https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Bronzetruck2016&feedformat=atom AoPS Wiki - User contributions [en] 2021-06-20T00:23:15Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2017_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_1&diff=151976 2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1 2021-04-19T22:21:12Z <p>Bronzetruck2016: /* Solution */</p> <hr /> <div><br /> <br /> == Problem ==<br /> <br /> What are the last two digits of &lt;math&gt;2017^{2017}&lt;/math&gt;?<br /> <br /> == Solution==<br /> <br /> In this problem, we will use the Chinese remainder theorem. This question is asking us to find &lt;math&gt;2017^{2017}\mod 100=17^{2017}\mod 100&lt;/math&gt;. By the Chinese remainder theorem, we can find &lt;math&gt;17^{2017}\mod 25&lt;/math&gt; and &lt;math&gt;17^{2017}\mod 4&lt;/math&gt;, and then &quot;combine&quot; them. &lt;math&gt;17^{2017}\mod4\equiv1^{2017}\mod4\equiv1\mod4&lt;/math&gt;. To find &lt;math&gt;17^{2017}\mod4&lt;/math&gt;, we'll look for a pattern. The pattern is &lt;math&gt;17, 14, 13, 21, 7, 19, 23, 16, 22, -1, -17, -14, -13, -21, -7, -19, -23, -16, -22, 1&lt;/math&gt;. Since there are 20 terms, and &lt;math&gt;2017\equiv17\mod20&lt;/math&gt;, we have that &lt;math&gt;17^{2017}\mod25\equiv2\mod25&lt;/math&gt;. Now, we want to find a number that's &lt;math&gt;2\mod25&lt;/math&gt; and &lt;math&gt;1\mod4&lt;/math&gt;. We find this number by clever guess and check. We know our number is odd, so it is &lt;math&gt;27\mod50&lt;/math&gt;. First, we try &lt;math&gt;27&lt;/math&gt;, but that doesn't work. Next, we try &lt;math&gt;77&lt;/math&gt;, and that works! So, our answer is &lt;cmath&gt;\boxed{77}&lt;/cmath&gt;<br /> &lt;i&gt;-bronzetruck2016&lt;i&gt;<br /> <br /> == See also ==<br /> {{UNM-PNM Math Contest box|year=2017|n=II|before=First question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2017_UNM-PNM_Statewide_High_School_Mathematics_Contest_II_Problems/Problem_1&diff=151974 2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 1 2021-04-19T22:20:27Z <p>Bronzetruck2016: /* Solution */</p> <hr /> <div><br /> <br /> == Problem ==<br /> <br /> What are the last two digits of &lt;math&gt;2017^{2017}&lt;/math&gt;?<br /> <br /> == Solution==<br /> <br /> In this problem, we will use the Chinese remainder theorem. This question is asking us to find &lt;math&gt;2017^{2017}\mod 100=17^{2017}\mod 100&lt;/math&gt;. By the Chinese remainder theorem, we can find &lt;math&gt;17^{2017}\mod 25&lt;/math&gt; and &lt;math&gt;17^{2017}\mod 4&lt;/math&gt;, and then &quot;combine&quot; them. &lt;math&gt;17^{2017}\mod4\equiv1^{2017}\mod4\equiv1\mod4&lt;/math&gt;. To find &lt;math&gt;17^{2017}\mod4&lt;/math&gt;, we'll look for a pattern. The pattern is &lt;math&gt;17, 14, 13, 21, 7, 19, 23, 16, 22, -1, -17, -14, -13, -21, -7, -19, -23, -16, -22, 1&lt;/math&gt;. Since there are 20 terms, and &lt;math&gt;2017\equiv17\mod20&lt;/math&gt;, we have that &lt;math&gt;17^{2017}\mod25\equiv2\mod25&lt;/math&gt;. Now, we want to find a number that's &lt;math&gt;2\mod25&lt;/math&gt; and &lt;math&gt;1\mod4&lt;/math&gt;. We find this number by clever guess and check. We know our number is odd, so it is &lt;math&gt;27\mod50&lt;/math&gt;. First, we try &lt;math&gt;27&lt;/math&gt;, but that doesn't work. Next, we try &lt;math&gt;77&lt;/math&gt;, and that works! So, our answer is &lt;math&gt;\boxed{77}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{UNM-PNM Math Contest box|year=2017|n=II|before=First question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_29&diff=151906 2008 iTest Problems/Problem 29 2021-04-19T00:17:16Z <p>Bronzetruck2016: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Find the number of ordered triplets &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers such that &lt;math&gt;abc=2008&lt;/math&gt; (the product of &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;2008&lt;/math&gt;). <br /> <br /> ==Solution 1==<br /> <br /> The number &lt;math&gt;2008&lt;/math&gt; can be factored into &lt;math&gt;2^3 \cdot 251&lt;/math&gt;. Use [[casework]] to organize the counting.<br /> <br /> * If two numbers are &lt;math&gt;1&lt;/math&gt;, then the third one must be &lt;math&gt;2008&lt;/math&gt;, and there are &lt;math&gt;3&lt;/math&gt; ways to write the ordered pairs.<br /> * If one number is a &lt;math&gt;1&lt;/math&gt;, then there are &lt;math&gt;\tfrac{8-2}{2} = 3&lt;/math&gt; possible pairs of numbers for the other two. Since the numbers are all different, there are &lt;math&gt;3 \cdot 6 = 18&lt;/math&gt; ways to write the ordered pairs.<br /> * If none of the numbers are &lt;math&gt;1&lt;/math&gt;, then since there are only four prime numbers being multiplied, one of the numbers must have two prime numbers being multiplied together. Thus, the two sets of numbers are &lt;math&gt;2,2,502&lt;/math&gt; and &lt;math&gt;2,4,251&lt;/math&gt;, and there are &lt;math&gt;3+6=9&lt;/math&gt; ways in this case.<br /> <br /> Altogether, there are &lt;math&gt;\boxed{30}&lt;/math&gt; ordered pairs that satisfy the criteria.<br /> <br /> ==Solution 2==<br /> <br /> &lt;math&gt;2008&lt;/math&gt; can be prime factorized into &lt;math&gt;2^3\cdot251&lt;/math&gt;. We can think of each ordered pair &lt;math&gt;(a,b,c)&lt;/math&gt; as a way to assign three 2s and one 251 to three distinct letters. You may now recognized this as a &quot;assign non-distinct objects to distinct piles&quot; problem. In problems like this, we should use [[stars and bars]]. There are &lt;math&gt;\binom{5}{2}=10&lt;/math&gt; ways to assign three 2s to three distinct letters, and there are &lt;math&gt;\binom{3}{2}=3&lt;/math&gt; ways to assign one 251 to three distinct letters. Multiplying, we get &lt;math&gt;3\cdot10=\boxed{30}&lt;/math&gt;.<br /> &lt;i&gt;-bronzetruck2016&lt;i&gt;<br /> <br /> ==See Also==<br /> {{2008 iTest box|num-b=28|num-a=30}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_29&diff=151905 2008 iTest Problems/Problem 29 2021-04-19T00:17:05Z <p>Bronzetruck2016: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Find the number of ordered triplets &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers such that &lt;math&gt;abc=2008&lt;/math&gt; (the product of &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;2008&lt;/math&gt;). <br /> <br /> ==Solution 1==<br /> <br /> The number &lt;math&gt;2008&lt;/math&gt; can be factored into &lt;math&gt;2^3 \cdot 251&lt;/math&gt;. Use [[casework]] to organize the counting.<br /> <br /> * If two numbers are &lt;math&gt;1&lt;/math&gt;, then the third one must be &lt;math&gt;2008&lt;/math&gt;, and there are &lt;math&gt;3&lt;/math&gt; ways to write the ordered pairs.<br /> * If one number is a &lt;math&gt;1&lt;/math&gt;, then there are &lt;math&gt;\tfrac{8-2}{2} = 3&lt;/math&gt; possible pairs of numbers for the other two. Since the numbers are all different, there are &lt;math&gt;3 \cdot 6 = 18&lt;/math&gt; ways to write the ordered pairs.