we place the values back into the expression: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.&lt;/cmath&gt; Plugging any positive integer value for &lt;math&gt;x&lt;/math&gt; into &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs &lt;math&gt;\left(a,b\right)&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> -fatant<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;a = 2x + 1&lt;/math&gt; and &lt;math&gt;b = 2x-1&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;4&lt;/math&gt;.We seek to show that &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt; because that will show that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> Claim 1: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4&lt;/math&gt;.<br /> We have that the remainder when &lt;math&gt;2x+1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt; and the remainder when &lt;math&gt;2x-1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always &lt;math&gt;1&lt;/math&gt;. Therefore, the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 2: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x&lt;/math&gt;<br /> We know that &lt;math&gt;(2x+1) \mod x \equiv 1&lt;/math&gt; and &lt;math&gt;(2x-1) \mod x \equiv 3&lt;/math&gt;, so the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 3: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt;<br /> Trivial given claim &lt;math&gt;1,2&lt;/math&gt;. &lt;math&gt;\boxed{}&lt;/math&gt;<br /> <br /> ~AopsUser101<br /> <br /> ==Solution 5==<br /> I claim &lt;math&gt;(a,b) = (2n-1,2n+1)&lt;/math&gt;, &lt;math&gt;n (\in \mathbb{N}) \geq 2&lt;/math&gt; always satisfies above conditions.<br /> <br /> Proof:<br /> <br /> Since there are infinitely many integers larger than or equal to 2, there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt;. <br /> <br /> We only need to prove:<br /> <br /> &lt;math&gt;a^b+b^a \equiv 0 \pmod{a+b}&lt;/math&gt;<br /> <br /> We can expand &lt;math&gt;a^b + b^a = (2n-1)^{2n+1} + (2n+1)^{2n-1}&lt;/math&gt; using binomial theorem. However, since &lt;math&gt;a + b = 2n-1 + 2n+1 = 4n&lt;/math&gt;, all the &lt;math&gt;2n&lt;/math&gt; terms (with more than 2 powers of) when evaluated modulo &lt;math&gt;4n&lt;/math&gt; equal to 0, and thus can be omitted. We are left with the terms: &lt;math&gt;(2n+1)(2n)^1-1+(2n-1)(2n)^1+1 = 4n \cdot 2n&lt;/math&gt;, which is divisible by &lt;math&gt;4n&lt;/math&gt;.<br /> <br /> &lt;math&gt;(2n-1)^{2n+1} + (2n+1)^{2n-1} \equiv (2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n \equiv 0 \pmod{4n}&lt;/math&gt;<br /> <br /> The proof is complete. &lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -AlexLikeMath<br /> <br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=124739 2017 USAJMO Problems/Problem 1 2020-06-10T03:10:28Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;a = 2n-1&lt;/math&gt; and &lt;math&gt;b = 2n+1&lt;/math&gt;. We see that &lt;math&gt;(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}&lt;/math&gt;. Therefore, we have &lt;math&gt;(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}&lt;/math&gt;, as desired. <br /> <br /> (Credits to mathmaster2012)<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;x&lt;/math&gt; be odd where &lt;math&gt;x&gt;1&lt;/math&gt;. We have &lt;math&gt;x^2-1=(x-1)(x+1),&lt;/math&gt; so &lt;math&gt;x^2-1 \equiv 0 \pmod{2x+2}.&lt;/math&gt; This means that &lt;math&gt;x^{x+2}-x^x \equiv 0 \pmod{2x+2},&lt;/math&gt; and since x is odd, &lt;math&gt;x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; or &lt;math&gt;x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; as desired.<br /> <br /> ==Solution 3==<br /> Because problems such as this usually are related to expressions along the lines of &lt;math&gt;x\pm1&lt;/math&gt;, it's tempting to try these. After a few cases, we see that &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; is convenient due to the repeated occurrence of &lt;math&gt;4x&lt;/math&gt; when squared and added. We rewrite the given expressions as: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.&lt;/cmath&gt; After repeatedly factoring the initial equation,we can get: &lt;cmath&gt;\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).&lt;/cmath&gt; Expanding each of the squares, we can compute each product independently then sum them: &lt;cmath&gt;\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},&lt;/cmath&gt; &lt;cmath&gt;\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.&lt;/cmath&gt; Now we place the values back into the expression: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.&lt;/cmath&gt; Plugging any positive integer value for &lt;math&gt;x&lt;/math&gt; into &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs &lt;math&gt;\left(a,b\right)&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> -fatant<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;a = 2x + 1&lt;/math&gt; and &lt;math&gt;b = 2x-1&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;4&lt;/math&gt;.We seek to show that &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt; because that will show that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> Claim 1: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4&lt;/math&gt;.<br /> We have that the remainder when &lt;math&gt;2x+1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt; and the remainder when &lt;math&gt;2x-1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always &lt;math&gt;1&lt;/math&gt;. Therefore, the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 2: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x&lt;/math&gt;<br /> We know that &lt;math&gt;(2x+1) \mod x \equiv 1&lt;/math&gt; and &lt;math&gt;(2x-1) \mod x \equiv 3&lt;/math&gt;, so the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 3: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt;<br /> Trivial given claim &lt;math&gt;1,2&lt;/math&gt;. &lt;math&gt;\boxed{}&lt;/math&gt;<br /> <br /> ~AopsUser101<br /> <br /> ==Solution 5==<br /> I claim &lt;math&gt;(a,b) = (2n-1,2n+1)&lt;/math&gt; always satisfies the conditions, n is integer, &lt;math&gt;n \geq 2&lt;/math&gt;, <br /> <br /> Proof:<br /> <br /> We can expand &lt;math&gt;(2n-1)^{2n+1} + (2n+1)^{2n-1}&lt;/math&gt; using binomial theorem. However, since &lt;math&gt;2n-1 + 2n+1 = 4n&lt;/math&gt;, all the terms with power &lt;math&gt;2n&lt;/math&gt; larger than &lt;math&gt;2&lt;/math&gt; modulo &lt;math&gt;4n&lt;/math&gt; evaluate to 0, and thus can be omitted. We are left with the terms: &lt;math&gt;(2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n&lt;/math&gt;, which is divisible by &lt;math&gt;4n&lt;/math&gt;.<br /> <br /> Since there are infinitely many integers larger than or equal to 2, the proof is complete .&lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> -AlexLikeMath<br /> <br /> <br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=124738 2017 USAJMO Problems/Problem 1 2020-06-10T03:09:58Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;a = 2n-1&lt;/math&gt; and &lt;math&gt;b = 2n+1&lt;/math&gt;. We see that &lt;math&gt;(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}&lt;/math&gt;. Therefore, we have &lt;math&gt;(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}&lt;/math&gt;, as desired. <br /> <br /> (Credits to mathmaster2012)<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;x&lt;/math&gt; be odd where &lt;math&gt;x&gt;1&lt;/math&gt;. We have &lt;math&gt;x^2-1=(x-1)(x+1),&lt;/math&gt; so &lt;math&gt;x^2-1 \equiv 0 \pmod{2x+2}.&lt;/math&gt; This means that &lt;math&gt;x^{x+2}-x^x \equiv 0 \pmod{2x+2},&lt;/math&gt; and since x is odd, &lt;math&gt;x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; or &lt;math&gt;x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; as desired.<br /> <br /> ==Solution 3==<br /> Because problems such as this usually are related to expressions along the lines of &lt;math&gt;x\pm1&lt;/math&gt;, it's tempting to try these. After a few cases, we see that &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; is convenient due to the repeated occurrence of &lt;math&gt;4x&lt;/math&gt; when squared and added. We rewrite the given expressions as: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.&lt;/cmath&gt; After repeatedly factoring the initial equation,we can get: &lt;cmath&gt;\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).&lt;/cmath&gt; Expanding each of the squares, we can compute each product independently then sum them: &lt;cmath&gt;\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},&lt;/cmath&gt; &lt;cmath&gt;\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.&lt;/cmath&gt; Now we place the values back into the expression: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.&lt;/cmath&gt; Plugging any positive integer value for &lt;math&gt;x&lt;/math&gt; into &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs &lt;math&gt;\left(a,b\right)&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> -fatant<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;a = 2x + 1&lt;/math&gt; and &lt;math&gt;b = 2x-1&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;4&lt;/math&gt;.We seek to show that &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt; because that will show that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> Claim 1: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4&lt;/math&gt;.<br /> We have that the remainder when &lt;math&gt;2x+1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt; and the remainder when &lt;math&gt;2x-1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always &lt;math&gt;1&lt;/math&gt;. Therefore, the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 2: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x&lt;/math&gt;<br /> We know that &lt;math&gt;(2x+1) \mod x \equiv 1&lt;/math&gt; and &lt;math&gt;(2x-1) \mod x \equiv 3&lt;/math&gt;, so the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 3: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt;<br /> Trivial given claim &lt;math&gt;1,2&lt;/math&gt;. &lt;math&gt;\boxed{}&lt;/math&gt;<br /> <br /> ~AopsUser101<br /> <br /> ==Solution 5==<br /> I claim &lt;math&gt;(a,b) = (2n-1,2n+1)&lt;/math&gt; always satisfies the conditions, n is integer, &lt;math&gt;n \geq 2&lt;/math&gt;, <br /> <br /> Proof:<br /> <br /> We can expand &lt;math&gt;(2n-1)^{2n+1} + (2n+1)^{2n-1}&lt;/math&gt; using binomial theorem. However, since &lt;math&gt;2n-1 + 2n+1 = 4n&lt;/math&gt;, all the terms with power &lt;math&gt;2n&lt;/math&gt; larger than &lt;math&gt;2&lt;/math&gt; modulo &lt;math&gt;4n&lt;/math&gt; evaluate to 0, and thus can be omitted. We are left with the terms: &lt;math&gt;(2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n&lt;/math&gt;, which is divisible by &lt;math&gt;4n&lt;/math&gt;.<br /> <br /> Since there are infinitely many integers larger than or equal to 2, the proof is complete .&lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> <br /> -AlexLikeMath<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=124736 2017 USAJMO Problems/Problem 1 2020-06-10T03:04:34Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;a = 2n-1&lt;/math&gt; and &lt;math&gt;b = 2n+1&lt;/math&gt;. We see that &lt;math&gt;(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}&lt;/math&gt;. Therefore, we have &lt;math&gt;(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}&lt;/math&gt;, as desired. <br /> <br /> (Credits to mathmaster2012)<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;x&lt;/math&gt; be odd where &lt;math&gt;x&gt;1&lt;/math&gt;. We have &lt;math&gt;x^2-1=(x-1)(x+1),&lt;/math&gt; so &lt;math&gt;x^2-1 \equiv 0 \pmod{2x+2}.&lt;/math&gt; This means that &lt;math&gt;x^{x+2}-x^x \equiv 0 \pmod{2x+2},&lt;/math&gt; and since x is odd, &lt;math&gt;x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; or &lt;math&gt;x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; as desired.<br /> <br /> ==Solution 3==<br /> Because problems such as this usually are related to expressions along the lines of &lt;math&gt;x\pm1&lt;/math&gt;, it's tempting to try these. After a few cases, we see that &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; is convenient due to the repeated occurrence of &lt;math&gt;4x&lt;/math&gt; when squared and added. We rewrite the given expressions as: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.&lt;/cmath&gt; After repeatedly factoring the initial equation,we can get: &lt;cmath&gt;\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).&lt;/cmath&gt; Expanding each of the squares, we can compute each product independently then sum them: &lt;cmath&gt;\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},&lt;/cmath&gt; &lt;cmath&gt;\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.&lt;/cmath&gt; Now we place the values back into the expression: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.&lt;/cmath&gt; Plugging any positive integer value for &lt;math&gt;x&lt;/math&gt; into &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs &lt;math&gt;\left(a,b\right)&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> -fatant<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;a = 2x + 1&lt;/math&gt; and &lt;math&gt;b = 2x-1&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;4&lt;/math&gt;.We seek to show that &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt; because that will show that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> Claim 1: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4&lt;/math&gt;.