<br /> * If none of the numbers are &lt;math&gt;1&lt;/math&gt;, then since there are only four prime numbers being multiplied, one of the numbers must have two prime numbers being multiplied together. Thus, the two sets of numbers are &lt;math&gt;2,2,502&lt;/math&gt; and &lt;math&gt;2,4,251&lt;/math&gt;, and there are &lt;math&gt;3+6=9&lt;/math&gt; ways in this case.<br /> <br /> Altogether, there are &lt;math&gt;\boxed{30}&lt;/math&gt; ordered pairs that satisfy the criteria.<br /> <br /> ==Solution 2==<br /> <br /> &lt;math&gt;2008&lt;/math&gt; can be prime factorized into &lt;math&gt;2^3\cdot251&lt;/math&gt;. We can think of each ordered pair &lt;math&gt;(a,b,c)&lt;/math&gt; as a way to assign three 2s and one 251 to three distinct letters. You may now recognized this as a &quot;assign non-distinct objects to distinct piles&quot; problem. In problems like this, we should use [[stars and bars]]. There are &lt;math&gt;\binom{5}{2}=10&lt;/math&gt; ways to assign three 2s to three distinct letters, and there are &lt;math&gt;\binom{3}{2}=3&lt;/math&gt; ways to assign one 251 to three distinct letters. Multiplying, we get &lt;math&gt;3\cdot10=\boxed{30}&lt;/math&gt;. &lt;math&gt;\n&lt;/math&gt;<br /> &lt;i&gt;-bronzetruck2016&lt;i&gt;<br /> <br /> ==See Also==<br /> {{2008 iTest box|num-b=28|num-a=30}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_29&diff=151904 2008 iTest Problems/Problem 29 2021-04-19T00:15:17Z <p>Bronzetruck2016: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> Find the number of ordered triplets &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers such that &lt;math&gt;abc=2008&lt;/math&gt; (the product of &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;2008&lt;/math&gt;). <br /> <br /> ==Solution 1==<br /> <br /> The number &lt;math&gt;2008&lt;/math&gt; can be factored into &lt;math&gt;2^3 \cdot 251&lt;/math&gt;. Use [[casework]] to organize the counting.<br /> <br /> * If two numbers are &lt;math&gt;1&lt;/math&gt;, then the third one must be &lt;math&gt;2008&lt;/math&gt;, and there are &lt;math&gt;3&lt;/math&gt; ways to write the ordered pairs.<br /> * If one number is a &lt;math&gt;1&lt;/math&gt;, then there are &lt;math&gt;\tfrac{8-2}{2} = 3&lt;/math&gt; possible pairs of numbers for the other two. Since the numbers are all different, there are &lt;math&gt;3 \cdot 6 = 18&lt;/math&gt; ways to write the ordered pairs.<br /> * If none of the numbers are &lt;math&gt;1&lt;/math&gt;, then since there are only four prime numbers being multiplied, one of the numbers must have two prime numbers being multiplied together. Thus, the two sets of numbers are &lt;math&gt;2,2,502&lt;/math&gt; and &lt;math&gt;2,4,251&lt;/math&gt;, and there are &lt;math&gt;3+6=9&lt;/math&gt; ways in this case.<br /> <br /> Altogether, there are &lt;math&gt;\boxed{30}&lt;/math&gt; ordered pairs that satisfy the criteria.<br /> <br /> ==Solution 2==<br /> <br /> &lt;math&gt;2008&lt;/math&gt; can be prime factorized into &lt;math&gt;2^3\cdot251&lt;/math&gt;. We can think of each ordered pair &lt;math&gt;(a,b,c)&lt;/math&gt; as a way to assign three 2s and one 251 to three distinct letters. You may now recognized this as a &quot;assign non-distinct objects to distinct piles&quot; problem. In problems like this, we should use [[stars and bars]]. There are &lt;math&gt;\binom{5}{2}=10&lt;/math&gt; ways to assign three 2s to three distinct letters, and there are &lt;math&gt;\binom{3}{2}=3&lt;/math&gt; ways to assign one 251 to three distinct letters. Multiplying, we get &lt;math&gt;3\cdot10=\boxed{30}&lt;/math&gt;.<br /> <br /> ==See Also==<br /> {{2008 iTest box|num-b=28|num-a=30}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_29&diff=151902 2008 iTest Problems/Problem 29 2021-04-18T23:56:05Z <p>Bronzetruck2016: </p> <hr /> <div>==Problem==<br /> <br /> Find the number of ordered triplets &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers such that &lt;math&gt;abc=2008&lt;/math&gt; (the product of &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;2008&lt;/math&gt;). <br /> <br /> ==Solution 1==<br /> <br /> The number &lt;math&gt;2008&lt;/math&gt; can be factored into &lt;math&gt;2^3 \cdot 251&lt;/math&gt;. Use [[casework]] to organize the counting.<br /> <br /> * If two numbers are &lt;math&gt;1&lt;/math&gt;, then the third one must be &lt;math&gt;2008&lt;/math&gt;, and there are &lt;math&gt;3&lt;/math&gt; ways to write the ordered pairs.<br /> * If one number is a &lt;math&gt;1&lt;/math&gt;, then there are &lt;math&gt;\tfrac{8-2}{2} = 3&lt;/math&gt; possible pairs of numbers for the other two. Since the numbers are all different, there are &lt;math&gt;3 \cdot 6 = 18&lt;/math&gt; ways to write the ordered pairs.<br /> * If none of the numbers are &lt;math&gt;1&lt;/math&gt;, then since there are only four prime numbers being multiplied, one of the numbers must have two prime numbers being multiplied together. Thus, the two sets of numbers are &lt;math&gt;2,2,502&lt;/math&gt; and &lt;math&gt;2,4,251&lt;/math&gt;, and there are &lt;math&gt;3+6=9&lt;/math&gt; ways in this case.<br /> <br /> Altogether, there are &lt;math&gt;\boxed{30}&lt;/math&gt; ordered pairs that satisfy the criteria.<br /> <br /> ==Solution 2==<br /> <br /> <br /> ==See Also==<br /> {{2008 iTest box|num-b=28|num-a=30}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2008_iTest_Problems/Problem_29&diff=151901 2008 iTest Problems/Problem 29 2021-04-18T23:55:18Z <p>Bronzetruck2016: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> Find the number of ordered triplets &lt;math&gt;(a,b,c)&lt;/math&gt; of positive integers such that &lt;math&gt;abc=2008&lt;/math&gt; (the product of &lt;math&gt;a, b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; is &lt;math&gt;2008&lt;/math&gt;). <br /> <br /> ==Solution 1==<br /> <br /> The number &lt;math&gt;2008&lt;/math&gt; can be factored into &lt;math&gt;2^3 \cdot 251&lt;/math&gt;. Use [[casework]] to organize the counting.<br /> <br /> * If two numbers are &lt;math&gt;1&lt;/math&gt;, then the third one must be &lt;math&gt;2008&lt;/math&gt;, and there are &lt;math&gt;3&lt;/math&gt; ways to write the ordered pairs.<br /> * If one number is a &lt;math&gt;1&lt;/math&gt;, then there are &lt;math&gt;\tfrac{8-2}{2} = 3&lt;/math&gt; possible pairs of numbers for the other two. Since the numbers are all different, there are &lt;math&gt;3 \cdot 6 = 18&lt;/math&gt; ways to write the ordered pairs.<br /> * If none of the numbers are &lt;math&gt;1&lt;/math&gt;, then since there are only four prime numbers being multiplied, one of the numbers must have two prime numbers being multiplied together. Thus, the two sets of numbers are &lt;math&gt;2,2,502&lt;/math&gt; and &lt;math&gt;2,4,251&lt;/math&gt;, and there are &lt;math&gt;3+6=9&lt;/math&gt; ways in this case.<br /> <br /> Altogether, there are &lt;math&gt;\boxed{30}&lt;/math&gt; ordered pairs that satisfy the criteria.