<br /> We have that the remainder when &lt;math&gt;2x+1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt; and the remainder when &lt;math&gt;2x-1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always &lt;math&gt;1&lt;/math&gt;. Therefore, the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 2: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x&lt;/math&gt;<br /> We know that &lt;math&gt;(2x+1) \mod x \equiv 1&lt;/math&gt; and &lt;math&gt;(2x-1) \mod x \equiv 3&lt;/math&gt;, so the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 3: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt;<br /> Trivial given claim &lt;math&gt;1,2&lt;/math&gt;. &lt;math&gt;\boxed{}&lt;/math&gt;<br /> <br /> ~AopsUser101<br /> <br /> ==Solution 5==<br /> I claim (a,b) = (2n-1,2n+1) always satisfies the conditions, n is integer, &lt;math&gt;n \geq 2&lt;/math&gt;, <br /> <br /> Proof:<br /> We can expand &lt;math&gt;(2n-1)^{2n+1} + (2n+1)^{2n-1}&lt;/math&gt; using binomial theorem. However, since 2n-1 + 2n+1 = 4n, all the terms with power 2n larger than 2 modulo 4n evaluate to 0, and thus can be omitted. We are left with the terms: &lt;math&gt;(2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n&lt;/math&gt;, which is divisible by 4n.<br /> <br /> Since there are infinitely many integers larger than or equal to 2, the proof is complete .&lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> <br /> -AlexLikeMath<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=124735 2017 USAJMO Problems/Problem 1 2020-06-10T03:04:03Z <p>Alexlikemath: New solution</p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;a = 2n-1&lt;/math&gt; and &lt;math&gt;b = 2n+1&lt;/math&gt;. We see that &lt;math&gt;(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}&lt;/math&gt;. Therefore, we have &lt;math&gt;(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}&lt;/math&gt;, as desired. <br /> <br /> (Credits to mathmaster2012)<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;x&lt;/math&gt; be odd where &lt;math&gt;x&gt;1&lt;/math&gt;. We have &lt;math&gt;x^2-1=(x-1)(x+1),&lt;/math&gt; so &lt;math&gt;x^2-1 \equiv 0 \pmod{2x+2}.&lt;/math&gt; This means that &lt;math&gt;x^{x+2}-x^x \equiv 0 \pmod{2x+2},&lt;/math&gt; and since x is odd, &lt;math&gt;x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; or &lt;math&gt;x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; as desired.<br /> <br /> ==Solution 3==<br /> Because problems such as this usually are related to expressions along the lines of &lt;math&gt;x\pm1&lt;/math&gt;, it's tempting to try these. After a few cases, we see that &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; is convenient due to the repeated occurrence of &lt;math&gt;4x&lt;/math&gt; when squared and added. We rewrite the given expressions as: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.&lt;/cmath&gt; After repeatedly factoring the initial equation,we can get: &lt;cmath&gt;\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).&lt;/cmath&gt; Expanding each of the squares, we can compute each product independently then sum them: &lt;cmath&gt;\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},&lt;/cmath&gt; &lt;cmath&gt;\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.&lt;/cmath&gt; Now we place the values back into the expression: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.&lt;/cmath&gt; Plugging any positive integer value for &lt;math&gt;x&lt;/math&gt; into &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs &lt;math&gt;\left(a,b\right)&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> -fatant<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;a = 2x + 1&lt;/math&gt; and &lt;math&gt;b = 2x-1&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;4&lt;/math&gt;.We seek to show that &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt; because that will show that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> Claim 1: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4&lt;/math&gt;.<br /> We have that the remainder when &lt;math&gt;2x+1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt; and the remainder when &lt;math&gt;2x-1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always &lt;math&gt;1&lt;/math&gt;. Therefore, the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 2: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x&lt;/math&gt;<br /> We know that &lt;math&gt;(2x+1) \mod x \equiv 1&lt;/math&gt; and &lt;math&gt;(2x-1) \mod x \equiv 3&lt;/math&gt;, so the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 3: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt;<br /> Trivial given claim &lt;math&gt;1,2&lt;/math&gt;. &lt;math&gt;\boxed{}&lt;/math&gt;<br /> <br /> ~AopsUser101<br /> <br /> ==Solution 5==<br /> I claim (a,b) = (2n-1,2n+1) always satisfies the conditions, n is integer, &lt;math&gt;n \geq 2&lt;/math&gt;, <br /> <br /> Proof:<br /> We can expand &lt;math&gt;(2n-1)^{2n+1} + (2n+1)^{2n-1}&lt;/math&gt; using binomial theorem. However, since 2n-1 + 2n+1 = 4n, all the terms with power 2n larger than 2 modulo 4n evaluate to 0, and thus can be omitted. We are left with the terms: &lt;math&gt;(2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n&lt;/math&gt;, which is divisible by 4n.<br /> <br /> Since there are infinitely many integers larger than 2n, the proof is complete .&lt;math&gt;\blacksquare&lt;/math&gt;<br /> <br /> <br /> -AlexLikeMath<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_USAJMO_Problems/Problem_1&diff=124734 2017 USAJMO Problems/Problem 1 2020-06-10T02:59:49Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> <br /> Prove that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Let &lt;math&gt;a = 2n-1&lt;/math&gt; and &lt;math&gt;b = 2n+1&lt;/math&gt;. We see that &lt;math&gt;(2n \pm 1)^2 = 4n^2\pm4n+1 \equiv 1 \pmod{4n}&lt;/math&gt;. Therefore, we have &lt;math&gt;(2n+1)^{2n-1} + (2n-1)^{2n+1} \equiv 2n + 1 + 2n - 1 = 4n \equiv 0 \pmod{4n}&lt;/math&gt;, as desired. <br /> <br /> (Credits to mathmaster2012)<br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;x&lt;/math&gt; be odd where &lt;math&gt;x&gt;1&lt;/math&gt;. We have &lt;math&gt;x^2-1=(x-1)(x+1),&lt;/math&gt; so &lt;math&gt;x^2-1 \equiv 0 \pmod{2x+2}.&lt;/math&gt; This means that &lt;math&gt;x^{x+2}-x^x \equiv 0 \pmod{2x+2},&lt;/math&gt; and since x is odd, &lt;math&gt;x^{x+2}+(-x)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; or &lt;math&gt;x^{x+2}+(x+2)^x \equiv 0 \pmod{2x+2},&lt;/math&gt; as desired.<br /> <br /> ==Solution 3==<br /> Because problems such as this usually are related to expressions along the lines of &lt;math&gt;x\pm1&lt;/math&gt;, it's tempting to try these. After a few cases, we see that &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; is convenient due to the repeated occurrence of &lt;math&gt;4x&lt;/math&gt; when squared and added. We rewrite the given expressions as: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}, \left(2x-1\right)+\left(2x+1\right)=4x.&lt;/cmath&gt; After repeatedly factoring the initial equation,we can get: &lt;cmath&gt;\left(2x-1\right)^{2}\left(2x-1\right)^{2}...\left(2x-1\right)+\left(2x+1\right)^{2}\left(2x+1\right)^{2}\left(2x+1\right)^{2}...\left(2x+1\right).&lt;/cmath&gt; Expanding each of the squares, we can compute each product independently then sum them: &lt;cmath&gt;\left(4x^{2}-4x+1\right)\left(4x^{2}-4x+1\right)...\left(2x-1\right)\equiv\left(1\right)\left(1\right)...\left(2x-1\right)\equiv2x-1\mod{4x},&lt;/cmath&gt; &lt;cmath&gt;\left(4x^{2}+4x+1\right)\left(4x^{2}+4x+1\right)...\left(2x+1\right)\equiv\left(1\right)\left(1\right)...\left(2x+1\right)\equiv2x+1\mod{4x}.&lt;/cmath&gt; Now we place the values back into the expression: &lt;cmath&gt;\left(2x-1\right)^{2x+1}+\left(2x+1\right)^{2x-1}\equiv\left(2x-1\right)+\left(2x+1\right)\equiv0\mod{4x}.&lt;/cmath&gt; Plugging any positive integer value for &lt;math&gt;x&lt;/math&gt; into &lt;math&gt;\left(a,b\right)=\left(2x-1,2x+1\right)&lt;/math&gt; yields a valid solution, because there is an infinite number of positive integers, there is an infinite number of distinct pairs &lt;math&gt;\left(a,b\right)&lt;/math&gt;. &lt;math&gt;\square&lt;/math&gt;<br /> <br /> -fatant<br /> <br /> ==Solution 4==<br /> Let &lt;math&gt;a = 2x + 1&lt;/math&gt; and &lt;math&gt;b = 2x-1&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; leaves a remainder of &lt;math&gt;1&lt;/math&gt; when divided by &lt;math&gt;4&lt;/math&gt;.We seek to show that &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt; because that will show that there are infinitely many distinct pairs &lt;math&gt;(a,b)&lt;/math&gt; of relatively prime integers &lt;math&gt;a&gt;1&lt;/math&gt; and &lt;math&gt;b&gt;1&lt;/math&gt; such that &lt;math&gt;a^b+b^a&lt;/math&gt; is divisible by &lt;math&gt;a+b&lt;/math&gt;.<br /> <br /> Claim 1: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4&lt;/math&gt;.<br /> We have that the remainder when &lt;math&gt;2x+1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;3&lt;/math&gt; and the remainder when &lt;math&gt;2x-1&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always &lt;math&gt;1&lt;/math&gt;. Therefore, the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 2: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod x&lt;/math&gt;<br /> We know that &lt;math&gt;(2x+1) \mod x \equiv 1&lt;/math&gt; and &lt;math&gt;(2x-1) \mod x \equiv 3&lt;/math&gt;, so the remainder when &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1}&lt;/math&gt; is divided by &lt;math&gt;4&lt;/math&gt; is always going to be &lt;math&gt;(-1)^{2x-1} + 1^{2x+1} = 0&lt;/math&gt;.<br /> <br /> Claim 3: &lt;math&gt;(2x+1)^{2x-1} + (2x-1)^{2x+1} \equiv 0 \mod 4x&lt;/math&gt;<br /> Trivial given claim &lt;math&gt;1,2&lt;/math&gt;. &lt;math&gt;\boxed{}&lt;/math&gt;<br /> <br /> ~AopsUser101<br /> <br /> ==Solution 5==<br /> I claim (a,b) = (2n-1,2n+1) always satisfies the conditions, n is integer, &lt;math&gt;n \geq 2&lt;/math&gt;, <br /> <br /> Proof:<br /> We can expand &lt;math&gt;(2n-1)^{2n+1} + (2n+1)^{2n-1}&lt;/math&gt; using binomial theorem. However, since 2n-1 + 2n+1 = 4n, all the terms with power 2n larger than 2 cancel out, and we are left with the terms: &lt;math&gt;(2n+1)(2n)-1+(2n-1)(2n)+1 = 4n \cdot 2n&lt;/math&gt;, which is divisible by 4n.<br /> <br /> Since there are infinitely many integers larger than 2n, the proof is complete<br /> <br /> <br /> -AlexLikeMath<br /> <br /> {{MAA Notice}}<br /> <br /> ==See also==<br /> {{USAJMO newbox|year=2017|beforetext=|before=First Problem|num-a=2}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_1&diff=124313 2010 USAJMO Problems/Problem 1 2020-06-07T23:26:42Z <p>Alexlikemath: Solution 5</p> <hr /> <div>== Problem ==<br /> A permutation of the set of positive integers &lt;math&gt;[n] = \{1, 2, \ldots, n\}&lt;/math&gt; is a sequence &lt;math&gt;(a_1, a_2, \ldots, a_n)&lt;/math&gt; such that each element of &lt;math&gt;[n]&lt;/math&gt; appears precisely one time as a term of the sequence. For example, &lt;math&gt;(3, 5, 1, 2, 4)&lt;/math&gt; is a permutation of &lt;math&gt;&lt;/math&gt;. Let &lt;math&gt;P(n)&lt;/math&gt; be the number of permutations of &lt;math&gt;[n]&lt;/math&gt; for which &lt;math&gt;ka_k&lt;/math&gt; is a perfect square for all &lt;math&gt;1\leq k\leq n&lt;/math&gt;. Find with proof the smallest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;P(n)&lt;/math&gt; is a multiple of &lt;math&gt;2010&lt;/math&gt;.<br /> <br /> == Solutions ==<br /> We claim that the smallest &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;67^2 = \boxed{4489}&lt;/math&gt;.<br /> <br /> === Solution 1 ===<br /> Let &lt;math&gt;S = \{1, 4, 9, \ldots\}&lt;/math&gt; be the set of positive perfect squares. We claim that the relation &lt;math&gt;R = \{(j, k)\in [n]\times[n]\mid jk\in S\}&lt;/math&gt; is an equivalence relation on &lt;math&gt;[n]&lt;/math&gt;.<br /> * It is reflexive because &lt;math&gt;k^2\in S&lt;/math&gt; for all &lt;math&gt;k\in [n]&lt;/math&gt;.<br /> * It is symmetric because &lt;math&gt;jk\in S\implies kj = jk\in S&lt;/math&gt;.<br /> * It is transitive because if &lt;math&gt;jk\in S&lt;/math&gt; and &lt;math&gt;kl\in S&lt;/math&gt;, then &lt;math&gt;jk\cdot kl = jlk^2\in S\implies jl\in S&lt;/math&gt;, since &lt;math&gt;S&lt;/math&gt; is closed under multiplication and a non-square times a square is always a non-square.<br /> <br /> We are restricted to permutations for which &lt;math&gt;ka_k \in S&lt;/math&gt;, in other words to permutations that send each element of &lt;math&gt;[n]&lt;/math&gt; into its equivalence class. Suppose there are &lt;math&gt;N&lt;/math&gt; equivalence classes: &lt;math&gt;C_1, \ldots, C_N&lt;/math&gt;. Let &lt;math&gt;n_i&lt;/math&gt; be the number of elements of &lt;math&gt;C_i&lt;/math&gt;, then &lt;math&gt;P(n) = \prod_{i=1}^{N} n_i!&lt;/math&gt;.<br /> <br /> Now &lt;math&gt;2010 = 2\cdot 3\cdot 5\cdot 67&lt;/math&gt;. In order that &lt;math&gt;2010\mid P(n)&lt;/math&gt;, we must have &lt;math&gt;67\mid n_m!&lt;/math&gt; for the class &lt;math&gt;C_m&lt;/math&gt; with the most elements. This means &lt;math&gt;n_m\geq 67&lt;/math&gt;, since no smaller factorial will have &lt;math&gt;67&lt;/math&gt; as a factor. This condition is sufficient, since &lt;math&gt;n_m!&lt;/math&gt; will be divisible by &lt;math&gt;30&lt;/math&gt; for &lt;math&gt;n_m\geq 5&lt;/math&gt;, and even more so &lt;math&gt;n_m\geq 67&lt;/math&gt;.<br /> <br /> The smallest element &lt;math&gt;g_m&lt;/math&gt; of the equivalence class &lt;math&gt;C_m&lt;/math&gt; is square-free, since if it were divisible by the square of a prime, the quotient would be a smaller element of &lt;math&gt;C_m&lt;/math&gt;. Also, each prime &lt;math&gt;p&lt;/math&gt; that divides &lt;math&gt;g_m&lt;/math&gt; divides all the other elements &lt;math&gt;k&lt;/math&gt; of &lt;math&gt;C_m&lt;/math&gt;, since &lt;math&gt;p^2\mid kg_m&lt;/math&gt; and thus &lt;math&gt;p\mid k&lt;/math&gt;. Therefore &lt;math&gt;g_m\mid k&lt;/math&gt; for all &lt;math&gt;k\in C_m&lt;/math&gt;. The primes that are not in &lt;math&gt;g_m&lt;/math&gt; occur an even number of times in each &lt;math&gt;k\in C_m&lt;/math&gt;.