<br /> <br /> ==See Also==<br /> {{2008 iTest box|num-b=28|num-a=30}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_1&diff=151722 2001 AIME I Problems/Problem 1 2021-04-16T21:14:03Z <p>Bronzetruck2016: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Find the sum of all positive two-digit integers that are divisible by each of their digits.<br /> <br /> == Solution 1 ==<br /> Let our number be &lt;math&gt;10a + b&lt;/math&gt;, &lt;math&gt;a,b \neq 0&lt;/math&gt;. Then we have two conditions: &lt;math&gt;10a + b \equiv 10a \equiv 0 \pmod{b}&lt;/math&gt; and &lt;math&gt;10a + b \equiv b \pmod{a}&lt;/math&gt;, or &lt;math&gt;a&lt;/math&gt; divides into &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; divides into &lt;math&gt;10a&lt;/math&gt;. Thus &lt;math&gt;b = a, 2a,&lt;/math&gt; or &lt;math&gt;5a&lt;/math&gt; (note that if &lt;math&gt;b = 10a&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; would not be a digit). <br /> <br /> *For &lt;math&gt;b = a&lt;/math&gt;, we have &lt;math&gt;n = 11a&lt;/math&gt; for nine possibilities, giving us a sum of &lt;math&gt;11 \cdot \frac {9(10)}{2} = 495&lt;/math&gt;. <br /> *For &lt;math&gt;b = 2a&lt;/math&gt;, we have &lt;math&gt;n = 12a&lt;/math&gt; for four possibilities (the higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;12 \cdot \frac {4(5)}{2} = 120&lt;/math&gt;.<br /> *For &lt;math&gt;b = 5a&lt;/math&gt;, we have &lt;math&gt;n = 15a&lt;/math&gt; for one possibility (again, higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;15&lt;/math&gt;. <br /> If we ignore the case &lt;math&gt;b = 0&lt;/math&gt; as we have been doing so far, then the sum is &lt;math&gt;495 + 120 + 15 = \boxed{630}&lt;/math&gt;. <br /> <br /> == Solution 2 ==<br /> <br /> Using casework, we can list out all of these numbers: &lt;cmath&gt;11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.&lt;/cmath&gt;<br /> <br /> == Solution 3 ==<br /> <br /> To further expand on solution 2, it would be tedious to test all &lt;math&gt;90&lt;/math&gt; two-digit numbers. We can reduce the amount to look at by focusing on the tens digit.<br /> First, we cannot have any number that is a multiple of &lt;math&gt;10&lt;/math&gt;. We also note that any number with the same digits is a number that satisfies this problem. This gives &lt;cmath&gt;11, 22, 33, ... 99.&lt;/cmath&gt; We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different tens digit. For example, we look at numbers &lt;math&gt;11, 12, 13, ... 19&lt;/math&gt; and numbers &lt;math&gt;22, 24, 26, 28&lt;/math&gt;. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of &lt;math&gt;5&lt;/math&gt; or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2.<br /> <br /> == Solution 4 ==<br /> <br /> In this solution, we will do casework on the ones digit.<br /> Before we start, let's make some variables. Let &lt;math&gt;a&lt;/math&gt; be the ones digit, and &lt;math&gt;b&lt;/math&gt; be the tens digit. Let &lt;math&gt;n&lt;/math&gt; equal our number. Our number can be expressed as &lt;math&gt;10b+a&lt;/math&gt;. We can easily see that &lt;math&gt;b|a&lt;/math&gt;, since &lt;math&gt;b|n&lt;/math&gt;, and &lt;math&gt;b|10b&lt;/math&gt;. Therefore, &lt;math&gt;b|(n-10b)&lt;/math&gt;. <br /> Now, let's start with the casework.<br /> <br /> Case 1: &lt;math&gt;a=1&lt;/math&gt;<br /> Since &lt;math&gt;b|a&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt;. From this, we get that &lt;math&gt;n=11&lt;/math&gt; satisfies the condition.<br /> <br /> Case 2: &lt;math&gt;a=2&lt;/math&gt;<br /> We either have &lt;math&gt;b=1&lt;/math&gt;, or &lt;math&gt;b=2&lt;/math&gt;. From this, we get that &lt;math&gt;n=12&lt;/math&gt; and &lt;math&gt;n=22&lt;/math&gt; satisfy the condition.<br /> <br /> Case 3: &lt;math&gt;a=3&lt;/math&gt;<br /> We have &lt;math&gt;b=3&lt;/math&gt;. From this, we get that &lt;math&gt;n=33&lt;/math&gt; satisfies the condition. Note that &lt;math&gt;b=1&lt;/math&gt; was not included because &lt;math&gt;3&lt;/math&gt; does not divide &lt;math&gt;13&lt;/math&gt;.<br /> <br /> Case 4: &lt;math&gt;a=4&lt;/math&gt;<br /> We either have &lt;math&gt;b=2&lt;/math&gt; or &lt;math&gt;b=4&lt;/math&gt;. From this, we get that &lt;math&gt;n=24&lt;/math&gt; and &lt;math&gt;n=44&lt;/math&gt; satisfy the condition. &lt;math&gt;b=1&lt;/math&gt; was not included for similar reasons as last time.<br /> <br /> Case 5: &lt;math&gt;a=5&lt;/math&gt;<br /> We either have &lt;math&gt;b=1&lt;/math&gt; or &lt;math&gt;b=5&lt;/math&gt;. From this, we get that &lt;math&gt;n=15&lt;/math&gt; and &lt;math&gt;n=55&lt;/math&gt; satisfy the condition.<br /> <br /> Continuing with this process up to &lt;math&gt;a=9&lt;/math&gt;, we get that &lt;math&gt;n&lt;/math&gt; could be &lt;math&gt;11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99&lt;/math&gt;. Summing, we get that the answer is &lt;math&gt;\boxed{630}&lt;/math&gt;. A clever way to sum would be to group the multiples of &lt;math&gt;11&lt;/math&gt; together to get &lt;math&gt;11+22+\dots+99=(45)(11)=495&lt;/math&gt;, and then add the remaining &lt;math&gt;12+24+15+36+48=135&lt;/math&gt;.<br /> <br /> &lt;i&gt;-bronzetruck2016&lt;i&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_1&diff=151721 2001 AIME I Problems/Problem 1 2021-04-16T21:13:44Z <p>Bronzetruck2016: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Find the sum of all positive two-digit integers that are divisible by each of their digits.<br /> <br /> == Solution 1 ==<br /> Let our number be &lt;math&gt;10a + b&lt;/math&gt;, &lt;math&gt;a,b \neq 0&lt;/math&gt;. Then we have two conditions: &lt;math&gt;10a + b \equiv 10a \equiv 0 \pmod{b}&lt;/math&gt; and &lt;math&gt;10a + b \equiv b \pmod{a}&lt;/math&gt;, or &lt;math&gt;a&lt;/math&gt; divides into &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; divides into &lt;math&gt;10a&lt;/math&gt;. Thus &lt;math&gt;b = a, 2a,&lt;/math&gt; or &lt;math&gt;5a&lt;/math&gt; (note that if &lt;math&gt;b = 10a&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; would not be a digit). <br /> <br /> *For &lt;math&gt;b = a&lt;/math&gt;, we have &lt;math&gt;n = 11a&lt;/math&gt; for nine possibilities, giving us a sum of &lt;math&gt;11 \cdot \frac {9(10)}{2} = 495&lt;/math&gt;. <br /> *For &lt;math&gt;b = 2a&lt;/math&gt;, we have &lt;math&gt;n = 12a&lt;/math&gt; for four possibilities (the higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;12 \cdot \frac {4(5)}{2} = 120&lt;/math&gt;.<br /> *For &lt;math&gt;b = 5a&lt;/math&gt;, we have &lt;math&gt;n = 15a&lt;/math&gt; for one possibility (again, higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;15&lt;/math&gt;. <br /> If we ignore the case &lt;math&gt;b = 0&lt;/math&gt; as we have been doing so far, then the sum is &lt;math&gt;495 + 120 + 15 = \boxed{630}&lt;/math&gt;. <br /> <br /> == Solution 2 ==<br /> <br /> Using casework, we can list out all of these numbers: &lt;cmath&gt;11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.&lt;/cmath&gt;<br /> <br /> == Solution 3 ==<br /> <br /> To further expand on solution 2, it would be tedious to test all &lt;math&gt;90&lt;/math&gt; two-digit numbers. We can reduce the amount to look at by focusing on the tens digit.<br /> First, we cannot have any number that is a multiple of &lt;math&gt;10&lt;/math&gt;. We also note that any number with the same digits is a number that satisfies this problem. This gives &lt;cmath&gt;11, 22, 33, ... 99.&lt;/cmath&gt; We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different tens digit. For example, we look at numbers &lt;math&gt;11, 12, 13, ... 19&lt;/math&gt; and numbers &lt;math&gt;22, 24, 26, 28&lt;/math&gt;. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of &lt;math&gt;5&lt;/math&gt; or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2.<br /> <br /> == Solution 4 ==<br /> <br /> In this solution, we will do casework on the ones digit.<br /> Before we start, let's make some variables. Let &lt;math&gt;a&lt;/math&gt; be the ones digit, and &lt;math&gt;b&lt;/math&gt; be the tens digit. Let &lt;math&gt;n&lt;/math&gt; equal our number. Our number can be expressed as &lt;math&gt;10b+a&lt;/math&gt;. We can easily see that &lt;math&gt;b|a&lt;/math&gt;, since &lt;math&gt;b|n&lt;/math&gt;, and &lt;math&gt;b|10b&lt;/math&gt;. Therefore, &lt;math&gt;b|(n-10b)&lt;/math&gt;. <br /> Now, let's start with the casework.<br /> <br /> Case 1: &lt;math&gt;a=1&lt;/math&gt;<br /> Since &lt;math&gt;b|a&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt;. From this, we get that &lt;math&gt;n=11&lt;/math&gt; satisfies the condition.<br /> <br /> Case 2: &lt;math&gt;a=2&lt;/math&gt;<br /> We either have &lt;math&gt;b=1&lt;/math&gt;, or &lt;math&gt;b=2&lt;/math&gt;. From this, we get that &lt;math&gt;n=12&lt;/math&gt; and &lt;math&gt;n=22&lt;/math&gt; satisfy the condition.<br /> <br /> Case 3: &lt;math&gt;a=3&lt;/math&gt;<br /> We have &lt;math&gt;b=3&lt;/math&gt;. From this, we get that &lt;math&gt;n=33&lt;/math&gt; satisfies the condition. Note that &lt;math&gt;b=1&lt;/math&gt; was not included because &lt;math&gt;3&lt;/math&gt; does not divide &lt;math&gt;13&lt;/math&gt;.<br /> <br /> Case 4: &lt;math&gt;a=4&lt;/math&gt;<br /> We either have &lt;math&gt;b=2&lt;/math&gt; or &lt;math&gt;b=4&lt;/math&gt;. From this, we get that &lt;math&gt;n=24&lt;/math&gt; and &lt;math&gt;n=44&lt;/math&gt; satisfy the condition. &lt;math&gt;b=1&lt;/math&gt; was not included for similar reasons as last time.<br /> <br /> Case 5: &lt;math&gt;a=5&lt;/math&gt;<br /> We either have &lt;math&gt;b=1&lt;/math&gt; or &lt;math&gt;b=5&lt;/math&gt;. From this, we get that &lt;math&gt;n=15&lt;/math&gt; and &lt;math&gt;n=55&lt;/math&gt; satisfy the condition.<br /> <br /> Continuing with this process up to &lt;math&gt;a=9&lt;/math&gt;, we get that &lt;math&gt;n&lt;/math&gt; could be &lt;math&gt;11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99&lt;/math&gt;. Summing, we get that the answer is &lt;math&gt;\boxed{630}&lt;/math&gt;. A clever way to sum would be to group the multiples of &lt;math&gt;11&lt;/math&gt; together to get &lt;math&gt;11+22+\dots+99=(45)(11)=495&lt;/math&gt; and then add the remaining &lt;math&gt;12+24+15+36+48=135&lt;/math&gt;.<br /> <br /> &lt;i&gt;-bronzetruck2016&lt;i&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_1&diff=151717 2001 AIME I Problems/Problem 1 2021-04-16T21:10:47Z <p>Bronzetruck2016: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Find the sum of all positive two-digit integers that are divisible by each of their digits.<br /> <br /> == Solution 1 ==<br /> Let our number be &lt;math&gt;10a + b&lt;/math&gt;, &lt;math&gt;a,b \neq 0&lt;/math&gt;. Then we have two conditions: &lt;math&gt;10a + b \equiv 10a \equiv 0 \pmod{b}&lt;/math&gt; and &lt;math&gt;10a + b \equiv b \pmod{a}&lt;/math&gt;, or &lt;math&gt;a&lt;/math&gt; divides into &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; divides into &lt;math&gt;10a&lt;/math&gt;. Thus &lt;math&gt;b = a, 2a,&lt;/math&gt; or &lt;math&gt;5a&lt;/math&gt; (note that if &lt;math&gt;b = 10a&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; would not be a digit). <br /> <br /> *For &lt;math&gt;b = a&lt;/math&gt;, we have &lt;math&gt;n = 11a&lt;/math&gt; for nine possibilities, giving us a sum of &lt;math&gt;11 \cdot \frac {9(10)}{2} = 495&lt;/math&gt;. <br /> *For &lt;math&gt;b = 2a&lt;/math&gt;, we have &lt;math&gt;n = 12a&lt;/math&gt; for four possibilities (the higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;12 \cdot \frac {4(5)}{2} = 120&lt;/math&gt;.<br /> *For &lt;math&gt;b = 5a&lt;/math&gt;, we have &lt;math&gt;n = 15a&lt;/math&gt; for one possibility (again, higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;15&lt;/math&gt;. <br /> If we ignore the case &lt;math&gt;b = 0&lt;/math&gt; as we have been doing so far, then the sum is &lt;math&gt;495 + 120 + 15 = \boxed{630}&lt;/math&gt;. <br /> <br /> == Solution 2 ==<br /> <br /> Using casework, we can list out all of these numbers: &lt;cmath&gt;11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.&lt;/cmath&gt;<br /> <br /> == Solution 3 ==<br /> <br /> To further expand on solution 2, it would be tedious to test all &lt;math&gt;90&lt;/math&gt; two-digit numbers. We can reduce the amount to look at by focusing on the tens digit.<br /> First, we cannot have any number that is a multiple of &lt;math&gt;10&lt;/math&gt;. We also note that any number with the same digits is a number that satisfies this problem. This gives &lt;cmath&gt;11, 22, 33, ... 99.&lt;/cmath&gt; We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different tens digit. For example, we look at numbers &lt;math&gt;11, 12, 13, ... 19&lt;/math&gt; and numbers &lt;math&gt;22, 24, 26, 28&lt;/math&gt;. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of &lt;math&gt;5&lt;/math&gt; or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2.<br /> <br /> == Solution 4 ==<br /> <br /> In this solution, we will do casework on the ones digit.<br /> Before we start, let's make some variables. Let &lt;math&gt;a&lt;/math&gt; be the ones digit, and &lt;math&gt;b&lt;/math&gt; be the tens digit. Let &lt;math&gt;n&lt;/math&gt; equal our number. Our number can be expressed as &lt;math&gt;10b+a&lt;/math&gt;. We can easily see that &lt;math&gt;b|a&lt;/math&gt;, since &lt;math&gt;b|n&lt;/math&gt;, and &lt;math&gt;b|10b&lt;/math&gt;. Therefore, &lt;math&gt;b|(n-10b)&lt;/math&gt;. <br /> Now, let's start with the casework.<br /> <br /> Case 1: &lt;math&gt;a=1&lt;/math&gt;<br /> Since &lt;math&gt;b|a&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt;. From this, we get that &lt;math&gt;n=11&lt;/math&gt; satisfies the condition.<br /> <br /> Case 2: &lt;math&gt;a=2&lt;/math&gt;<br /> We either have &lt;math&gt;b=1&lt;/math&gt;, or &lt;math&gt;b=2&lt;/math&gt;. From this, we get that &lt;math&gt;n=12&lt;/math&gt; and &lt;math&gt;n=22&lt;/math&gt; satisfy the condition.<br /> <br /> Case 3: &lt;math&gt;a=3&lt;/math&gt;<br /> We have &lt;math&gt;b=3&lt;/math&gt;. From this, we get that &lt;math&gt;n=33&lt;/math&gt; satisfies the condition. Note that &lt;math&gt;b=1&lt;/math&gt; was not included because &lt;math&gt;3&lt;/math&gt; does not divide &lt;math&gt;13&lt;/math&gt;.