<br /> <br /> Thus the equivalence class &lt;math&gt;C_m = \{g_mk^2\leq n\}&lt;/math&gt;. With &lt;math&gt;g_m = 1&lt;/math&gt;, we get the largest possible &lt;math&gt;n_m&lt;/math&gt;. This is just the set of squares in &lt;math&gt;[n]&lt;/math&gt;, of which we need at least &lt;math&gt;67&lt;/math&gt;, so &lt;math&gt;n\geq 67^2&lt;/math&gt;. This condition is necessary and sufficient.<br /> <br /> === Solution 2 ===<br /> This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as &quot;equivalence relation&quot;:<br /> <br /> It is possible to write all positive integers &lt;math&gt;n&lt;/math&gt; in the form &lt;math&gt;p\cdot m^2&lt;/math&gt;, where &lt;math&gt;m^2&lt;/math&gt; is the largest perfect square dividing &lt;math&gt;n&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Obviously, one working permutation of &lt;math&gt;[n]&lt;/math&gt; is simply &lt;math&gt;(1, 2, \ldots, n)&lt;/math&gt;; this is acceptable, as &lt;math&gt;ka_k&lt;/math&gt; is always &lt;math&gt;k^2&lt;/math&gt; in this sequence.<br /> <br /> '''Lemma 1.''' We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities &lt;math&gt;p&lt;/math&gt;.<br /> <br /> ''Proof.'' Let &lt;math&gt;p_k&lt;/math&gt; and &lt;math&gt;m_k&lt;/math&gt; be the values of &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt;, respectively, for a given &lt;math&gt;k&lt;/math&gt; as defined above, such that &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. We can obviously permute two numbers which have the same &lt;math&gt;p&lt;/math&gt;, since if &lt;math&gt;p_j = p_w&lt;/math&gt; where &lt;math&gt;j&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; are 2 values of &lt;math&gt;k&lt;/math&gt;, then &lt;math&gt;j\cdot w = p_j^2\cdot m_j^2\cdot m_w^2&lt;/math&gt;, which is a perfect square. This proves that we can permute any numbers with the same value of &lt;math&gt;p&lt;/math&gt;.<br /> <br /> '''End Lemma'''<br /> <br /> '''Lemma 2.''' We will prove the converse of Lemma 1: Let one number have a &lt;math&gt;p&lt;/math&gt; value of &lt;math&gt;\phi&lt;/math&gt; and another, &lt;math&gt;\gamma&lt;/math&gt;. &lt;math&gt;\phi\cdot f&lt;/math&gt; and &lt;math&gt;\gamma\cdot g&lt;/math&gt; are both perfect squares.<br /> <br /> ''Proof.'' &lt;math&gt;\phi\cdot f&lt;/math&gt; and &lt;math&gt;\gamma\cdot g&lt;/math&gt; are both perfect squares, so for &lt;math&gt;\phi\cdot \gamma&lt;/math&gt; to be a perfect square, if &lt;math&gt;g&lt;/math&gt; is greater than or equal to &lt;math&gt;f&lt;/math&gt;, &lt;math&gt;g/f&lt;/math&gt; must be a perfect square, too. Thus &lt;math&gt;g&lt;/math&gt; is &lt;math&gt;f&lt;/math&gt; times a square, but &lt;math&gt;g&lt;/math&gt; cannot divide any squares besides &lt;math&gt;1&lt;/math&gt;, so &lt;math&gt;g = 1f&lt;/math&gt;; &lt;math&gt;g = f&lt;/math&gt;. Similarly, if &lt;math&gt;f\geq g&lt;/math&gt;, then &lt;math&gt;f = g&lt;/math&gt; for our rules to keep working.<br /> <br /> '''End Lemma'''<br /> <br /> We can permute &lt;math&gt;l&lt;/math&gt; numbers with the same &lt;math&gt;p&lt;/math&gt; in &lt;math&gt;l!&lt;/math&gt; ways. We must have at least 67 numbers with a certain &lt;math&gt;p&lt;/math&gt; so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as &lt;math&gt;h&lt;/math&gt;, in general, we need numbers all the way up to &lt;math&gt;h\cdot 67^2&lt;/math&gt;, so obviously, &lt;math&gt;67^2&lt;/math&gt; is the smallest such number such that we can get a &lt;math&gt;67!&lt;/math&gt; term; here 67 &lt;math&gt;p&lt;/math&gt; terms are 1. Thus we need the integers &lt;math&gt;1, 2, \ldots, 67^2&lt;/math&gt;, so &lt;math&gt;67^2&lt;/math&gt;, or &lt;math&gt;\boxed{4489}&lt;/math&gt;, is the answer.<br /> <br /> ==Solution Number Sense==<br /> We have to somehow calculate the number of permutations for a given &lt;math&gt;n&lt;/math&gt;. How in the world do we do this? Because we want squares, why not call a number &lt;math&gt;k=m*s^2&lt;/math&gt;, where &lt;math&gt;s&lt;/math&gt; is the largest square that allows &lt;math&gt;m&lt;/math&gt; to be non-square? &lt;math&gt;m=1&lt;/math&gt; is the only square &lt;math&gt;m&lt;/math&gt; can be, which only happens if &lt;math&gt;k&lt;/math&gt; is a perfect square.<br /> <br /> For example, &lt;math&gt;126 = 14 * 3^2&lt;/math&gt;, therefore in this case &lt;math&gt;k=126, m = 14, s = 3&lt;/math&gt;.<br /> <br /> I will call a permutation of the numbers &lt;math&gt;P&lt;/math&gt;, while the original &lt;math&gt;1, 2, 3, 4, ...&lt;/math&gt; I will call &lt;math&gt;S&lt;/math&gt;.<br /> <br /> Note that essentially we are looking at &quot;pairing up&quot; elements between &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;S&lt;/math&gt; such that the product of &lt;math&gt;P_k&lt;/math&gt; and &lt;math&gt;S_k&lt;/math&gt; is a perfect square. How do we do this? Using the representation above.<br /> <br /> Each square has to have an even exponent of every prime represented in its prime factorization. Therefore, we can just take all exponents of the primes &lt;math&gt;mod 2&lt;/math&gt; and if there are any odd numbers, those are the ones we have to match- in effect, they are the &lt;math&gt;m&lt;/math&gt; numbers mentioned at the beginning.<br /> <br /> By listing the &lt;math&gt;m&lt;/math&gt; values, in my search for &quot;dumb&quot; or &quot;obvious&quot; ideas I am pretty confident that only values with identical &lt;math&gt;m&lt;/math&gt;s can be matched together. With such a solid idea let me prove it.<br /> <br /> If we were to &quot;pair up&quot; numbers with different &lt;math&gt;m&lt;/math&gt;s, take for example &lt;math&gt;S_{18}&lt;/math&gt; with an &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;P_{18}=26&lt;/math&gt; with an &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;26&lt;/math&gt;, note that their product gives a supposed &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;13&lt;/math&gt; because the &lt;math&gt;2&lt;/math&gt; values cancel out. But then, what happens to the extra &lt;math&gt;13&lt;/math&gt; left? It doesn't make a square, contradiction. To finish up this easy proof, note that if a &quot;pair&quot; has different &lt;math&gt;m&lt;/math&gt; values, and the smaller one is &lt;math&gt;m_1&lt;/math&gt;, in order for the product to leave a square, the larger &lt;math&gt;m&lt;/math&gt; value has to have not just &lt;math&gt;m_1&lt;/math&gt; but another square inside it, which is absurd because we stipulated at the beginning that &lt;math&gt;m&lt;/math&gt; was square-free except for the trivial multiplication identity, 1.<br /> <br /> Now, how many ways are there to do this? If there are &lt;math&gt;c_1&lt;/math&gt; numbers with &lt;math&gt;m=1&lt;/math&gt;, there are clearly &lt;math&gt;(c_1)!&lt;/math&gt; ways of sorting them. The same goes for &lt;math&gt;m=2, 3, etc.&lt;/math&gt; by this logic. Note that the &lt;math&gt;P(n)&lt;/math&gt; as stated by the problem requires a &lt;math&gt;67&lt;/math&gt; thrown in there because &lt;math&gt;2010=2*3*5*67&lt;/math&gt;, so there has to be a &lt;math&gt;S_n&lt;/math&gt; with 67 elements with the same &lt;math&gt;m&lt;/math&gt;. It is evident that the smallest &lt;math&gt;n&lt;/math&gt; will occur when &lt;math&gt;m=1&lt;/math&gt;, because if &lt;math&gt;m&lt;/math&gt; is bigger we would have to expand &lt;math&gt;n&lt;/math&gt; to get the same number of &lt;math&gt;m&lt;/math&gt; values. Finally, realize that the only numbers with &lt;math&gt;m=1&lt;/math&gt; are square numbers! So our smallest &lt;math&gt;n=67^2=4489&lt;/math&gt;, and we are done.<br /> <br /> I relied on looking for patterns a lot in this problem. When faced with combo/number theory, it is always good to draw a sketch. Never be scared to try a problem on the USAJMO. It takes about 45 minutes. Well, it's 2010 and a number 1. Cheers!<br /> <br /> -expiLnCalc<br /> <br /> ==Solution Easy==<br /> <br /> Consider the set of numbers &lt;math&gt;m*k^2&lt;/math&gt; such that &lt;math&gt;m&lt;/math&gt; is not divisible by any squares other than 1 and &lt;math&gt;k = 1, 2, ...&lt;/math&gt; By changing &lt;math&gt;m&lt;/math&gt; we can encompass all numbers less than or equal to &lt;math&gt;n&lt;/math&gt;. Now notice that for a working arrangement these numbers can be permutated in any way to create a new one; for instance, the numbers &lt;math&gt;2,8,18&lt;/math&gt; have &lt;math&gt;m = 2&lt;/math&gt; and they can be arranged in &lt;math&gt;3! = 6&lt;/math&gt; ways. Thus, since &lt;math&gt;2010 = 2*3*5*67&lt;/math&gt; we need a permutation to have at least 67 elements (since 67 is prime). To minimize &lt;math&gt;n&lt;/math&gt;, we let &lt;math&gt;m = 1&lt;/math&gt; and we have &lt;math&gt;1^2,2^2,3^2,...67^2&lt;/math&gt; and we stop at &lt;math&gt;67^2&lt;/math&gt; to get &lt;math&gt;\boxed{4489}&lt;/math&gt;. ~Leonard_my_dude~<br /> <br /> ==Solution 5==<br /> <br /> It's well known that there exists &lt;math&gt;f(n)&lt;/math&gt; and &lt;math&gt;g(n)&lt;/math&gt; such that &lt;math&gt;n = f(n) \cdot g(n)&lt;/math&gt;, no square divides &lt;math&gt;f(n)&lt;/math&gt; other than 1, and &lt;math&gt;g(n)&lt;/math&gt; is a perfect square.<br /> <br /> Lemma: &lt;math&gt;k \cdot a_k&lt;/math&gt; is a perfect square if and only if &lt;math&gt;f(k) = f(a_k)&lt;/math&gt; <br /> <br /> We prove first: If &lt;math&gt;f(k) = f(a_k)&lt;/math&gt;, &lt;math&gt;k \cdot a_k&lt;/math&gt; is a perfect square. <br /> <br /> &lt;math&gt;k \cdot a_k = f(k) \cdot g(k) \cdot f(a_k) \cdot g(a_k) = f(k)^2 \cdot g(k) \cdot g(a_k)&lt;/math&gt;, which is a perfect square. <br /> <br /> We will now prove: If &lt;math&gt;k \cdot a_k&lt;/math&gt; is a perfect square, &lt;math&gt;f(k) = f(a_k)&lt;/math&gt;.<br /> <br /> We do proof by contrapositive: If &lt;math&gt;f(k) \neq f(a_k)&lt;/math&gt;, &lt;math&gt;k \cdot a_k&lt;/math&gt; is not a perfect square. <br /> <br /> &lt;math&gt;v_p(k)&lt;/math&gt; is the p-adic valuation of k. (Basically how many factors of p you can take out of k)<br /> <br /> Note that if &lt;math&gt;f(k) \neq f(a_k)&lt;/math&gt;, By the Fundamental Theorem of Arithmetic, &lt;math&gt;f(k)&lt;/math&gt; and &lt;math&gt;f(a_k)&lt;/math&gt;'s prime factorization are different, and thus there exists a prime p, such that &lt;math&gt;v_p(f(k)) \neq v_p(f(a_k))&lt;/math&gt;. Also, since &lt;math&gt;f(k)&lt;/math&gt; and &lt;math&gt;f(a_k)&lt;/math&gt; is squarefree, &lt;math&gt;v_p(k), v_p(a_k) \leq 1&lt;/math&gt;. Thus, &lt;math&gt;v_p(k \cdot a_k) = 1&lt;/math&gt;, making &lt;math&gt;k \cdot a_k&lt;/math&gt; not a square.<br /> <br /> End Lemma<br /> <br /> Thus, we can only match k with &lt;math&gt;a_k&lt;/math&gt; if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the a_k with f value 1, then 2, ... Thus, our answer is:<br /> <br /> &lt;math&gt;P(n) = \prod_{1 \leq i \leq n, g(i) = 1} \left\lfloor \sqrt{\frac{n}{i}} \right \rfloor !&lt;/math&gt; <br /> <br /> For all &lt;math&gt;n &lt; 67^2&lt;/math&gt;, &lt;math&gt;P(n)&lt;/math&gt; doesn't have a factor of 67. However, if &lt;math&gt;n = 67^2&lt;/math&gt;, the first term will be a multiple of 2010, and thus the answer is &lt;math&gt;67^2 = \boxed{4489}&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> <br /> {{alternate solutions}}<br /> <br /> == See Also ==<br /> * &lt;url&gt;viewtopic.php?t=347303 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> {{USAJMO newbox|year=2010|before=First Question|num-a=2}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_1&diff=124275 2010 USAJMO Problems/Problem 1 2020-06-07T20:01:55Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> A permutation of the set of positive integers &lt;math&gt;[n] = \{1, 2, \ldots, n\}&lt;/math&gt; is a sequence &lt;math&gt;(a_1, a_2, \ldots, a_n)&lt;/math&gt; such that each element of &lt;math&gt;[n]&lt;/math&gt; appears precisely one time as a term of the sequence. For example, &lt;math&gt;(3, 5, 1, 2, 4)&lt;/math&gt; is a permutation of &lt;math&gt;&lt;/math&gt;. Let &lt;math&gt;P(n)&lt;/math&gt; be the number of permutations of &lt;math&gt;[n]&lt;/math&gt; for which &lt;math&gt;ka_k&lt;/math&gt; is a perfect square for all &lt;math&gt;1\leq k\leq n&lt;/math&gt;. Find with proof the smallest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;P(n)&lt;/math&gt; is a multiple of &lt;math&gt;2010&lt;/math&gt;.<br /> <br /> == Solutions ==<br /> We claim that the smallest &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;67^2 = \boxed{4489}&lt;/math&gt;.<br /> <br /> === Solution 1 ===<br /> Let &lt;math&gt;S = \{1, 4, 9, \ldots\}&lt;/math&gt; be the set of positive perfect squares. We claim that the relation &lt;math&gt;R = \{(j, k)\in [n]\times[n]\mid jk\in S\}&lt;/math&gt; is an equivalence relation on &lt;math&gt;[n]&lt;/math&gt;.<br /> * It is reflexive because &lt;math&gt;k^2\in S&lt;/math&gt; for all &lt;math&gt;k\in [n]&lt;/math&gt;.<br /> * It is symmetric because &lt;math&gt;jk\in S\implies kj = jk\in S&lt;/math&gt;.<br /> * It is transitive because if &lt;math&gt;jk\in S&lt;/math&gt; and &lt;math&gt;kl\in S&lt;/math&gt;, then &lt;math&gt;jk\cdot kl = jlk^2\in S\implies jl\in S&lt;/math&gt;, since &lt;math&gt;S&lt;/math&gt; is closed under multiplication and a non-square times a square is always a non-square.<br /> <br /> We are restricted to permutations for which &lt;math&gt;ka_k \in S&lt;/math&gt;, in other words to permutations that send each element of &lt;math&gt;[n]&lt;/math&gt; into its equivalence class. Suppose there are &lt;math&gt;N&lt;/math&gt; equivalence classes: &lt;math&gt;C_1, \ldots, C_N&lt;/math&gt;. Let &lt;math&gt;n_i&lt;/math&gt; be the number of elements of &lt;math&gt;C_i&lt;/math&gt;, then &lt;math&gt;P(n) = \prod_{i=1}^{N} n_i!&lt;/math&gt;.<br /> <br /> Now &lt;math&gt;2010 = 2\cdot 3\cdot 5\cdot 67&lt;/math&gt;. In order that &lt;math&gt;2010\mid P(n)&lt;/math&gt;, we must have &lt;math&gt;67\mid n_m!&lt;/math&gt; for the class &lt;math&gt;C_m&lt;/math&gt; with the most elements. This means &lt;math&gt;n_m\geq 67&lt;/math&gt;, since no smaller factorial will have &lt;math&gt;67&lt;/math&gt; as a factor. This condition is sufficient, since &lt;math&gt;n_m!&lt;/math&gt; will be divisible by &lt;math&gt;30&lt;/math&gt; for &lt;math&gt;n_m\geq 5&lt;/math&gt;, and even more so &lt;math&gt;n_m\geq 67&lt;/math&gt;.<br /> <br /> The smallest element &lt;math&gt;g_m&lt;/math&gt; of the equivalence class &lt;math&gt;C_m&lt;/math&gt; is square-free, since if it were divisible by the square of a prime, the quotient would be a smaller element of &lt;math&gt;C_m&lt;/math&gt;. Also, each prime &lt;math&gt;p&lt;/math&gt; that divides &lt;math&gt;g_m&lt;/math&gt; divides all the other elements &lt;math&gt;k&lt;/math&gt; of &lt;math&gt;C_m&lt;/math&gt;, since &lt;math&gt;p^2\mid kg_m&lt;/math&gt; and thus &lt;math&gt;p\mid k&lt;/math&gt;. Therefore &lt;math&gt;g_m\mid k&lt;/math&gt; for all &lt;math&gt;k\in C_m&lt;/math&gt;. The primes that are not in &lt;math&gt;g_m&lt;/math&gt; occur an even number of times in each &lt;math&gt;k\in C_m&lt;/math&gt;.<br /> <br /> Thus the equivalence class &lt;math&gt;C_m = \{g_mk^2\leq n\}&lt;/math&gt;. With &lt;math&gt;g_m = 1&lt;/math&gt;, we get the largest possible &lt;math&gt;n_m&lt;/math&gt;. This is just the set of squares in &lt;math&gt;[n]&lt;/math&gt;, of which we need at least &lt;math&gt;67&lt;/math&gt;, so &lt;math&gt;n\geq 67^2&lt;/math&gt;. This condition is necessary and sufficient.<br /> <br /> === Solution 2 ===<br /> This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as &quot;equivalence relation&quot;:<br /> <br /> It is possible to write all positive integers &lt;math&gt;n&lt;/math&gt; in the form &lt;math&gt;p\cdot m^2&lt;/math&gt;, where &lt;math&gt;m^2&lt;/math&gt; is the largest perfect square dividing &lt;math&gt;n&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Obviously, one working permutation of &lt;math&gt;[n]&lt;/math&gt; is simply &lt;math&gt;(1, 2, \ldots, n)&lt;/math&gt;; this is acceptable, as &lt;math&gt;ka_k&lt;/math&gt; is always &lt;math&gt;k^2&lt;/math&gt; in this sequence.<br /> <br /> '''Lemma 1.''' We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities &lt;math&gt;p&lt;/math&gt;.<br /> <br /> ''Proof.'' Let &lt;math&gt;p_k&lt;/math&gt; and &lt;math&gt;m_k&lt;/math&gt; be the values of &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt;, respectively, for a given &lt;math&gt;k&lt;/math&gt; as defined above, such that &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. We can obviously permute two numbers which have the same &lt;math&gt;p&lt;/math&gt;, since if &lt;math&gt;p_j = p_w&lt;/math&gt; where &lt;math&gt;j&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; are 2 values of &lt;math&gt;k&lt;/math&gt;, then &lt;math&gt;j\cdot w = p_j^2\cdot m_j^2\cdot m_w^2&lt;/math&gt;, which is a perfect square. This proves that we can permute any numbers with the same value of &lt;math&gt;p&lt;/math&gt;.<br /> <br /> '''End Lemma'''<br /> <br /> '''Lemma 2.''' We will prove the converse of Lemma 1: Let one number have a &lt;math&gt;p&lt;/math&gt; value of &lt;math&gt;\phi&lt;/math&gt; and another, &lt;math&gt;\gamma&lt;/math&gt;. &lt;math&gt;\phi\cdot f&lt;/math&gt; and &lt;math&gt;\gamma\cdot g&lt;/math&gt; are both perfect squares.<br /> <br /> ''Proof.'' &lt;math&gt;\phi\cdot f&lt;/math&gt; and &lt;math&gt;\gamma\cdot g&lt;/math&gt; are both perfect squares, so for &lt;math&gt;\phi\cdot \gamma&lt;/math&gt; to be a perfect square, if &lt;math&gt;g&lt;/math&gt; is greater than or equal to &lt;math&gt;f&lt;/math&gt;, &lt;math&gt;g/f&lt;/math&gt; must be a perfect square, too. Thus &lt;math&gt;g&lt;/math&gt; is &lt;math&gt;f&lt;/math&gt; times a square, but &lt;math&gt;g&lt;/math&gt; cannot divide any squares besides &lt;math&gt;1&lt;/math&gt;, so &lt;math&gt;g = 1f&lt;/math&gt;; &lt;math&gt;g = f&lt;/math&gt;. Similarly, if &lt;math&gt;f\geq g&lt;/math&gt;, then &lt;math&gt;f = g&lt;/math&gt; for our rules to keep working.<br /> <br /> '''End Lemma'''<br /> <br /> We can permute &lt;math&gt;l&lt;/math&gt; numbers with the same &lt;math&gt;p&lt;/math&gt; in &lt;math&gt;l!&lt;/math&gt; ways. We must have at least 67 numbers with a certain &lt;math&gt;p&lt;/math&gt; so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as &lt;math&gt;h&lt;/math&gt;, in general, we need numbers all the way up to &lt;math&gt;h\cdot 67^2&lt;/math&gt;, so obviously, &lt;math&gt;67^2&lt;/math&gt; is the smallest such number such that we can get a &lt;math&gt;67!&lt;/math&gt; term; here 67 &lt;math&gt;p&lt;/math&gt; terms are 1. Thus we need the integers &lt;math&gt;1, 2, \ldots, 67^2&lt;/math&gt;, so &lt;math&gt;67^2&lt;/math&gt;, or &lt;math&gt;\boxed{4489}&lt;/math&gt;, is the answer.<br /> <br /> ==Solution Number Sense==<br /> We have to somehow calculate the number of permutations for a given &lt;math&gt;n&lt;/math&gt;. How in the world do we do this? Because we want squares, why not call a number &lt;math&gt;k=m*s^2&lt;/math&gt;, where &lt;math&gt;s&lt;/math&gt; is the largest square that allows &lt;math&gt;m&lt;/math&gt; to be non-square? &lt;math&gt;m=1&lt;/math&gt; is the only square &lt;math&gt;m&lt;/math&gt; can be, which only happens if &lt;math&gt;k&lt;/math&gt; is a perfect square.<br /> <br /> For example, &lt;math&gt;126 = 14 * 3^2&lt;/math&gt;, therefore in this case &lt;math&gt;k=126, m = 14, s = 3&lt;/math&gt;.<br /> <br /> I will call a permutation of the numbers &lt;math&gt;P&lt;/math&gt;, while the original &lt;math&gt;1, 2, 3, 4, ...&lt;/math&gt; I will call &lt;math&gt;S&lt;/math&gt;.<br /> <br /> Note that essentially we are looking at &quot;pairing up&quot; elements between &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;S&lt;/math&gt; such that the product of &lt;math&gt;P_k&lt;/math&gt; and &lt;math&gt;S_k&lt;/math&gt; is a perfect square. How do we do this? Using the representation above.<br /> <br /> Each square has to have an even exponent of every prime represented in its prime factorization. Therefore, we can just take all exponents of the primes &lt;math&gt;mod 2&lt;/math&gt; and if there are any odd numbers, those are the ones we have to match- in effect, they are the &lt;math&gt;m&lt;/math&gt; numbers mentioned at the beginning.<br /> <br /> By listing the &lt;math&gt;m&lt;/math&gt; values, in my search for &quot;dumb&quot; or &quot;obvious&quot; ideas I am pretty confident that only values with identical &lt;math&gt;m&lt;/math&gt;s can be matched together. With such a solid idea let me prove it.<br /> <br /> If we were to &quot;pair up&quot; numbers with different &lt;math&gt;m&lt;/math&gt;s, take for example &lt;math&gt;S_{18}&lt;/math&gt; with an &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;P_{18}=26&lt;/math&gt; with an &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;26&lt;/math&gt;, note that their product gives a supposed &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;13&lt;/math&gt; because the &lt;math&gt;2&lt;/math&gt; values cancel out. But then, what happens to the extra &lt;math&gt;13&lt;/math&gt; left? It doesn't make a square, contradiction. To finish up this easy proof, note that if a &quot;pair&quot; has different &lt;math&gt;m&lt;/math&gt; values, and the smaller one is &lt;math&gt;m_1&lt;/math&gt;, in order for the product to leave a square, the larger &lt;math&gt;m&lt;/math&gt; value has to have not just &lt;math&gt;m_1&lt;/math&gt; but another square inside it, which is absurd because we stipulated at the beginning that &lt;math&gt;m&lt;/math&gt; was square-free except for the trivial multiplication identity, 1.<br /> <br /> Now, how many ways are there to do this? If there are &lt;math&gt;c_1&lt;/math&gt; numbers with &lt;math&gt;m=1&lt;/math&gt;, there are clearly &lt;math&gt;(c_1)!&lt;/math&gt; ways of sorting them. The same goes for &lt;math&gt;m=2, 3, etc.&lt;/math&gt; by this logic. Note that the &lt;math&gt;P(n)&lt;/math&gt; as stated by the problem requires a &lt;math&gt;67&lt;/math&gt; thrown in there because &lt;math&gt;2010=2*3*5*67&lt;/math&gt;, so there has to be a &lt;math&gt;S_n&lt;/math&gt; with 67 elements with the same &lt;math&gt;m&lt;/math&gt;. It is evident that the smallest &lt;math&gt;n&lt;/math&gt; will occur when &lt;math&gt;m=1&lt;/math&gt;, because if &lt;math&gt;m&lt;/math&gt; is bigger we would have to expand &lt;math&gt;n&lt;/math&gt; to get the same number of &lt;math&gt;m&lt;/math&gt; values. Finally, realize that the only numbers with &lt;math&gt;m=1&lt;/math&gt; are square numbers! So our smallest &lt;math&gt;n=67^2=4489&lt;/math&gt;, and we are done.<br /> <br /> I relied on looking for patterns a lot in this problem. When faced with combo/number theory, it is always good to draw a sketch. Never be scared to try a problem on the USAJMO. It takes about 45 minutes. Well, it's 2010 and a number 1. Cheers!<br /> <br /> -expiLnCalc<br /> <br /> ==Solution Easy==<br /> <br /> Consider the set of numbers &lt;math&gt;m*k^2&lt;/math&gt; such that &lt;math&gt;m&lt;/math&gt; is not divisible by any squares other than 1 and &lt;math&gt;k = 1, 2, ...&lt;/math&gt; By changing &lt;math&gt;m&lt;/math&gt; we can encompass all numbers less than or equal to &lt;math&gt;n&lt;/math&gt;. Now notice that for a working arrangement these numbers can be permutated in any way to create a new one; for instance, the numbers &lt;math&gt;2,8,18&lt;/math&gt; have &lt;math&gt;m = 2&lt;/math&gt; and they can be arranged in &lt;math&gt;3! = 6&lt;/math&gt; ways. Thus, since &lt;math&gt;2010 = 2*3*5*67&lt;/math&gt; we need a permutation to have at least 67 elements (since 67 is prime). To minimize &lt;math&gt;n&lt;/math&gt;, we let &lt;math&gt;m = 1&lt;/math&gt; and we have &lt;math&gt;1^2,2^2,3^2,...67^2&lt;/math&gt; and we stop at &lt;math&gt;67^2&lt;/math&gt; to get &lt;math&gt;\boxed{4489}&lt;/math&gt;. ~Leonard_my_dude~<br /> <br /> ==Solution 5==<br /> <br /> It's well known that there exists &lt;math&gt;f(n)&lt;/math&gt; and &lt;math&gt;g(n)&lt;/math&gt; such that &lt;math&gt;n = f(n) \cdot g(n)&lt;/math&gt;, no square divides &lt;math&gt;f(n)&lt;/math&gt; other than 1, and &lt;math&gt;g(n)&lt;/math&gt; is a perfect square.<br /> <br /> Lemma: &lt;math&gt;k \cdot a_k&lt;/math&gt; is a perfect square if and only if &lt;math&gt;f(k) = f(a_k)&lt;/math&gt; <br /> <br /> We prove first: If &lt;math&gt;f(k) = f(a_k)&lt;/math&gt;, &lt;math&gt;k \cdot a_k&lt;/math&gt; is a perfect square. <br /> <br /> &lt;math&gt;k \cdot a_k = f(k) \cdot g(k) \cdot f(a_k) \cdot g(a_k) = f(k)^2 \cdot g(k) \cdot g(a_k)&lt;/math&gt;, which is a perfect square. <br /> <br /> We will now prove: If &lt;math&gt;k \cdot a_k&lt;/math&gt; is a perfect square, &lt;math&gt;f(k) = f(a_k)&lt;/math&gt;.<br /> <br /> We do proof by contrapositive: If &lt;math&gt;f(k) \neq f(a_k)&lt;/math&gt;, &lt;math&gt;k \cdot a_k&lt;/math&gt; is not a perfect square. <br /> <br /> Note that if &lt;math&gt;f(k) \neq f(a_k)&lt;/math&gt;, There exists a prime p, such that &lt;math&gt;v_p(k) \neq v_p(a_k)&lt;/math&gt;. Also, &lt;math&gt;v_p(k), v_p(a_k) \leq 1&lt;/math&gt;. Thus, &lt;math&gt;v_p(k \cdot a_k) = 1&lt;/math&gt;, making &lt;math&gt;k \cdot a_k&lt;/math&gt; not a square.<br /> <br /> End Lemma<br /> <br /> Thus, we can only match k with &lt;math&gt;a_k&lt;/math&gt; if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the a_k with f value 1, then 2, ... Thus, our answer is:<br /> <br /> &lt;math&gt;P(n) = \prod_{1 \leq i \leq n, g(i) = 1} \left\lfloor \sqrt{\frac{n}{i}} \right \rfloor !&lt;/math&gt; <br /> <br /> For all &lt;math&gt;n &lt; 67^2&lt;/math&gt;, &lt;math&gt;P(n)&lt;/math&gt; doesn't have a factor of 67. However, if &lt;math&gt;n = 67^2&lt;/math&gt;, the first term will be a multiple of 2010, and thus the answer is &lt;math&gt;67^2 = \boxed{4489}&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> <br /> {{alternate solutions}}<br /> <br /> == See Also ==<br /> * &lt;url&gt;viewtopic.php?t=347303 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> {{USAJMO newbox|year=2010|before=First Question|num-a=2}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_1&diff=124273 2010 USAJMO Problems/Problem 1 2020-06-07T19:58:18Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> A permutation of the set of positive integers &lt;math&gt;[n] = \{1, 2, \ldots, n\}&lt;/math&gt; is a sequence &lt;math&gt;(a_1, a_2, \ldots, a_n)&lt;/math&gt; such that each element of &lt;math&gt;[n]&lt;/math&gt; appears precisely one time as a term of the sequence. For example, &lt;math&gt;(3, 5, 1, 2, 4)&lt;/math&gt; is a permutation of &lt;math&gt;&lt;/math&gt;. Let &lt;math&gt;P(n)&lt;/math&gt; be the number of permutations of &lt;math&gt;[n]&lt;/math&gt; for which &lt;math&gt;ka_k&lt;/math&gt; is a perfect square for all &lt;math&gt;1\leq k\leq n&lt;/math&gt;. Find with proof the smallest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;P(n)&lt;/math&gt; is a multiple of &lt;math&gt;2010&lt;/math&gt;.<br /> <br /> == Solutions ==<br /> We claim that the smallest &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;67^2 = \boxed{4489}&lt;/math&gt;.<br /> <br /> === Solution 1 ===<br /> Let &lt;math&gt;S = \{1, 4, 9, \ldots\}&lt;/math&gt; be the set of positive perfect squares. We claim that the relation &lt;math&gt;R = \{(j, k)\in [n]\times[n]\mid jk\in S\}&lt;/math&gt; is an equivalence relation on &lt;math&gt;[n]&lt;/math&gt;.