<br /> <br /> Case 4: &lt;math&gt;a=4&lt;/math&gt;<br /> We either have &lt;math&gt;b=2&lt;/math&gt; or &lt;math&gt;b=4&lt;/math&gt;. From this, we get that &lt;math&gt;n=24&lt;/math&gt; and &lt;math&gt;n=44&lt;/math&gt; satisfy the condition. &lt;math&gt;b=1&lt;/math&gt; was not included for similar reasons as last time.<br /> <br /> Case 5: &lt;math&gt;a=5&lt;/math&gt;<br /> We either have &lt;math&gt;b=1&lt;/math&gt; or &lt;math&gt;b=5&lt;/math&gt;. From this, we get that &lt;math&gt;n=15&lt;/math&gt; and &lt;math&gt;n=55&lt;/math&gt; satisfy the condition.<br /> <br /> Continuing with this process up to &lt;math&gt;a=9&lt;/math&gt;, we get that &lt;math&gt;n&lt;/math&gt; could be &lt;math&gt;11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99&lt;/math&gt;. Summing, we get that the answer is &lt;math&gt;\boxed{630}&lt;/math&gt;. A clever way to sum would be to group the multiples of &lt;math&gt;11&lt;/math&gt; together.<br /> <br /> &lt;i&gt;-bronzetruck2016&lt;i&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_1&diff=151716 2001 AIME I Problems/Problem 1 2021-04-16T21:09:53Z <p>Bronzetruck2016: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Find the sum of all positive two-digit integers that are divisible by each of their digits.<br /> <br /> == Solution 1 ==<br /> Let our number be &lt;math&gt;10a + b&lt;/math&gt;, &lt;math&gt;a,b \neq 0&lt;/math&gt;. Then we have two conditions: &lt;math&gt;10a + b \equiv 10a \equiv 0 \pmod{b}&lt;/math&gt; and &lt;math&gt;10a + b \equiv b \pmod{a}&lt;/math&gt;, or &lt;math&gt;a&lt;/math&gt; divides into &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; divides into &lt;math&gt;10a&lt;/math&gt;. Thus &lt;math&gt;b = a, 2a,&lt;/math&gt; or &lt;math&gt;5a&lt;/math&gt; (note that if &lt;math&gt;b = 10a&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; would not be a digit). <br /> <br /> *For &lt;math&gt;b = a&lt;/math&gt;, we have &lt;math&gt;n = 11a&lt;/math&gt; for nine possibilities, giving us a sum of &lt;math&gt;11 \cdot \frac {9(10)}{2} = 495&lt;/math&gt;. <br /> *For &lt;math&gt;b = 2a&lt;/math&gt;, we have &lt;math&gt;n = 12a&lt;/math&gt; for four possibilities (the higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;12 \cdot \frac {4(5)}{2} = 120&lt;/math&gt;.<br /> *For &lt;math&gt;b = 5a&lt;/math&gt;, we have &lt;math&gt;n = 15a&lt;/math&gt; for one possibility (again, higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;15&lt;/math&gt;. <br /> If we ignore the case &lt;math&gt;b = 0&lt;/math&gt; as we have been doing so far, then the sum is &lt;math&gt;495 + 120 + 15 = \boxed{630}&lt;/math&gt;. <br /> <br /> == Solution 2 ==<br /> <br /> Using casework, we can list out all of these numbers: &lt;cmath&gt;11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.&lt;/cmath&gt;<br /> <br /> == Solution 3 ==<br /> <br /> To further expand on solution 2, it would be tedious to test all &lt;math&gt;90&lt;/math&gt; two-digit numbers. We can reduce the amount to look at by focusing on the tens digit.<br /> First, we cannot have any number that is a multiple of &lt;math&gt;10&lt;/math&gt;. We also note that any number with the same digits is a number that satisfies this problem. This gives &lt;cmath&gt;11, 22, 33, ... 99.&lt;/cmath&gt; We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different tens digit. For example, we look at numbers &lt;math&gt;11, 12, 13, ... 19&lt;/math&gt; and numbers &lt;math&gt;22, 24, 26, 28&lt;/math&gt;. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of &lt;math&gt;5&lt;/math&gt; or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2.<br /> <br /> == Solution 4 ==<br /> <br /> In this solution, we will do casework on the ones digit.<br /> Before we start, let's make some variables. Let &lt;math&gt;a&lt;/math&gt; be the ones digit, and &lt;math&gt;b&lt;/math&gt; be the tens digit. Let &lt;math&gt;n&lt;/math&gt; equal our number. Our number can be expressed as &lt;math&gt;10b+a&lt;/math&gt;. We can easily see that &lt;math&gt;b|a&lt;/math&gt;, since &lt;math&gt;b|n&lt;/math&gt;, and &lt;math&gt;b|10b&lt;/math&gt;. Therefore, &lt;math&gt;b|(n-10b)&lt;/math&gt;. <br /> Now, let's start with the casework.<br /> <br /> Case 1: &lt;math&gt;a=1&lt;/math&gt;<br /> Since &lt;math&gt;b|a&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt;. From this, we get that &lt;math&gt;n=11&lt;/math&gt; satisfies the condition.<br /> <br /> Case 2: &lt;math&gt;a=2&lt;/math&gt;<br /> We either have &lt;math&gt;b=1&lt;/math&gt;, or &lt;math&gt;b=2&lt;/math&gt;. From this, we get that &lt;math&gt;n=12&lt;/math&gt; and &lt;math&gt;n=22&lt;/math&gt; satisfy the condition.<br /> <br /> Case 3: &lt;math&gt;a=3&lt;/math&gt;<br /> We have &lt;math&gt;b=3&lt;/math&gt;. From this, we get that &lt;math&gt;n=33&lt;/math&gt; satisfies the condition. Note that &lt;math&gt;b=1&lt;/math&gt; was not included because &lt;math&gt;3&lt;/math&gt; does not divide &lt;math&gt;13&lt;/math&gt;.<br /> <br /> Case 4: &lt;math&gt;a=4&lt;/math&gt;<br /> We either have &lt;math&gt;b=2&lt;/math&gt; or &lt;math&gt;b=4&lt;/math&gt;. From this, we get that &lt;math&gt;n=24&lt;/math&gt; and &lt;math&gt;n=44&lt;/math&gt; satisfy the condition. &lt;math&gt;b=1&lt;/math&gt; was not included for similar reasons as last time.<br /> <br /> Case 5: &lt;math&gt;a=5&lt;/math&gt;<br /> We either have &lt;math&gt;b=1&lt;/math&gt; or &lt;math&gt;b=5&lt;/math&gt;. From this, we get that &lt;math&gt;n=15&lt;/math&gt; and &lt;math&gt;n=55&lt;/math&gt; satisfy the condition.<br /> <br /> Continuing with this process up to &lt;math&gt;a=9&lt;/math&gt;, we get that &lt;math&gt;n&lt;/math&gt; could be &lt;math&gt;11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99&lt;/math&gt;. Summing, we get that the answer is \boxed{630}. A clever way to sum would be to group the multiples of &lt;math&gt;11&lt;/math&gt; together.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_1&diff=151715 2001 AIME I Problems/Problem 1 2021-04-16T21:08:21Z <p>Bronzetruck2016: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Find the sum of all positive two-digit integers that are divisible by each of their digits.<br /> <br /> == Solution 1 ==<br /> Let our number be &lt;math&gt;10a + b&lt;/math&gt;, &lt;math&gt;a,b \neq 0&lt;/math&gt;. Then we have two conditions: &lt;math&gt;10a + b \equiv 10a \equiv 0 \pmod{b}&lt;/math&gt; and &lt;math&gt;10a + b \equiv b \pmod{a}&lt;/math&gt;, or &lt;math&gt;a&lt;/math&gt; divides into &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; divides into &lt;math&gt;10a&lt;/math&gt;. Thus &lt;math&gt;b = a, 2a,&lt;/math&gt; or &lt;math&gt;5a&lt;/math&gt; (note that if &lt;math&gt;b = 10a&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; would not be a digit). <br /> <br /> *For &lt;math&gt;b = a&lt;/math&gt;, we have &lt;math&gt;n = 11a&lt;/math&gt; for nine possibilities, giving us a sum of &lt;math&gt;11 \cdot \frac {9(10)}{2} = 495&lt;/math&gt;. <br /> *For &lt;math&gt;b = 2a&lt;/math&gt;, we have &lt;math&gt;n = 12a&lt;/math&gt; for four possibilities (the higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;12 \cdot \frac {4(5)}{2} = 120&lt;/math&gt;.