<br /> * It is reflexive because &lt;math&gt;k^2\in S&lt;/math&gt; for all &lt;math&gt;k\in [n]&lt;/math&gt;.<br /> * It is symmetric because &lt;math&gt;jk\in S\implies kj = jk\in S&lt;/math&gt;.<br /> * It is transitive because if &lt;math&gt;jk\in S&lt;/math&gt; and &lt;math&gt;kl\in S&lt;/math&gt;, then &lt;math&gt;jk\cdot kl = jlk^2\in S\implies jl\in S&lt;/math&gt;, since &lt;math&gt;S&lt;/math&gt; is closed under multiplication and a non-square times a square is always a non-square.<br /> <br /> We are restricted to permutations for which &lt;math&gt;ka_k \in S&lt;/math&gt;, in other words to permutations that send each element of &lt;math&gt;[n]&lt;/math&gt; into its equivalence class. Suppose there are &lt;math&gt;N&lt;/math&gt; equivalence classes: &lt;math&gt;C_1, \ldots, C_N&lt;/math&gt;. Let &lt;math&gt;n_i&lt;/math&gt; be the number of elements of &lt;math&gt;C_i&lt;/math&gt;, then &lt;math&gt;P(n) = \prod_{i=1}^{N} n_i!&lt;/math&gt;.<br /> <br /> Now &lt;math&gt;2010 = 2\cdot 3\cdot 5\cdot 67&lt;/math&gt;. In order that &lt;math&gt;2010\mid P(n)&lt;/math&gt;, we must have &lt;math&gt;67\mid n_m!&lt;/math&gt; for the class &lt;math&gt;C_m&lt;/math&gt; with the most elements. This means &lt;math&gt;n_m\geq 67&lt;/math&gt;, since no smaller factorial will have &lt;math&gt;67&lt;/math&gt; as a factor. This condition is sufficient, since &lt;math&gt;n_m!&lt;/math&gt; will be divisible by &lt;math&gt;30&lt;/math&gt; for &lt;math&gt;n_m\geq 5&lt;/math&gt;, and even more so &lt;math&gt;n_m\geq 67&lt;/math&gt;.<br /> <br /> The smallest element &lt;math&gt;g_m&lt;/math&gt; of the equivalence class &lt;math&gt;C_m&lt;/math&gt; is square-free, since if it were divisible by the square of a prime, the quotient would be a smaller element of &lt;math&gt;C_m&lt;/math&gt;. Also, each prime &lt;math&gt;p&lt;/math&gt; that divides &lt;math&gt;g_m&lt;/math&gt; divides all the other elements &lt;math&gt;k&lt;/math&gt; of &lt;math&gt;C_m&lt;/math&gt;, since &lt;math&gt;p^2\mid kg_m&lt;/math&gt; and thus &lt;math&gt;p\mid k&lt;/math&gt;. Therefore &lt;math&gt;g_m\mid k&lt;/math&gt; for all &lt;math&gt;k\in C_m&lt;/math&gt;. The primes that are not in &lt;math&gt;g_m&lt;/math&gt; occur an even number of times in each &lt;math&gt;k\in C_m&lt;/math&gt;.<br /> <br /> Thus the equivalence class &lt;math&gt;C_m = \{g_mk^2\leq n\}&lt;/math&gt;. With &lt;math&gt;g_m = 1&lt;/math&gt;, we get the largest possible &lt;math&gt;n_m&lt;/math&gt;. This is just the set of squares in &lt;math&gt;[n]&lt;/math&gt;, of which we need at least &lt;math&gt;67&lt;/math&gt;, so &lt;math&gt;n\geq 67^2&lt;/math&gt;. This condition is necessary and sufficient.<br /> <br /> === Solution 2 ===<br /> This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as &quot;equivalence relation&quot;:<br /> <br /> It is possible to write all positive integers &lt;math&gt;n&lt;/math&gt; in the form &lt;math&gt;p\cdot m^2&lt;/math&gt;, where &lt;math&gt;m^2&lt;/math&gt; is the largest perfect square dividing &lt;math&gt;n&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Obviously, one working permutation of &lt;math&gt;[n]&lt;/math&gt; is simply &lt;math&gt;(1, 2, \ldots, n)&lt;/math&gt;; this is acceptable, as &lt;math&gt;ka_k&lt;/math&gt; is always &lt;math&gt;k^2&lt;/math&gt; in this sequence.<br /> <br /> '''Lemma 1.''' We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities &lt;math&gt;p&lt;/math&gt;.<br /> <br /> ''Proof.'' Let &lt;math&gt;p_k&lt;/math&gt; and &lt;math&gt;m_k&lt;/math&gt; be the values of &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt;, respectively, for a given &lt;math&gt;k&lt;/math&gt; as defined above, such that &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. We can obviously permute two numbers which have the same &lt;math&gt;p&lt;/math&gt;, since if &lt;math&gt;p_j = p_w&lt;/math&gt; where &lt;math&gt;j&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; are 2 values of &lt;math&gt;k&lt;/math&gt;, then &lt;math&gt;j\cdot w = p_j^2\cdot m_j^2\cdot m_w^2&lt;/math&gt;, which is a perfect square. This proves that we can permute any numbers with the same value of &lt;math&gt;p&lt;/math&gt;.<br /> <br /> '''End Lemma'''<br /> <br /> '''Lemma 2.''' We will prove the converse of Lemma 1: Let one number have a &lt;math&gt;p&lt;/math&gt; value of &lt;math&gt;\phi&lt;/math&gt; and another, &lt;math&gt;\gamma&lt;/math&gt;. &lt;math&gt;\phi\cdot f&lt;/math&gt; and &lt;math&gt;\gamma\cdot g&lt;/math&gt; are both perfect squares.<br /> <br /> ''Proof.'' &lt;math&gt;\phi\cdot f&lt;/math&gt; and &lt;math&gt;\gamma\cdot g&lt;/math&gt; are both perfect squares, so for &lt;math&gt;\phi\cdot \gamma&lt;/math&gt; to be a perfect square, if &lt;math&gt;g&lt;/math&gt; is greater than or equal to &lt;math&gt;f&lt;/math&gt;, &lt;math&gt;g/f&lt;/math&gt; must be a perfect square, too. Thus &lt;math&gt;g&lt;/math&gt; is &lt;math&gt;f&lt;/math&gt; times a square, but &lt;math&gt;g&lt;/math&gt; cannot divide any squares besides &lt;math&gt;1&lt;/math&gt;, so &lt;math&gt;g = 1f&lt;/math&gt;; &lt;math&gt;g = f&lt;/math&gt;. Similarly, if &lt;math&gt;f\geq g&lt;/math&gt;, then &lt;math&gt;f = g&lt;/math&gt; for our rules to keep working.<br /> <br /> '''End Lemma'''<br /> <br /> We can permute &lt;math&gt;l&lt;/math&gt; numbers with the same &lt;math&gt;p&lt;/math&gt; in &lt;math&gt;l!&lt;/math&gt; ways. We must have at least 67 numbers with a certain &lt;math&gt;p&lt;/math&gt; so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as &lt;math&gt;h&lt;/math&gt;, in general, we need numbers all the way up to &lt;math&gt;h\cdot 67^2&lt;/math&gt;, so obviously, &lt;math&gt;67^2&lt;/math&gt; is the smallest such number such that we can get a &lt;math&gt;67!&lt;/math&gt; term; here 67 &lt;math&gt;p&lt;/math&gt; terms are 1. Thus we need the integers &lt;math&gt;1, 2, \ldots, 67^2&lt;/math&gt;, so &lt;math&gt;67^2&lt;/math&gt;, or &lt;math&gt;\boxed{4489}&lt;/math&gt;, is the answer.<br /> <br /> ==Solution Number Sense==<br /> We have to somehow calculate the number of permutations for a given &lt;math&gt;n&lt;/math&gt;. How in the world do we do this? Because we want squares, why not call a number &lt;math&gt;k=m*s^2&lt;/math&gt;, where &lt;math&gt;s&lt;/math&gt; is the largest square that allows &lt;math&gt;m&lt;/math&gt; to be non-square? &lt;math&gt;m=1&lt;/math&gt; is the only square &lt;math&gt;m&lt;/math&gt; can be, which only happens if &lt;math&gt;k&lt;/math&gt; is a perfect square.<br /> <br /> For example, &lt;math&gt;126 = 14 * 3^2&lt;/math&gt;, therefore in this case &lt;math&gt;k=126, m = 14, s = 3&lt;/math&gt;.<br /> <br /> I will call a permutation of the numbers &lt;math&gt;P&lt;/math&gt;, while the original &lt;math&gt;1, 2, 3, 4, ...&lt;/math&gt; I will call &lt;math&gt;S&lt;/math&gt;.<br /> <br /> Note that essentially we are looking at &quot;pairing up&quot; elements between &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;S&lt;/math&gt; such that the product of &lt;math&gt;P_k&lt;/math&gt; and &lt;math&gt;S_k&lt;/math&gt; is a perfect square. How do we do this? Using the representation above.<br /> <br /> Each square has to have an even exponent of every prime represented in its prime factorization. Therefore, we can just take all exponents of the primes &lt;math&gt;mod 2&lt;/math&gt; and if there are any odd numbers, those are the ones we have to match- in effect, they are the &lt;math&gt;m&lt;/math&gt; numbers mentioned at the beginning.<br /> <br /> By listing the &lt;math&gt;m&lt;/math&gt; values, in my search for &quot;dumb&quot; or &quot;obvious&quot; ideas I am pretty confident that only values with identical &lt;math&gt;m&lt;/math&gt;s can be matched together. With such a solid idea let me prove it.<br /> <br /> If we were to &quot;pair up&quot; numbers with different &lt;math&gt;m&lt;/math&gt;s, take for example &lt;math&gt;S_{18}&lt;/math&gt; with an &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;P_{18}=26&lt;/math&gt; with an &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;26&lt;/math&gt;, note that their product gives a supposed &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;13&lt;/math&gt; because the &lt;math&gt;2&lt;/math&gt; values cancel out. But then, what happens to the extra &lt;math&gt;13&lt;/math&gt; left? It doesn't make a square, contradiction. To finish up this easy proof, note that if a &quot;pair&quot; has different &lt;math&gt;m&lt;/math&gt; values, and the smaller one is &lt;math&gt;m_1&lt;/math&gt;, in order for the product to leave a square, the larger &lt;math&gt;m&lt;/math&gt; value has to have not just &lt;math&gt;m_1&lt;/math&gt; but another square inside it, which is absurd because we stipulated at the beginning that &lt;math&gt;m&lt;/math&gt; was square-free except for the trivial multiplication identity, 1.<br /> <br /> Now, how many ways are there to do this? If there are &lt;math&gt;c_1&lt;/math&gt; numbers with &lt;math&gt;m=1&lt;/math&gt;, there are clearly &lt;math&gt;(c_1)!&lt;/math&gt; ways of sorting them. The same goes for &lt;math&gt;m=2, 3, etc.&lt;/math&gt; by this logic. Note that the &lt;math&gt;P(n)&lt;/math&gt; as stated by the problem requires a &lt;math&gt;67&lt;/math&gt; thrown in there because &lt;math&gt;2010=2*3*5*67&lt;/math&gt;, so there has to be a &lt;math&gt;S_n&lt;/math&gt; with 67 elements with the same &lt;math&gt;m&lt;/math&gt;. It is evident that the smallest &lt;math&gt;n&lt;/math&gt; will occur when &lt;math&gt;m=1&lt;/math&gt;, because if &lt;math&gt;m&lt;/math&gt; is bigger we would have to expand &lt;math&gt;n&lt;/math&gt; to get the same number of &lt;math&gt;m&lt;/math&gt; values. Finally, realize that the only numbers with &lt;math&gt;m=1&lt;/math&gt; are square numbers! So our smallest &lt;math&gt;n=67^2=4489&lt;/math&gt;, and we are done.<br /> <br /> I relied on looking for patterns a lot in this problem. When faced with combo/number theory, it is always good to draw a sketch. Never be scared to try a problem on the USAJMO. It takes about 45 minutes. Well, it's 2010 and a number 1. Cheers!<br /> <br /> -expiLnCalc<br /> <br /> ==Solution Easy==<br /> <br /> Consider the set of numbers &lt;math&gt;m*k^2&lt;/math&gt; such that &lt;math&gt;m&lt;/math&gt; is not divisible by any squares other than 1 and &lt;math&gt;k = 1, 2, ...&lt;/math&gt; By changing &lt;math&gt;m&lt;/math&gt; we can encompass all numbers less than or equal to &lt;math&gt;n&lt;/math&gt;. Now notice that for a working arrangement these numbers can be permutated in any way to create a new one; for instance, the numbers &lt;math&gt;2,8,18&lt;/math&gt; have &lt;math&gt;m = 2&lt;/math&gt; and they can be arranged in &lt;math&gt;3! = 6&lt;/math&gt; ways. Thus, since &lt;math&gt;2010 = 2*3*5*67&lt;/math&gt; we need a permutation to have at least 67 elements (since 67 is prime). To minimize &lt;math&gt;n&lt;/math&gt;, we let &lt;math&gt;m = 1&lt;/math&gt; and we have &lt;math&gt;1^2,2^2,3^2,...67^2&lt;/math&gt; and we stop at &lt;math&gt;67^2&lt;/math&gt; to get &lt;math&gt;\boxed{4489}&lt;/math&gt;. ~Leonard_my_dude~<br /> <br /> ==Solution 5==<br /> <br /> It's well known that there exists &lt;math&gt;f(n)&lt;/math&gt; and &lt;math&gt;g(n)&lt;/math&gt; such that &lt;math&gt;n = f(n) \cdot g(n)&lt;/math&gt;, no square divides &lt;math&gt;f(n)&lt;/math&gt; other than 1, and &lt;math&gt;g(n)&lt;/math&gt; is a perfect square.<br /> <br /> Lemma: &lt;math&gt;k \cdot a_k&lt;/math&gt; is a perfect square if and only if &lt;math&gt;f(k) = f(a_k)&lt;/math&gt; <br /> <br /> We prove the backwards direction first. If &lt;math&gt;f(k) = f(a_k)&lt;/math&gt;, &lt;math&gt;k \cdot a_k = f(k) \cdot g(k) \cdot f(a_k) \cdot g(a_k) = f(k)^2 \cdot g(k) \cdot g(a_k)&lt;/math&gt;, which is a perfect square. <br /> <br /> We will now prove the forwards direction. We will prove the contrapositive: If &lt;math&gt;f(k) \neq f(a_k)&lt;/math&gt;, &lt;math&gt;k \cdot a_k&lt;/math&gt; is not a perfect square. Note that if &lt;math&gt;f(k) \neq f(a_k)&lt;/math&gt;, There exists a prime p, such that &lt;math&gt;v_p(k) \neq v_p(a_k)&lt;/math&gt;. Also, &lt;math&gt;v_p(k), v_p(a_k) \leq 1&lt;/math&gt;. Thus, &lt;math&gt;v_p(k \cdot a_k) = 1&lt;/math&gt;, making &lt;math&gt;k \cdot a_k&lt;/math&gt; not a square.<br /> <br /> End Lemma<br /> <br /> Thus, we can only match k with &lt;math&gt;a_k&lt;/math&gt; if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the a_k with f value 1, then 2, ... Thus, our answer is:<br /> <br /> &lt;math&gt;P(n) = \prod_{1 \leq i \leq n, g(i) = 1} \left\lfloor \sqrt{\frac{n}{i}} \right \rfloor !&lt;/math&gt; <br /> <br /> For all &lt;math&gt;n &lt; 67^2&lt;/math&gt;, &lt;math&gt;P(n)&lt;/math&gt; doesn't have a factor of 67. However, if &lt;math&gt;n = 67^2&lt;/math&gt;, the first term will be a multiple of 2010, and thus the answer is &lt;math&gt;67^2 = \boxed{4489}&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> <br /> {{alternate solutions}}<br /> <br /> == See Also ==<br /> * &lt;url&gt;viewtopic.php?t=347303 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> {{USAJMO newbox|year=2010|before=First Question|num-a=2}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_1&diff=124249 2010 USAJMO Problems/Problem 1 2020-06-07T18:17:38Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> A permutation of the set of positive integers &lt;math&gt;[n] = \{1, 2, \ldots, n\}&lt;/math&gt; is a sequence &lt;math&gt;(a_1, a_2, \ldots, a_n)&lt;/math&gt; such that each element of &lt;math&gt;[n]&lt;/math&gt; appears precisely one time as a term of the sequence. For example, &lt;math&gt;(3, 5, 1, 2, 4)&lt;/math&gt; is a permutation of &lt;math&gt;&lt;/math&gt;. Let &lt;math&gt;P(n)&lt;/math&gt; be the number of permutations of &lt;math&gt;[n]&lt;/math&gt; for which &lt;math&gt;ka_k&lt;/math&gt; is a perfect square for all &lt;math&gt;1\leq k\leq n&lt;/math&gt;. Find with proof the smallest &lt;math&gt;n&lt;/math&gt; such that &lt;math&gt;P(n)&lt;/math&gt; is a multiple of &lt;math&gt;2010&lt;/math&gt;.<br /> <br /> == Solutions ==<br /> We claim that the smallest &lt;math&gt;n&lt;/math&gt; is &lt;math&gt;67^2 = \boxed{4489}&lt;/math&gt;.<br /> <br /> === Solution 1 ===<br /> Let &lt;math&gt;S = \{1, 4, 9, \ldots\}&lt;/math&gt; be the set of positive perfect squares. We claim that the relation &lt;math&gt;R = \{(j, k)\in [n]\times[n]\mid jk\in S\}&lt;/math&gt; is an equivalence relation on &lt;math&gt;[n]&lt;/math&gt;.<br /> * It is reflexive because &lt;math&gt;k^2\in S&lt;/math&gt; for all &lt;math&gt;k\in [n]&lt;/math&gt;.<br /> * It is symmetric because &lt;math&gt;jk\in S\implies kj = jk\in S&lt;/math&gt;.<br /> * It is transitive because if &lt;math&gt;jk\in S&lt;/math&gt; and &lt;math&gt;kl\in S&lt;/math&gt;, then &lt;math&gt;jk\cdot kl = jlk^2\in S\implies jl\in S&lt;/math&gt;, since &lt;math&gt;S&lt;/math&gt; is closed under multiplication and a non-square times a square is always a non-square.<br /> <br /> We are restricted to permutations for which &lt;math&gt;ka_k \in S&lt;/math&gt;, in other words to permutations that send each element of &lt;math&gt;[n]&lt;/math&gt; into its equivalence class. Suppose there are &lt;math&gt;N&lt;/math&gt; equivalence classes: &lt;math&gt;C_1, \ldots, C_N&lt;/math&gt;. Let &lt;math&gt;n_i&lt;/math&gt; be the number of elements of &lt;math&gt;C_i&lt;/math&gt;, then &lt;math&gt;P(n) = \prod_{i=1}^{N} n_i!&lt;/math&gt;.<br /> <br /> Now &lt;math&gt;2010 = 2\cdot 3\cdot 5\cdot 67&lt;/math&gt;. In order that &lt;math&gt;2010\mid P(n)&lt;/math&gt;, we must have &lt;math&gt;67\mid n_m!&lt;/math&gt; for the class &lt;math&gt;C_m&lt;/math&gt; with the most elements. This means &lt;math&gt;n_m\geq 67&lt;/math&gt;, since no smaller factorial will have &lt;math&gt;67&lt;/math&gt; as a factor. This condition is sufficient, since &lt;math&gt;n_m!&lt;/math&gt; will be divisible by &lt;math&gt;30&lt;/math&gt; for &lt;math&gt;n_m\geq 5&lt;/math&gt;, and even more so &lt;math&gt;n_m\geq 67&lt;/math&gt;.<br /> <br /> The smallest element &lt;math&gt;g_m&lt;/math&gt; of the equivalence class &lt;math&gt;C_m&lt;/math&gt; is square-free, since if it were divisible by the square of a prime, the quotient would be a smaller element of &lt;math&gt;C_m&lt;/math&gt;. Also, each prime &lt;math&gt;p&lt;/math&gt; that divides &lt;math&gt;g_m&lt;/math&gt; divides all the other elements &lt;math&gt;k&lt;/math&gt; of &lt;math&gt;C_m&lt;/math&gt;, since &lt;math&gt;p^2\mid kg_m&lt;/math&gt; and thus &lt;math&gt;p\mid k&lt;/math&gt;. Therefore &lt;math&gt;g_m\mid k&lt;/math&gt; for all &lt;math&gt;k\in C_m&lt;/math&gt;. The primes that are not in &lt;math&gt;g_m&lt;/math&gt; occur an even number of times in each &lt;math&gt;k\in C_m&lt;/math&gt;.<br /> <br /> Thus the equivalence class &lt;math&gt;C_m = \{g_mk^2\leq n\}&lt;/math&gt;. With &lt;math&gt;g_m = 1&lt;/math&gt;, we get the largest possible &lt;math&gt;n_m&lt;/math&gt;. This is just the set of squares in &lt;math&gt;[n]&lt;/math&gt;, of which we need at least &lt;math&gt;67&lt;/math&gt;, so &lt;math&gt;n\geq 67^2&lt;/math&gt;. This condition is necessary and sufficient.<br /> <br /> === Solution 2 ===<br /> This proof can also be rephrased as follows, in a longer way, but with fewer highly technical words such as &quot;equivalence relation&quot;:<br /> <br /> It is possible to write all positive integers &lt;math&gt;n&lt;/math&gt; in the form &lt;math&gt;p\cdot m^2&lt;/math&gt;, where &lt;math&gt;m^2&lt;/math&gt; is the largest perfect square dividing &lt;math&gt;n&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. Obviously, one working permutation of &lt;math&gt;[n]&lt;/math&gt; is simply &lt;math&gt;(1, 2, \ldots, n)&lt;/math&gt;; this is acceptable, as &lt;math&gt;ka_k&lt;/math&gt; is always &lt;math&gt;k^2&lt;/math&gt; in this sequence.<br /> <br /> '''Lemma 1.''' We can permute any numbers that, when each divided by the largest perfect square that divides it, yield equal quantities &lt;math&gt;p&lt;/math&gt;.<br /> <br /> ''Proof.'' Let &lt;math&gt;p_k&lt;/math&gt; and &lt;math&gt;m_k&lt;/math&gt; be the values of &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;m&lt;/math&gt;, respectively, for a given &lt;math&gt;k&lt;/math&gt; as defined above, such that &lt;math&gt;p&lt;/math&gt; is not divisible by the square of any prime. We can obviously permute two numbers which have the same &lt;math&gt;p&lt;/math&gt;, since if &lt;math&gt;p_j = p_w&lt;/math&gt; where &lt;math&gt;j&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt; are 2 values of &lt;math&gt;k&lt;/math&gt;, then &lt;math&gt;j\cdot w = p_j^2\cdot m_j^2\cdot m_w^2&lt;/math&gt;, which is a perfect square. This proves that we can permute any numbers with the same value of &lt;math&gt;p&lt;/math&gt;.<br /> <br /> '''End Lemma'''<br /> <br /> '''Lemma 2.''' We will prove the converse of Lemma 1: Let one number have a &lt;math&gt;p&lt;/math&gt; value of &lt;math&gt;\phi&lt;/math&gt; and another, &lt;math&gt;\gamma&lt;/math&gt;. &lt;math&gt;\phi\cdot f&lt;/math&gt; and &lt;math&gt;\gamma\cdot g&lt;/math&gt; are both perfect squares.<br /> <br /> ''Proof.'' &lt;math&gt;\phi\cdot f&lt;/math&gt; and &lt;math&gt;\gamma\cdot g&lt;/math&gt; are both perfect squares, so for &lt;math&gt;\phi\cdot \gamma&lt;/math&gt; to be a perfect square, if &lt;math&gt;g&lt;/math&gt; is greater than or equal to &lt;math&gt;f&lt;/math&gt;, &lt;math&gt;g/f&lt;/math&gt; must be a perfect square, too. Thus &lt;math&gt;g&lt;/math&gt; is &lt;math&gt;f&lt;/math&gt; times a square, but &lt;math&gt;g&lt;/math&gt; cannot divide any squares besides &lt;math&gt;1&lt;/math&gt;, so &lt;math&gt;g = 1f&lt;/math&gt;; &lt;math&gt;g = f&lt;/math&gt;. Similarly, if &lt;math&gt;f\geq g&lt;/math&gt;, then &lt;math&gt;f = g&lt;/math&gt; for our rules to keep working.<br /> <br /> '''End Lemma'''<br /> <br /> We can permute &lt;math&gt;l&lt;/math&gt; numbers with the same &lt;math&gt;p&lt;/math&gt; in &lt;math&gt;l!&lt;/math&gt; ways. We must have at least 67 numbers with a certain &lt;math&gt;p&lt;/math&gt; so our product will be divisible by 67. Obviously, then it will also be divisible by 2, 3, and 5, and thus 2010, as well. Toms as &lt;math&gt;h&lt;/math&gt;, in general, we need numbers all the way up to &lt;math&gt;h\cdot 67^2&lt;/math&gt;, so obviously, &lt;math&gt;67^2&lt;/math&gt; is the smallest such number such that we can get a &lt;math&gt;67!&lt;/math&gt; term; here 67 &lt;math&gt;p&lt;/math&gt; terms are 1. Thus we need the integers &lt;math&gt;1, 2, \ldots, 67^2&lt;/math&gt;, so &lt;math&gt;67^2&lt;/math&gt;, or &lt;math&gt;\boxed{4489}&lt;/math&gt;, is the answer.<br /> <br /> ==Solution Number Sense==<br /> We have to somehow calculate the number of permutations for a given &lt;math&gt;n&lt;/math&gt;. How in the world do we do this? Because we want squares, why not call a number &lt;math&gt;k=m*s^2&lt;/math&gt;, where &lt;math&gt;s&lt;/math&gt; is the largest square that allows &lt;math&gt;m&lt;/math&gt; to be non-square? &lt;math&gt;m=1&lt;/math&gt; is the only square &lt;math&gt;m&lt;/math&gt; can be, which only happens if &lt;math&gt;k&lt;/math&gt; is a perfect square.<br /> <br /> For example, &lt;math&gt;126 = 14 * 3^2&lt;/math&gt;, therefore in this case &lt;math&gt;k=126, m = 14, s = 3&lt;/math&gt;.<br /> <br /> I will call a permutation of the numbers &lt;math&gt;P&lt;/math&gt;, while the original &lt;math&gt;1, 2, 3, 4, ...&lt;/math&gt; I will call &lt;math&gt;S&lt;/math&gt;.<br /> <br /> Note that essentially we are looking at &quot;pairing up&quot; elements between &lt;math&gt;P&lt;/math&gt; and &lt;math&gt;S&lt;/math&gt; such that the product of &lt;math&gt;P_k&lt;/math&gt; and &lt;math&gt;S_k&lt;/math&gt; is a perfect square. How do we do this? Using the representation above.<br /> <br /> Each square has to have an even exponent of every prime represented in its prime factorization. Therefore, we can just take all exponents of the primes &lt;math&gt;mod 2&lt;/math&gt; and if there are any odd numbers, those are the ones we have to match- in effect, they are the &lt;math&gt;m&lt;/math&gt; numbers mentioned at the beginning.<br /> <br /> By listing the &lt;math&gt;m&lt;/math&gt; values, in my search for &quot;dumb&quot; or &quot;obvious&quot; ideas I am pretty confident that only values with identical &lt;math&gt;m&lt;/math&gt;s can be matched together. With such a solid idea let me prove it.<br /> <br /> If we were to &quot;pair up&quot; numbers with different &lt;math&gt;m&lt;/math&gt;s, take for example &lt;math&gt;S_{18}&lt;/math&gt; with an &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;P_{18}=26&lt;/math&gt; with an &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;26&lt;/math&gt;, note that their product gives a supposed &lt;math&gt;m&lt;/math&gt; of &lt;math&gt;13&lt;/math&gt; because the &lt;math&gt;2&lt;/math&gt; values cancel out. But then, what happens to the extra &lt;math&gt;13&lt;/math&gt; left? It doesn't make a square, contradiction. To finish up this easy proof, note that if a &quot;pair&quot; has different &lt;math&gt;m&lt;/math&gt; values, and the smaller one is &lt;math&gt;m_1&lt;/math&gt;, in order for the product to leave a square, the larger &lt;math&gt;m&lt;/math&gt; value has to have not just &lt;math&gt;m_1&lt;/math&gt; but another square inside it, which is absurd because we stipulated at the beginning that &lt;math&gt;m&lt;/math&gt; was square-free except for the trivial multiplication identity, 1.<br /> <br /> Now, how many ways are there to do this? If there are &lt;math&gt;c_1&lt;/math&gt; numbers with &lt;math&gt;m=1&lt;/math&gt;, there are clearly &lt;math&gt;(c_1)!&lt;/math&gt; ways of sorting them. The same goes for &lt;math&gt;m=2, 3, etc.&lt;/math&gt; by this logic. Note that the &lt;math&gt;P(n)&lt;/math&gt; as stated by the problem requires a &lt;math&gt;67&lt;/math&gt; thrown in there because &lt;math&gt;2010=2*3*5*67&lt;/math&gt;, so there has to be a &lt;math&gt;S_n&lt;/math&gt; with 67 elements with the same &lt;math&gt;m&lt;/math&gt;. It is evident that the smallest &lt;math&gt;n&lt;/math&gt; will occur when &lt;math&gt;m=1&lt;/math&gt;, because if &lt;math&gt;m&lt;/math&gt; is bigger we would have to expand &lt;math&gt;n&lt;/math&gt; to get the same number of &lt;math&gt;m&lt;/math&gt; values. Finally, realize that the only numbers with &lt;math&gt;m=1&lt;/math&gt; are square numbers! So our smallest &lt;math&gt;n=67^2=4489&lt;/math&gt;, and we are done.<br /> <br /> I relied on looking for patterns a lot in this problem. When faced with combo/number theory, it is always good to draw a sketch. Never be scared to try a problem on the USAJMO. It takes about 45 minutes. Well, it's 2010 and a number 1. Cheers!<br /> <br /> -expiLnCalc<br /> <br /> ==Solution Easy==<br /> <br /> Consider the set of numbers &lt;math&gt;m*k^2&lt;/math&gt; such that &lt;math&gt;m&lt;/math&gt; is not divisible by any squares other than 1 and &lt;math&gt;k = 1, 2, ...&lt;/math&gt; By changing &lt;math&gt;m&lt;/math&gt; we can encompass all numbers less than or equal to &lt;math&gt;n&lt;/math&gt;. Now notice that for a working arrangement these numbers can be permutated in any way to create a new one; for instance, the numbers &lt;math&gt;2,8,18&lt;/math&gt; have &lt;math&gt;m = 2&lt;/math&gt; and they can be arranged in &lt;math&gt;3! = 6&lt;/math&gt; ways. Thus, since &lt;math&gt;2010 = 2*3*5*67&lt;/math&gt; we need a permutation to have at least 67 elements (since 67 is prime). To minimize &lt;math&gt;n&lt;/math&gt;, we let &lt;math&gt;m = 1&lt;/math&gt; and we have &lt;math&gt;1^2,2^2,3^2,...67^2&lt;/math&gt; and we stop at &lt;math&gt;67^2&lt;/math&gt; to get &lt;math&gt;\boxed{4489}&lt;/math&gt;. ~Leonard_my_dude~<br /> <br /> ==Solution 5==<br /> <br /> It's well known that there exists &lt;math&gt;f(n)&lt;/math&gt; and &lt;math&gt;g(n)&lt;/math&gt; s.t. &lt;math&gt;n = f(n) \cdot g(n)&lt;/math&gt;, s.t. no square divides &lt;math&gt;f(n)&lt;/math&gt; other than 1, and &lt;math&gt;g(n)&lt;/math&gt; is a perfect square.<br /> <br /> Lemma: &lt;math&gt;k \cdot a_k&lt;/math&gt; is a perfect square if and only if &lt;math&gt;f(k) = f(a_k)&lt;/math&gt; <br /> We prove the backwards direction first. If &lt;math&gt;f(k) = f(a_k)&lt;/math&gt;, &lt;math&gt;k \cdot a_k = f(k) \cdot g(k) \cdot f(a_k) \cdot g(a_k) = f(k)^2 \cdot g(k) \cdot g(a_k)&lt;/math&gt;, which is a perfect square. <br /> <br /> We will now prove the forwards direction. We will prove the contrapositive: If &lt;math&gt;f(k) \neq f(a_k)&lt;/math&gt;, &lt;math&gt;k \cdot a_k&lt;/math&gt; is not a perfect square. Note that if &lt;math&gt;f(k) \neq f(a_k)&lt;/math&gt;, There exists a prime p, s.t. &lt;math&gt;v_p(k) \neq v_p(a_k)&lt;/math&gt;. Also, &lt;math&gt;v_p(k), v_p(a_k) \leq 1&lt;/math&gt;. Thus, &lt;math&gt;v_p(k \cdot a_k) = 1&lt;/math&gt;, making &lt;math&gt;k \cdot a_k&lt;/math&gt; not a square.<br /> <br /> Thus, we can only match k with a_k if they have the same f value. Thus, to find P(k), we can do it by f value, permuting the a_k with f value 1, then 2, ... Thus, our answer is:<br /> <br /> &lt;math&gt;P(n) = \prod_{1 \leq i \leq n, g(i) = 1} \left\lfloor \sqrt{\frac{n}{i}} \right \rfloor !