<br /> *For &lt;math&gt;b = 5a&lt;/math&gt;, we have &lt;math&gt;n = 15a&lt;/math&gt; for one possibility (again, higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;15&lt;/math&gt;. <br /> If we ignore the case &lt;math&gt;b = 0&lt;/math&gt; as we have been doing so far, then the sum is &lt;math&gt;495 + 120 + 15 = \boxed{630}&lt;/math&gt;. <br /> <br /> == Solution 2 ==<br /> <br /> Using casework, we can list out all of these numbers: &lt;cmath&gt;11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.&lt;/cmath&gt;<br /> <br /> == Solution 3 ==<br /> <br /> To further expand on solution 2, it would be tedious to test all &lt;math&gt;90&lt;/math&gt; two-digit numbers. We can reduce the amount to look at by focusing on the tens digit.<br /> First, we cannot have any number that is a multiple of &lt;math&gt;10&lt;/math&gt;. We also note that any number with the same digits is a number that satisfies this problem. This gives &lt;cmath&gt;11, 22, 33, ... 99.&lt;/cmath&gt; We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different tens digit. For example, we look at numbers &lt;math&gt;11, 12, 13, ... 19&lt;/math&gt; and numbers &lt;math&gt;22, 24, 26, 28&lt;/math&gt;. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of &lt;math&gt;5&lt;/math&gt; or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2.<br /> <br /> == Solution 4 ==<br /> <br /> In this solution, we will do casework on the ones digit.<br /> Before we start, let's make some variables. Let &lt;math&gt;a&lt;/math&gt; be the ones digit, and &lt;math&gt;b&lt;/math&gt; be the tens digit. Let &lt;math&gt;n&lt;/math&gt; equal our number. Our number can be expressed as &lt;math&gt;10b+a&lt;/math&gt;. We can easily see that &lt;math&gt;b|a&lt;/math&gt;, since &lt;math&gt;b|n&lt;/math&gt;, and &lt;math&gt;b|10b&lt;/math&gt;. Therefore, &lt;math&gt;b|(n-10b)&lt;/math&gt;. <br /> Now, let's start with the casework.<br /> <br /> Case 1: &lt;math&gt;a=1&lt;/math&gt;<br /> Since &lt;math&gt;b|a&lt;/math&gt;, &lt;math&gt;b=1&lt;/math&gt;. From this, we get that &lt;math&gt;n=11&lt;/math&gt; satisfies the condition.<br /> <br /> Case 2: &lt;math&gt;a=2&lt;/math&gt;<br /> We either have &lt;math&gt;b=1&lt;/math&gt;, or &lt;math&gt;b=2&lt;/math&gt;. From this, we get that &lt;math&gt;n=12&lt;/math&gt; and &lt;math&gt;n=22&lt;/math&gt; satisfy the condition.<br /> <br /> Case 3: &lt;math&gt;a=3&lt;/math&gt;<br /> We have &lt;math&gt;b=3&lt;/math&gt;. From this, we get that &lt;math&gt;n=33&lt;/math&gt; satisfies the condition. Note that &lt;math&gt;b=1&lt;/math&gt; was not included because &lt;math&gt;3&lt;/math&gt; does not divide &lt;math&gt;13&lt;/math&gt;.<br /> <br /> Case 4: &lt;math&gt;a=4&lt;/math&gt;<br /> We either have &lt;math&gt;b=2&lt;/math&gt; or &lt;math&gt;b=4&lt;/math&gt;. From this, we get that &lt;math&gt;n=24&lt;/math&gt; and &lt;math&gt;n=44&lt;/math&gt; satisfy the condition. &lt;math&gt;b=1&lt;/math&gt; was not included for similar reasons as last time.<br /> <br /> Case 5: &lt;math&gt;a=5&lt;/math&gt;<br /> We either have &lt;math&gt;b=1&lt;/math&gt; or &lt;math&gt;b=5&lt;/math&gt;. From this, we get that &lt;math&gt;n=15&lt;/math&gt; and &lt;math&gt;n=55&lt;/math&gt; satisfy the condition.<br /> <br /> Continuing with this process up to &lt;math&gt;a=9&lt;/math&gt;, we get that &lt;math&gt;n&lt;/math&gt; could be &lt;math&gt;11, 12, 22, 33, 24, 44, 15, 55, 36, 66, 77, 48, 88, 99&lt;/math&gt;. Summing, we get that the answer is &lt;math&gt;630&lt;/math&gt;. A clever way to sum would be to group the multiples of 11 together.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_1&diff=151714 2001 AIME I Problems/Problem 1 2021-04-16T20:42:37Z <p>Bronzetruck2016: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Find the sum of all positive two-digit integers that are divisible by each of their digits.<br /> <br /> == Solution 1 ==<br /> Let our number be &lt;math&gt;10a + b&lt;/math&gt;, &lt;math&gt;a,b \neq 0&lt;/math&gt;. Then we have two conditions: &lt;math&gt;10a + b \equiv 10a \equiv 0 \pmod{b}&lt;/math&gt; and &lt;math&gt;10a + b \equiv b \pmod{a}&lt;/math&gt;, or &lt;math&gt;a&lt;/math&gt; divides into &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; divides into &lt;math&gt;10a&lt;/math&gt;. Thus &lt;math&gt;b = a, 2a,&lt;/math&gt; or &lt;math&gt;5a&lt;/math&gt; (note that if &lt;math&gt;b = 10a&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; would not be a digit). <br /> <br /> *For &lt;math&gt;b = a&lt;/math&gt;, we have &lt;math&gt;n = 11a&lt;/math&gt; for nine possibilities, giving us a sum of &lt;math&gt;11 \cdot \frac {9(10)}{2} = 495&lt;/math&gt;. <br /> *For &lt;math&gt;b = 2a&lt;/math&gt;, we have &lt;math&gt;n = 12a&lt;/math&gt; for four possibilities (the higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;12 \cdot \frac {4(5)}{2} = 120&lt;/math&gt;.<br /> *For &lt;math&gt;b = 5a&lt;/math&gt;, we have &lt;math&gt;n = 15a&lt;/math&gt; for one possibility (again, higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;15&lt;/math&gt;. <br /> If we ignore the case &lt;math&gt;b = 0&lt;/math&gt; as we have been doing so far, then the sum is &lt;math&gt;495 + 120 + 15 = \boxed{630}&lt;/math&gt;. <br /> <br /> == Solution 2 ==<br /> <br /> Using casework, we can list out all of these numbers: &lt;cmath&gt;11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.&lt;/cmath&gt;<br /> <br /> == Solution 3 ==<br /> <br /> To further expand on solution 2, it would be tedious to test all &lt;math&gt;90&lt;/math&gt; two-digit numbers. We can reduce the amount to look at by focusing on the tens digit.<br /> First, we cannot have any number that is a multiple of &lt;math&gt;10&lt;/math&gt;. We also note that any number with the same digits is a number that satisfies this problem. This gives &lt;cmath&gt;11, 22, 33, ... 99.&lt;/cmath&gt; We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different tens digit. For example, we look at numbers &lt;math&gt;11, 12, 13, ... 19&lt;/math&gt; and numbers &lt;math&gt;22, 24, 26, 28&lt;/math&gt;. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of &lt;math&gt;5&lt;/math&gt; or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2.<br /> <br /> == Solution 4 ==<br /> <br /> In this solution, we will do casework on the ones digit.<br /> Before we start, let's make some variables. Let &lt;math&gt;a&lt;/math&gt; be the ones digit.<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2001_AIME_I_Problems/Problem_1&diff=151713 2001 AIME I Problems/Problem 1 2021-04-16T20:40:13Z <p>Bronzetruck2016: </p> <hr /> <div>== Problem ==<br /> Find the sum of all positive two-digit integers that are divisible by each of their digits.