&lt;/math&gt; <br /> <br /> For all &lt;math&gt;n &lt; 67^2&lt;/math&gt;, &lt;math&gt;P(n)&lt;/math&gt; doesn't have a factor of 67. However, if &lt;math&gt;n = 67^2&lt;/math&gt;, the first term will be a multiple of 2010, and thus the answer is &lt;math&gt;67^2 = \boxed{4489}&lt;/math&gt;<br /> <br /> {{alternate solutions}}<br /> <br /> == See Also ==<br /> * &lt;url&gt;viewtopic.php?t=347303 Discussion on AoPS/MathLinks&lt;/url&gt;<br /> <br /> {{USAJMO newbox|year=2010|before=First Question|num-a=2}}<br /> <br /> [[Category:Olympiad Number Theory Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_2&diff=123542 1998 AIME Problems/Problem 2 2020-06-03T04:30:55Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> Find the number of [[ordered pair]]s &lt;math&gt;(x,y)&lt;/math&gt; of positive integers that satisfy &lt;math&gt;x \le 2y \le 60&lt;/math&gt; and &lt;math&gt;y \le 2x \le 60&lt;/math&gt;.<br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> [[Image:AIME_1998-2.png|thumb|300px|right]]<br /> [[Pick's theorem]] states that:<br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;A = I + \frac B2 - 1&lt;/math&gt;&lt;/div&gt;<br /> <br /> The conditions give us four [[inequality|inequalities]]: &lt;math&gt;x \le 30&lt;/math&gt;, &lt;math&gt;y\le 30&lt;/math&gt;, &lt;math&gt;x \le 2y&lt;/math&gt;, &lt;math&gt;y \le 2x&lt;/math&gt;. These create a [[quadrilateral]], whose area is &lt;math&gt;\frac 12&lt;/math&gt; of the 30 by 30 [[square]] it is in. A simple way to see this is to note that the two triangles outside of the quadrilateral form half of the area of the 30 by 30 square.<br /> <br /> So &lt;math&gt;A = \frac 12 \cdot 30^2 = 450&lt;/math&gt;. &lt;math&gt;B&lt;/math&gt; we can calculate by just counting. Ignoring the vertices, the top and right sides have 14 [[lattice point]]s, and the two diagonals each have 14 lattice points (for the top diagonal, every value of &lt;math&gt;x&lt;/math&gt; corresponds with an integer value of &lt;math&gt;y&lt;/math&gt; as &lt;math&gt;y = 2x&lt;/math&gt; and vice versa for the bottom, so and there are 14 values for x not counting vertices). Adding the four vertices, there are 60 points on the borders.<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;450 = I + \frac {60}2 - 1&lt;/math&gt;&lt;br /&gt;&lt;math&gt;I = 421&lt;/math&gt;&lt;/div&gt;<br /> <br /> Since the inequalities also include the equals case, we include the boundaries, which gives us &lt;math&gt;421 + 60 = 481&lt;/math&gt; ordered pairs. However, the question asks us for positive integers, so &lt;math&gt;(0,0)&lt;/math&gt; doesn't count; hence, the answer is &lt;math&gt;480&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> First, note that all pairs of the form &lt;math&gt;(a,a)&lt;/math&gt;, &lt;math&gt;1\le a\le30&lt;/math&gt; work.<br /> <br /> Now, considered the ordered pairs with &lt;math&gt;x &lt; y&lt;/math&gt;, so that &lt;math&gt;x &lt; 2y&lt;/math&gt; is automatically satisfied. Since &lt;math&gt;x &lt; y\le 2x&lt;/math&gt;, there are &lt;math&gt;2x - x = x&lt;/math&gt; possible values of &lt;math&gt;y&lt;/math&gt;. Hence, given &lt;math&gt;x&lt;/math&gt;, there are &lt;math&gt;x&lt;/math&gt; values of possible &lt;math&gt;y&lt;/math&gt; for which &lt;math&gt;x &lt; y&lt;/math&gt; and the above conditions are satisfied. But &lt;math&gt;2y\le60&lt;/math&gt;, so this only works for &lt;math&gt;x\le15&lt;/math&gt;. Thus, there are<br /> <br /> &lt;math&gt;\sum_{i=1}^{15} i=\frac{(15)(16)}{2}&lt;/math&gt;<br /> <br /> ordered pairs. For &lt;math&gt;x &gt; 15&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; must follow &lt;math&gt;x &lt; y\le 30&lt;/math&gt;. Hence, there are &lt;math&gt;30 - x&lt;/math&gt; possibilities for &lt;math&gt;y&lt;/math&gt;, and there are<br /> <br /> &lt;math&gt;\sum_{i=16}^{30}(30-i)=\sum_{i=0}^{14}i=\frac{(14)(15)}{2}&lt;/math&gt;<br /> <br /> ordered pairs.<br /> <br /> By symmetry, there are also &lt;math&gt;\frac {(15)(16)}{2} + \frac {(14)(15)}{2}&lt;/math&gt; ordered pairs with &lt;math&gt;x &gt; y&lt;/math&gt; and the above criteria satisfied.<br /> <br /> Hence, the total is<br /> <br /> &lt;math&gt;\frac{(15)(16)}{2}+\frac{(14)(15)}{2}+\frac{(15)(16)}{2}+\frac{(14)(15)}{2}+30=480.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> &lt;math&gt; y\le2x\le60&lt;/math&gt;<br /> <br /> Multiplying both sides by 2 yields:<br /> <br /> &lt;math&gt; 2y\le4x\le120&lt;/math&gt;<br /> <br /> Then the two inequalities can be merged to form the following inequality:<br /> <br /> &lt;math&gt; x\le2y\le4x\le120&lt;/math&gt;<br /> <br /> Additionally, we must ensure that &lt;math&gt;2y&lt;60&lt;/math&gt;<br /> <br /> Therefore we must find pairs &lt;math&gt;(x,y)&lt;/math&gt; that satisfy the inequality above. <br /> A bit of trial and error and observing patterns leads to the answer &lt;math&gt;480&lt;/math&gt;.<br /> <br /> It should be noted that the cases for &lt;math&gt;x\le15&lt;/math&gt; and &lt;math&gt;x&gt;15&lt;/math&gt; should be considered separately in order to ensure that &lt;math&gt;2y &lt; 60&lt;/math&gt;.<br /> <br /> ===Solution 4 - Unrigorous engineers induction solution===<br /> <br /> We will try out small cases.<br /> <br /> By replacing 60 in this problem with 2, we count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. With 6, we get 7 pairs. With 8 we get 12. By continuing on, and then finding the difference between adjacent terms (1,3,3,5,5,...). We suspect that if 60 was replaced with 2n, we will find 1+3+3+5+5+7+7 ...., where there will be n terms. Thus, our answer is 1+3+3+5+5.... 29+29+31 = 16*30 = 480.<br /> <br /> -Alexlikemath<br /> <br /> == See also ==<br /> {{AIME box|year=1998|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_2&diff=123541 1998 AIME Problems/Problem 2 2020-06-03T04:27:19Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> Find the number of [[ordered pair]]s &lt;math&gt;(x,y)&lt;/math&gt; of positive integers that satisfy &lt;math&gt;x \le 2y \le 60&lt;/math&gt; and &lt;math&gt;y \le 2x \le 60&lt;/math&gt;.<br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> [[Image:AIME_1998-2.png|thumb|300px|right]]<br /> [[Pick's theorem]] states that:<br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;A = I + \frac B2 - 1&lt;/math&gt;&lt;/div&gt;<br /> <br /> The conditions give us four [[inequality|inequalities]]: &lt;math&gt;x \le 30&lt;/math&gt;, &lt;math&gt;y\le 30&lt;/math&gt;, &lt;math&gt;x \le 2y&lt;/math&gt;, &lt;math&gt;y \le 2x&lt;/math&gt;. These create a [[quadrilateral]], whose area is &lt;math&gt;\frac 12&lt;/math&gt; of the 30 by 30 [[square]] it is in. A simple way to see this is to note that the two triangles outside of the quadrilateral form half of the area of the 30 by 30 square.<br /> <br /> So &lt;math&gt;A = \frac 12 \cdot 30^2 = 450&lt;/math&gt;. &lt;math&gt;B&lt;/math&gt; we can calculate by just counting. Ignoring the vertices, the top and right sides have 14 [[lattice point]]s, and the two diagonals each have 14 lattice points (for the top diagonal, every value of &lt;math&gt;x&lt;/math&gt; corresponds with an integer value of &lt;math&gt;y&lt;/math&gt; as &lt;math&gt;y = 2x&lt;/math&gt; and vice versa for the bottom, so and there are 14 values for x not counting vertices). Adding the four vertices, there are 60 points on the borders.<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;450 = I + \frac {60}2 - 1&lt;/math&gt;&lt;br /&gt;&lt;math&gt;I = 421&lt;/math&gt;&lt;/div&gt;<br /> <br /> Since the inequalities also include the equals case, we include the boundaries, which gives us &lt;math&gt;421 + 60 = 481&lt;/math&gt; ordered pairs. However, the question asks us for positive integers, so &lt;math&gt;(0,0)&lt;/math&gt; doesn't count; hence, the answer is &lt;math&gt;480&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> First, note that all pairs of the form &lt;math&gt;(a,a)&lt;/math&gt;, &lt;math&gt;1\le a\le30&lt;/math&gt; work.<br /> <br /> Now, considered the ordered pairs with &lt;math&gt;x &lt; y&lt;/math&gt;, so that &lt;math&gt;x &lt; 2y&lt;/math&gt; is automatically satisfied. Since &lt;math&gt;x &lt; y\le 2x&lt;/math&gt;, there are &lt;math&gt;2x - x = x&lt;/math&gt; possible values of &lt;math&gt;y&lt;/math&gt;. Hence, given &lt;math&gt;x&lt;/math&gt;, there are &lt;math&gt;x&lt;/math&gt; values of possible &lt;math&gt;y&lt;/math&gt; for which &lt;math&gt;x &lt; y&lt;/math&gt; and the above conditions are satisfied. But &lt;math&gt;2y\le60&lt;/math&gt;, so this only works for &lt;math&gt;x\le15&lt;/math&gt;. Thus, there are<br /> <br /> &lt;math&gt;\sum_{i=1}^{15} i=\frac{(15)(16)}{2}&lt;/math&gt;<br /> <br /> ordered pairs. For &lt;math&gt;x &gt; 15&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; must follow &lt;math&gt;x &lt; y\le 30&lt;/math&gt;. Hence, there are &lt;math&gt;30 - x&lt;/math&gt; possibilities for &lt;math&gt;y&lt;/math&gt;, and there are<br /> <br /> &lt;math&gt;\sum_{i=16}^{30}(30-i)=\sum_{i=0}^{14}i=\frac{(14)(15)}{2}&lt;/math&gt;<br /> <br /> ordered pairs.<br /> <br /> By symmetry, there are also &lt;math&gt;\frac {(15)(16)}{2} + \frac {(14)(15)}{2}&lt;/math&gt; ordered pairs with &lt;math&gt;x &gt; y&lt;/math&gt; and the above criteria satisfied.<br /> <br /> Hence, the total is<br /> <br /> &lt;math&gt;\frac{(15)(16)}{2}+\frac{(14)(15)}{2}+\frac{(15)(16)}{2}+\frac{(14)(15)}{2}+30=480.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> &lt;math&gt; y\le2x\le60&lt;/math&gt;<br /> <br /> Multiplying both sides by 2 yields:<br /> <br /> &lt;math&gt; 2y\le4x\le120&lt;/math&gt;<br /> <br /> Then the two inequalities can be merged to form the following inequality:<br /> <br /> &lt;math&gt; x\le2y\le4x\le120&lt;/math&gt;<br /> <br /> Additionally, we must ensure that &lt;math&gt;2y&lt;60&lt;/math&gt;<br /> <br /> Therefore we must find pairs &lt;math&gt;(x,y)&lt;/math&gt; that satisfy the inequality above. <br /> A bit of trial and error and observing patterns leads to the answer &lt;math&gt;480&lt;/math&gt;.<br /> <br /> It should be noted that the cases for &lt;math&gt;x\le15&lt;/math&gt; and &lt;math&gt;x&gt;15&lt;/math&gt; should be considered separately in order to ensure that &lt;math&gt;2y &lt; 60&lt;/math&gt;.<br /> <br /> ==Solution 4 - Unrigorous engineers induction solution==<br /> <br /> We will try out small cases.<br /> <br /> By replacing 60 in this problem with 2, we count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. By continuing on, we suspect that if 60 was replaced with 2n, we will find 1+3+3+5+5+7+7 ...., where there will be n terms. Thus, our answer is 1+3+3+5+5.... 29+29+31 = 16*30 = 480.<br /> <br /> == See also ==<br /> {{AIME box|year=1998|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=1998_AIME_Problems/Problem_2&diff=123540 1998 AIME Problems/Problem 2 2020-06-03T04:18:27Z <p>Alexlikemath: </p> <hr /> <div>== Problem ==<br /> Find the number of [[ordered pair]]s &lt;math&gt;(x,y)&lt;/math&gt; of positive integers that satisfy &lt;math&gt;x \le 2y \le 60&lt;/math&gt; and &lt;math&gt;y \le 2x \le 60&lt;/math&gt;.<br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> [[Image:AIME_1998-2.png|thumb|300px|right]]<br /> [[Pick's theorem]] states that:<br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;A = I + \frac B2 - 1&lt;/math&gt;&lt;/div&gt;<br /> <br /> The conditions give us four [[inequality|inequalities]]: &lt;math&gt;x \le 30&lt;/math&gt;, &lt;math&gt;y\le 30&lt;/math&gt;, &lt;math&gt;x \le 2y&lt;/math&gt;, &lt;math&gt;y \le 2x&lt;/math&gt;. These create a [[quadrilateral]], whose area is &lt;math&gt;\frac 12&lt;/math&gt; of the 30 by 30 [[square]] it is in. A simple way to see this is to note that the two triangles outside of the quadrilateral form half of the area of the 30 by 30 square.<br /> <br /> So &lt;math&gt;A = \frac 12 \cdot 30^2 = 450&lt;/math&gt;. &lt;math&gt;B&lt;/math&gt; we can calculate by just counting. Ignoring the vertices, the top and right sides have 14 [[lattice point]]s, and the two diagonals each have 14 lattice points (for the top diagonal, every value of &lt;math&gt;x&lt;/math&gt; corresponds with an integer value of &lt;math&gt;y&lt;/math&gt; as &lt;math&gt;y = 2x&lt;/math&gt; and vice versa for the bottom, so and there are 14 values for x not counting vertices). Adding the four vertices, there are 60 points on the borders.<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;450 = I + \frac {60}2 - 1&lt;/math&gt;&lt;br /&gt;&lt;math&gt;I = 421&lt;/math&gt;&lt;/div&gt;<br /> <br /> Since the inequalities also include the equals case, we include the boundaries, which gives us &lt;math&gt;421 + 60 = 481&lt;/math&gt; ordered pairs. However, the question asks us for positive integers, so &lt;math&gt;(0,0)&lt;/math&gt; doesn't count; hence, the answer is &lt;math&gt;480&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> First, note that all pairs of the form &lt;math&gt;(a,a)&lt;/math&gt;, &lt;math&gt;1\le a\le30&lt;/math&gt; work.<br /> <br /> Now, considered the ordered pairs with &lt;math&gt;x &lt; y&lt;/math&gt;, so that &lt;math&gt;x &lt; 2y&lt;/math&gt; is automatically satisfied. Since &lt;math&gt;x &lt; y\le 2x&lt;/math&gt;, there are &lt;math&gt;2x - x = x&lt;/math&gt; possible values of &lt;math&gt;y&lt;/math&gt;. Hence, given &lt;math&gt;x&lt;/math&gt;, there are &lt;math&gt;x&lt;/math&gt; values of possible &lt;math&gt;y&lt;/math&gt; for which &lt;math&gt;x &lt; y&lt;/math&gt; and the above conditions are satisfied. But &lt;math&gt;2y\le60&lt;/math&gt;, so this only works for &lt;math&gt;x\le15&lt;/math&gt;. Thus, there are<br /> <br /> &lt;math&gt;\sum_{i=1}^{15} i=\frac{(15)(16)}{2}&lt;/math&gt;<br /> <br /> ordered pairs. For &lt;math&gt;x &gt; 15&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; must follow &lt;math&gt;x &lt; y\le 30&lt;/math&gt;. Hence, there are &lt;math&gt;30 - x&lt;/math&gt; possibilities for &lt;math&gt;y&lt;/math&gt;, and there are<br /> <br /> &lt;math&gt;\sum_{i=16}^{30}(30-i)=\sum_{i=0}^{14}i=\frac{(14)(15)}{2}&lt;/math&gt;<br /> <br /> ordered pairs.<br /> <br /> By symmetry, there are also &lt;math&gt;\frac {(15)(16)}{2} + \frac {(14)(15)}{2}&lt;/math&gt; ordered pairs with &lt;math&gt;x &gt; y&lt;/math&gt; and the above criteria satisfied.<br /> <br /> Hence, the total is<br /> <br /> &lt;math&gt;\frac{(15)(16)}{2}+\frac{(14)(15)}{2}+\frac{(15)(16)}{2}+\frac{(14)(15)}{2}+30=480.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> &lt;math&gt; y\le2x\le60&lt;/math&gt;<br /> <br /> Multiplying both sides by 2 yields:<br /> <br /> &lt;math&gt; 2y\le4x\le120&lt;/math&gt;<br /> <br /> Then the two inequalities can be merged to form the following inequality:<br /> <br /> &lt;math&gt; x\le2y\le4x\le120&lt;/math&gt;<br /> <br /> Additionally, we must ensure that &lt;math&gt;2y&lt;60&lt;/math&gt;<br /> <br /> Therefore we must find pairs &lt;math&gt;(x,y)&lt;/math&gt; that satisfy the inequality above. <br /> A bit of trial and error and observing patterns leads to the answer &lt;math&gt;480&lt;/math&gt;.<br /> <br /> It should be noted that the cases for &lt;math&gt;x\le15&lt;/math&gt; and &lt;math&gt;x&gt;15&lt;/math&gt; should be considered separately in order to ensure that &lt;math&gt;2y &lt; 60&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> We will try out small cases.<br /> <br /> By replacing 60 in this problem with 2, we count only 1 ordered pair. By doing with 4, we count 4 ordered pairs. By continuing on, we suspect that if 60 was replaced with 2n, we will find 1+3+3+5+5+7+7 ...., where there will be n terms. Thus, our answer is 1+3+3+5+5.... 29+29+31 = 16*30 = 480.<br /> <br /> == See also ==<br /> {{AIME box|year=1998|num-b=1|num-a=3}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Alexlikemath https://artofproblemsolving.com/wiki/index.php?title=2017_AIME_II_Problems/Problem_9&diff=122942 2017 AIME II Problems/Problem 9 2020-05-25T19:34:44Z <p>Alexlikemath: Solution 6</p> <hr /> <div>==Problem==<br /> A special deck of cards contains &lt;math&gt;49&lt;/math&gt; cards, each labeled with a number from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;7&lt;/math&gt; and colored with one of seven colors. Each number-color combination appears on exactly one card. Sharon will select a set of eight cards from the deck at random. Given that she gets at least one card of each color and at least one card with each number, the probability that Sharon can discard one of her cards and &lt;math&gt;\textit{still}&lt;/math&gt; have at least one card of each color and at least one card with each number is &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. Find &lt;math&gt;p+q&lt;/math&gt;.<br /> <br /> ==Solution 1==<br /> Without loss of generality, assume that the &lt;math&gt;8&lt;/math&gt; numbers on Sharon's cards are &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;1&lt;/math&gt;, &lt;math&gt;2&lt;/math&gt;, &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;4&lt;/math&gt;, &lt;math&gt;5&lt;/math&gt;, &lt;math&gt;6&lt;/math&gt;, and &lt;math&gt;7&lt;/math&gt;, in that order, and assume the &lt;math&gt;8&lt;/math&gt; colors are red, red, and six different arbitrary colors. There are &lt;math&gt;{8\choose2}-1&lt;/math&gt; ways of assigning the two red cards to the &lt;math&gt;8&lt;/math&gt; numbers; we subtract &lt;math&gt;1&lt;/math&gt; because we cannot assign the two reds to the two &lt;math&gt;1&lt;/math&gt;'s.<br /> In order for Sharon to be able to remove at least one card and still have at least one card of each color, one of the reds have to be assigned with one of the &lt;math&gt;1&lt;/math&gt;s. The number of ways for this to happen is &lt;math&gt;2 \cdot 6 = 12&lt;/math&gt;. Each of these assignments is equally likely, so desired probability is &lt;math&gt;\frac{12}{{8\choose2}-1}=\frac{4}{9} \implies 4 + 9 = 13 = \boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> First note that out of the &lt;math&gt;8&lt;/math&gt; selected cards, one pair of cards have to share the same number and another pair of cards have to share the same color. Now, these &lt;math&gt;2&lt;/math&gt; pairs of cards can't be the same or else there will be &lt;math&gt;2&lt;/math&gt; cards which are completely same. Then, WLOG let the numbers be &lt;math&gt;1,1,2,3,4,5,6,&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; and the colors be &lt;math&gt;a,a,b,c,d,e,f,&lt;/math&gt; and &lt;math&gt;g&lt;/math&gt;. We therefore obtain only &lt;math&gt;2&lt;/math&gt; cases:<br /> <br /> Case One: &lt;math&gt;1a,1b,2a,3c,4d,5e,6f,&lt;/math&gt; and &lt;math&gt;7g&lt;/math&gt;<br /> In this case, we can discard &lt;math&gt;1a&lt;/math&gt;.<br /> There are &lt;math&gt;2*6=12&lt;/math&gt; situations in this case.<br /> <br /> Case Two: &lt;math&gt;1b,1c,2a,3a,4d,5e,6f,&lt;/math&gt; and &lt;math&gt;7g&lt;/math&gt;<br /> In this case, we can't discard.<br /> There are &lt;math&gt;\dbinom{6}{2}=15&lt;/math&gt; situations in this case.<br /> <br /> So the probability is &lt;math&gt;\frac{12}{12+15}=\frac{4}{9}&lt;/math&gt;, giving us the answer of &lt;math&gt;4+9=\boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> There are &lt;math&gt;7!&lt;/math&gt; ways to choose a set of 7 cards that have all the numbers from 1-7 and all 7 colors. There are then &lt;math&gt;42&lt;/math&gt; cards remaining. Thus, there are &lt;math&gt;7!(42)&lt;/math&gt; desired sets.<br /> <br /> Now, the next thing to find is the number of ways to choose 8 cards where there is not a set of 7 such cards. In this case, one color must have 2 cards and one number must have 2 cards, and they can't be the same number/color card. The number of ways to pick this is equal to a multiplication of &lt;math&gt;\binom{7}{2}&lt;/math&gt; ways to pick 2 numbers, &lt;math&gt;7&lt;/math&gt; colors to assign them to, &lt;math&gt;\binom{6}{2}&lt;/math&gt; ways to pick 2 nonchosen colors, &lt;math&gt;5&lt;/math&gt; ways to pick a number to assign them to, and &lt;math&gt;4!&lt;/math&gt; ways to assign the rest.<br /> <br /> Thus, the answer is &lt;math&gt;\frac{7!(42)}{7!(42) + 21(15)(7)(5!)}&lt;/math&gt;. Dividing out &lt;math&gt;5!&lt;/math&gt; yields &lt;math&gt;\frac{42(42)}{42(42) + 21(15)(7)}&lt;/math&gt; which is equal to &lt;math&gt;\frac{2(42)}{2(42) + 15(7)}&lt;/math&gt; which is equal to &lt;math&gt;\frac{12}{12 + 15}&lt;/math&gt; which is equal to &lt;math&gt;\frac{4}{9}&lt;/math&gt; giving a final answer of &lt;math&gt;\boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> We can rewrite the problem as &quot;What is the probability that, given 8 cards with numbers a, b, c, d, e, f, g, g and some assortment of seven colors A, B, C, D, E, F, G, G where G is repeated, one of the cards is Gg?&quot; Note that this is a valid restatement because Sharon has to have two of one number and two of one color. She needs to be able to take away one of the cards with the duplicate number, but this also has to have the duplicate color. There are two cases.<br /> <br /> Case I: One of the cards is Gg. This implies that the other card with color G can be placed in &lt;math&gt;6&lt;/math&gt; ways, and the rest of the colors can be paired with cards in &lt;math&gt;6!&lt;/math&gt; ways. <br /> <br /> Case II: None of the cards are Gg. This implies that the cards with color G can be chosen in &lt;math&gt;\dbinom{6}{2}=15&lt;/math&gt; ways, and the rest of the colors can be paired with cards in &lt;math&gt;\frac{6!}{2}&lt;/math&gt; ways, with the divide by 2 because of the double-g.<br /> <br /> Note that there is no case with Gg, Gg because all 49 cards are unique!<br /> <br /> Therefore our answer is &lt;math&gt;\frac{6\times6!}{\frac{15}{2}\times6! + 6\times6!}=\frac{4}{9}&lt;/math&gt;, so &lt;math&gt;\boxed{013}&lt;/math&gt;.<br /> <br /> ==Solution 5==<br /> We count the two entities: The number of sets of eight cards that contain all seven numbers and all seven colors; and The number of sets of eight cards that contain all seven numbers and colors, and one card can be removed and the property still holds.<br /> <br /> For the first: It is equivalent to counting:<br /> How many ways can (colors) &lt;math&gt;A, A, B, C, D, E, F, G&lt;/math&gt; be matched to (numbers) &lt;math&gt;1, 1, 2, 3, 4, 5, 6, 7&lt;/math&gt; such that the two &lt;math&gt;A's&lt;/math&gt; cannot both be matched to the two ones. (This would mean we've chosen two identical cards from the deck, both with color &lt;math&gt;A&lt;/math&gt; and number one.)<br /> We consider two cases: <br /> <br /> 1) Neither &lt;math&gt;A&lt;/math&gt; is matched with a one. Then there are &lt;math&gt;\binom{6}{2}&lt;/math&gt; ways to choose which two numbers are matched with the &lt;math&gt;A's,&lt;/math&gt; and &lt;math&gt;6! / 2!&lt;/math&gt; ways to order the remaining numbers. (There are two one's, so we divide by two.)<br /> <br /> 2) One of the &lt;math&gt;A&lt;/math&gt;'s is matched with a one. There are &lt;math&gt;6 \cdot 6!&lt;/math&gt; ways to arrange the remaining color<br /> <br /> Since there are seven ways to choose the extra color and likewise for the extra number, <br /> we have in total &lt;math&gt;7 \cdot 7 \cdot \left( 6C2 \cdot \frac{6!}{2!} + 6 \cdot 6! \right)&lt;/math&gt; ways to choose eight cards satisfying the first condition.<br /> <br /> For the second: View the cards as rooks on a seven by seven chess board. Seven cards that contain all colors and numbers is can be represented as placing seven rooks on the board so that no two rooks are attacking each other. There are &lt;math&gt;7!&lt;/math&gt; ways to do this. For each arrangement, we can choose any of the other 42 positions on the board (cards) to add an extra rook. Thus, there are &lt;math&gt;7! \cdot 42&lt;/math&gt; good sets of eight cards. Note that we are not overcounting.<br /> <br /> Thus, the probability of getting a good set given the first condition is &lt;math&gt;\frac{ 7! \cdot 42 }{ 7 \cdot 7 \cdot \left( 6C2 \cdot \frac{6!}{2!} + 6 \cdot 6! \right) } = \frac{4}{9},&lt;/math&gt; and the answer is &lt;math&gt;\boxed{013}.&lt;/math&gt;<br /> <br /> ==Solution 6==<br /> We recast this problem to coloring 8 cells in a 7x7 grid, such that each row and column has at least one colored cell. If a cell in row a and column b is colored, that means we drew a card with color a and number b. <br /> <br /> By pigeonhole principle, there exists a row that has 2 colored cells, and intuitively, that row is unique. There also exists a column with 2 colored cells. Call the unique row, &quot;Row A&quot;, and Call the unique column, &quot;Column B&quot;.<br /> <br /> If we are able to remove a card so that we still have a card of each color and number, that means, we can erase a cell, so each row still has at least one colored cell, and so does each column. If we erase a cell not in Row A, then the erased cell's row will be empty. If we erase a cell not in Column B, the erased cell's column will be empty. Thus the cell that we erase must be in both Row A and Column B. There is only one cell in both Row A and Column B, and thus that cell must be colored.<br /> <br /> We first count the number of successful outcomes. We must chose which row to be &quot;Row A&quot;, and which column to be &quot;Column B&quot;. There are 7 ways to chose &quot;Row A&quot;, and 7 ways to chose &quot;Column B&quot;. Because of the logic used above, the intersection of Row A and Column B must be colored. However, Row A needs to have 2 colored cells. There are 6 other uncolored cells to chose from. Similarly, there are 6 uncolored cells in Column B. <br /> <br /> Note that we can ignore all the rows and columns that contain colored cells. This is because Row A already contains 2 colored cells, and thus we can't color any more cells in Row A, similarly for Column B. The other row that contains a colored cell is not Row A, so it can only contain 1 colored cell, which it has, thus we can't color anymore cells in it. Similarly for the other column. <br /> <br /> We can actually remove these rows/columns. This leaves a 5x5 grid, full of uncolored cells. Each row can have 1 cell, and so can each column. We count the # of colorings from the perspective of the rows. We look at the first row, it has 5 choices. We can look at the 2nd row, it has 4 choices, as it's selected cell can't occupy the same column. We continue down the line, and see there are &lt;math&gt;5! = 120&lt;/math&gt; ways to color.<br /> <br /> Thus, the # of successful outcomes is &lt;math&gt;5! \cdot 6 \cdot 6 \cdot 7 \cdot 7&lt;/math&gt;.<br /> <br /> We can then count the # of possible outcomes. <br /> <br /> We can count the # of unsuccessful outcomes. We select &quot;Row A&quot; and &quot;Column B&quot; as usual, with &lt;math&gt;7 \cdot 7&lt;/math&gt; ways. We don't color the intersection, so we have to color 2 cells in each. There are 6 other cells that can be colored in each, so we have &lt;math&gt;{6 \choose 2} = 15&lt;/math&gt; ways to color each. Then, we can get rid of rows and columns using the method above and we are left with a 4x4 grid. There are &lt;math&gt;4! = 24&lt;/math&gt; ways to color. <br /> <br /> Thus, the # of unsuccessful outcomes is &lt;math&gt;4! \cdot 15 \cdot 15 \cdot 7 \cdot 7&lt;/math&gt;.<br /> <br /> After some calculation, we get &lt;math&gt;\frac{successful}{possible} = \frac{successful}{successful+unsuccessful} = \frac{5! \cdot 6^2 \cdot 7^2}{7^2 \cdot(5! \cdot 6^2 + 4! \cdot 15^2} = \frac{5! \cdot 6^2}{5! \cdot 6^2 +4! \cdot 15^2} = \frac{5 \cdot 6^2}{5 \cdot 6^2 + 15^2} = \frac{180}{180+225} = \frac{4}{9} =&gt; 4+9 = \boxed{13}&lt;/math&gt;<br /> <br /> -Alexlikemath<br /> <br /> =See Also=<br /> {{AIME box|year=2017|n=II|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Alexlikemath