<br /> <br /> == Solution 1 ==<br /> Let our number be &lt;math&gt;10a + b&lt;/math&gt;, &lt;math&gt;a,b \neq 0&lt;/math&gt;. Then we have two conditions: &lt;math&gt;10a + b \equiv 10a \equiv 0 \pmod{b}&lt;/math&gt; and &lt;math&gt;10a + b \equiv b \pmod{a}&lt;/math&gt;, or &lt;math&gt;a&lt;/math&gt; divides into &lt;math&gt;b&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; divides into &lt;math&gt;10a&lt;/math&gt;. Thus &lt;math&gt;b = a, 2a,&lt;/math&gt; or &lt;math&gt;5a&lt;/math&gt; (note that if &lt;math&gt;b = 10a&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; would not be a digit). <br /> <br /> *For &lt;math&gt;b = a&lt;/math&gt;, we have &lt;math&gt;n = 11a&lt;/math&gt; for nine possibilities, giving us a sum of &lt;math&gt;11 \cdot \frac {9(10)}{2} = 495&lt;/math&gt;. <br /> *For &lt;math&gt;b = 2a&lt;/math&gt;, we have &lt;math&gt;n = 12a&lt;/math&gt; for four possibilities (the higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;12 \cdot \frac {4(5)}{2} = 120&lt;/math&gt;.<br /> *For &lt;math&gt;b = 5a&lt;/math&gt;, we have &lt;math&gt;n = 15a&lt;/math&gt; for one possibility (again, higher ones give &lt;math&gt;b &gt; 9&lt;/math&gt;), giving us a sum of &lt;math&gt;15&lt;/math&gt;. <br /> If we ignore the case &lt;math&gt;b = 0&lt;/math&gt; as we have been doing so far, then the sum is &lt;math&gt;495 + 120 + 15 = \boxed{630}&lt;/math&gt;. <br /> <br /> == Solution 2 ==<br /> <br /> Using casework, we can list out all of these numbers: &lt;cmath&gt;11+12+15+22+24+33+36+44+48+55+66+77+88+99=\boxed{630}.&lt;/cmath&gt;<br /> <br /> == Solution 3 ==<br /> <br /> To further expand on solution 2, it would be tedious to test all &lt;math&gt;90&lt;/math&gt; two-digit numbers. We can reduce the amount to look at by focusing on the tens digit.<br /> First, we cannot have any number that is a multiple of &lt;math&gt;10&lt;/math&gt;. We also note that any number with the same digits is a number that satisfies this problem. This gives &lt;cmath&gt;11, 22, 33, ... 99.&lt;/cmath&gt; We start from each of these numbers and constantly add the digit of the tens number of the respective number until we get a different tens digit. For example, we look at numbers &lt;math&gt;11, 12, 13, ... 19&lt;/math&gt; and numbers &lt;math&gt;22, 24, 26, 28&lt;/math&gt;. This heavily reduces the numbers we need to check, as we can deduce that any number with a tens digit of &lt;math&gt;5&lt;/math&gt; or greater that does not have two of the same digits is not a valid number for this problem. This will give us the numbers from solution 2.<br /> <br /> == Solution 4 ==<br /> <br /> == See also ==<br /> {{AIME box|year=2001|n=I|before=First Question|num-a=2}}<br /> <br /> [[Category:Introductory Number Theory Problems]]<br /> {{MAA Notice}}</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_16&diff=151622 2018 AMC 10A Problems/Problem 16 2021-04-16T00:03:50Z <p>Bronzetruck2016: /* Solution 2 - Circles */</p> <hr /> <div>==Problem==<br /> <br /> Right triangle &lt;math&gt;ABC&lt;/math&gt; has leg lengths &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=21&lt;/math&gt;. Including &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, how many line segments with integer length can be drawn from vertex &lt;math&gt;B&lt;/math&gt; to a point on hypotenuse &lt;math&gt;\overline{AC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }5 \qquad<br /> \textbf{(B) }8 \qquad<br /> \textbf{(C) }12 \qquad<br /> \textbf{(D) }13 \qquad<br /> \textbf{(E) }15 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> &lt;/asy&gt;<br /> As the problem has no diagram, we draw a diagram. The hypotenuse has length &lt;math&gt;29&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the foot of the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Note that &lt;math&gt;BP&lt;/math&gt; is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for &lt;math&gt;BP=\dfrac{20\cdot 21}{29}&lt;/math&gt;, which is between &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. <br /> <br /> Let the line segment be &lt;math&gt;BX&lt;/math&gt;, with &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. As you move &lt;math&gt;X&lt;/math&gt; along the hypotenuse from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt;, the length of &lt;math&gt;BX&lt;/math&gt; strictly decreases, hitting all the integer values from &lt;math&gt;20, 19, \dots 15&lt;/math&gt; (IVT). Similarly, moving &lt;math&gt;X&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt; hits all the integer values from &lt;math&gt;15, 16, \dots, 21&lt;/math&gt;. This is a total of &lt;math&gt;\boxed{(D) 13}&lt;/math&gt; line segments.<br /> (asymptote diagram added by elements2015)<br /> <br /> ==Solution 2 - Circles==<br /> Note that if a circle with an integer radius &lt;math&gt;r&lt;/math&gt; centered at vertex &lt;math&gt;B&lt;/math&gt; intersects hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, the lines drawn from &lt;math&gt;B&lt;/math&gt; to the points of intersection are integer lengths. As in the previous solution, the shortest distance &lt;math&gt;14&lt;\overline{BP}&lt;15&lt;/math&gt;. As a result, a circle of &lt;math&gt;14&lt;/math&gt; will &lt;b&gt;not&lt;/b&gt; reach the hypotenuse and thus does not intersect it. We also know that a circle of radius &lt;math&gt;21&lt;/math&gt; intersects the hypotenuse once and a circle of radius &lt;math&gt;\{15, 16, 17, 18, 19, 20 \}&lt;/math&gt; intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.<br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> draw(arc((0,0),21, 90, 180));<br /> draw(arc((0,0),20, 90, 180));<br /> draw(arc((0,0),19, 90, 180));<br /> draw(arc((0,0),18, 90, 180));<br /> draw(arc((0,0),17, 90, 180));<br /> draw(arc((0,0),16, 90, 180));<br /> draw(arc((0,0),15, 90, 180));<br /> &lt;/asy&gt;<br /> It follows that we can draw circles of radii &lt;math&gt;15, 16, 17, 18, 19,&lt;/math&gt; and &lt;math&gt;20,&lt;/math&gt; that each contribute &lt;b&gt;two&lt;/b&gt; integer lengths from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;\overline{AC}&lt;/math&gt; and one circle of radius &lt;math&gt;21&lt;/math&gt; that contributes only one such segment. Our answer is then &lt;cmath&gt;6 \cdot 2 + 1 = 13 \implies \boxed{D}&lt;/cmath&gt; ~samrocksnature<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> == Video Solution 2 ==<br /> https://youtu.be/4_x1sgcQCp4?t=3790<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_16&diff=151621 2018 AMC 10A Problems/Problem 16 2021-04-16T00:03:17Z <p>Bronzetruck2016: /* Solution 2 - Circles */</p> <hr /> <div>==Problem==<br /> <br /> Right triangle &lt;math&gt;ABC&lt;/math&gt; has leg lengths &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=21&lt;/math&gt;. Including &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, how many line segments with integer length can be drawn from vertex &lt;math&gt;B&lt;/math&gt; to a point on hypotenuse &lt;math&gt;\overline{AC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }5 \qquad<br /> \textbf{(B) }8 \qquad<br /> \textbf{(C) }12 \qquad<br /> \textbf{(D) }13 \qquad<br /> \textbf{(E) }15 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> &lt;/asy&gt;<br /> As the problem has no diagram, we draw a diagram. The hypotenuse has length &lt;math&gt;29&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the foot of the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Note that &lt;math&gt;BP&lt;/math&gt; is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for &lt;math&gt;BP=\dfrac{20\cdot 21}{29}&lt;/math&gt;, which is between &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. <br /> <br /> Let the line segment be &lt;math&gt;BX&lt;/math&gt;, with &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. As you move &lt;math&gt;X&lt;/math&gt; along the hypotenuse from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt;, the length of &lt;math&gt;BX&lt;/math&gt; strictly decreases, hitting all the integer values from &lt;math&gt;20, 19, \dots 15&lt;/math&gt; (IVT). Similarly, moving &lt;math&gt;X&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt; hits all the integer values from &lt;math&gt;15, 16, \dots, 21&lt;/math&gt;. This is a total of &lt;math&gt;\boxed{(D) 13}&lt;/math&gt; line segments.<br /> (asymptote diagram added by elements2015)<br /> <br /> ==Solution 2 - Circles==<br /> Note that if a circle with an integer radius &lt;math&gt;r&lt;/math&gt; centered at vertex &lt;math&gt;B&lt;/math&gt; intersects hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, the lines drawn from &lt;math&gt;B&lt;/math&gt; to the points of intersection are integer lengths. As in the previous solution, the shortest distance &lt;math&gt;14&lt;\overline{BP}&lt;15&lt;/math&gt;. As a result, a circle of &lt;math&gt;14&lt;/math&gt; will &lt;math&gt;&lt;b&gt;not&lt;/b&gt;&lt;/math&gt; reach the hypotenuse and thus does not intersect it. We also know that a circle of radius &lt;math&gt;21&lt;/math&gt; intersects the hypotenuse once and a circle of radius &lt;math&gt;\{15, 16, 17, 18, 19, 20 \}&lt;/math&gt; intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.<br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> draw(arc((0,0),21, 90, 180));<br /> draw(arc((0,0),20, 90, 180));<br /> draw(arc((0,0),19, 90, 180));<br /> draw(arc((0,0),18, 90, 180));<br /> draw(arc((0,0),17, 90, 180));<br /> draw(arc((0,0),16, 90, 180));<br /> draw(arc((0,0),15, 90, 180));<br /> &lt;/asy&gt;<br /> It follows that we can draw circles of radii &lt;math&gt;15, 16, 17, 18, 19,&lt;/math&gt; and &lt;math&gt;20,&lt;/math&gt; that each contribute &lt;b&gt;two&lt;/b&gt; integer lengths from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;\overline{AC}&lt;/math&gt; and one circle of radius &lt;math&gt;21&lt;/math&gt; that contributes only one such segment. Our answer is then &lt;cmath&gt;6 \cdot 2 + 1 = 13 \implies \boxed{D}&lt;/cmath&gt; ~samrocksnature<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> == Video Solution 2 ==<br /> https://youtu.be/4_x1sgcQCp4?t=3790<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_16&diff=151620 2018 AMC 10A Problems/Problem 16 2021-04-16T00:02:37Z <p>Bronzetruck2016: /* Solution 2 - Circles */</p> <hr /> <div>==Problem==<br /> <br /> Right triangle &lt;math&gt;ABC&lt;/math&gt; has leg lengths &lt;math&gt;AB=20&lt;/math&gt; and &lt;math&gt;BC=21&lt;/math&gt;. Including &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{BC}&lt;/math&gt;, how many line segments with integer length can be drawn from vertex &lt;math&gt;B&lt;/math&gt; to a point on hypotenuse &lt;math&gt;\overline{AC}&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }5 \qquad<br /> \textbf{(B) }8 \qquad<br /> \textbf{(C) }12 \qquad<br /> \textbf{(D) }13 \qquad<br /> \textbf{(E) }15 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution==<br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> &lt;/asy&gt;<br /> As the problem has no diagram, we draw a diagram. The hypotenuse has length &lt;math&gt;29&lt;/math&gt;. Let &lt;math&gt;P&lt;/math&gt; be the foot of the altitude from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;AC&lt;/math&gt;. Note that &lt;math&gt;BP&lt;/math&gt; is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for &lt;math&gt;BP=\dfrac{20\cdot 21}{29}&lt;/math&gt;, which is between &lt;math&gt;14&lt;/math&gt; and &lt;math&gt;15&lt;/math&gt;. <br /> <br /> Let the line segment be &lt;math&gt;BX&lt;/math&gt;, with &lt;math&gt;X&lt;/math&gt; on &lt;math&gt;AC&lt;/math&gt;. As you move &lt;math&gt;X&lt;/math&gt; along the hypotenuse from &lt;math&gt;A&lt;/math&gt; to &lt;math&gt;P&lt;/math&gt;, the length of &lt;math&gt;BX&lt;/math&gt; strictly decreases, hitting all the integer values from &lt;math&gt;20, 19, \dots 15&lt;/math&gt; (IVT). Similarly, moving &lt;math&gt;X&lt;/math&gt; from &lt;math&gt;P&lt;/math&gt; to &lt;math&gt;C&lt;/math&gt; hits all the integer values from &lt;math&gt;15, 16, \dots, 21&lt;/math&gt;. This is a total of &lt;math&gt;\boxed{(D) 13}&lt;/math&gt; line segments.<br /> (asymptote diagram added by elements2015)<br /> <br /> ==Solution 2 - Circles==<br /> Note that if a circle with an integer radius &lt;math&gt;r&lt;/math&gt; centered at vertex &lt;math&gt;B&lt;/math&gt; intersects hypotenuse &lt;math&gt;\overline{AB}&lt;/math&gt;, the lines drawn from &lt;math&gt;B&lt;/math&gt; to the points of intersection are integer lengths. As in the previous solution, the shortest distance &lt;math&gt;14&lt;\overline{BP}&lt;15&lt;/math&gt;. As a result, a circle of &lt;math&gt;14&lt;/math&gt; will &lt;math&gt;[b]not[/b]&lt;/math&gt; reach the hypotenuse and thus does not intersect it. We also know that a circle of radius &lt;math&gt;21&lt;/math&gt; intersects the hypotenuse once and a circle of radius &lt;math&gt;\{15, 16, 17, 18, 19, 20 \}&lt;/math&gt; intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.<br /> &lt;asy&gt;<br /> unitsize(4);<br /> pair A, B, C, E, P;<br /> A=(-20, 0);<br /> B=origin;<br /> C=(0,21);<br /> E=(-21, 20);<br /> P=extension(B,E, A, C);<br /> draw(A--B--C--cycle);<br /> draw(B--P);<br /> dot(&quot;$A$&quot;, A, SW);<br /> dot(&quot;$B$&quot;, B, SE);<br /> dot(&quot;$C$&quot;, C, NE);<br /> dot(&quot;$P$&quot;, P, S);<br /> draw(arc((0,0),21, 90, 180));<br /> draw(arc((0,0),20, 90, 180));<br /> draw(arc((0,0),19, 90, 180));<br /> draw(arc((0,0),18, 90, 180));<br /> draw(arc((0,0),17, 90, 180));<br /> draw(arc((0,0),16, 90, 180));<br /> draw(arc((0,0),15, 90, 180));<br /> &lt;/asy&gt;<br /> It follows that we can draw circles of radii &lt;math&gt;15, 16, 17, 18, 19,&lt;/math&gt; and &lt;math&gt;20,&lt;/math&gt; that each contribute &lt;b&gt;two&lt;/b&gt; integer lengths from &lt;math&gt;B&lt;/math&gt; to &lt;math&gt;\overline{AC}&lt;/math&gt; and one circle of radius &lt;math&gt;21&lt;/math&gt; that contributes only one such segment. Our answer is then &lt;cmath&gt;6 \cdot 2 + 1 = 13 \implies \boxed{D}&lt;/cmath&gt; ~samrocksnature<br /> <br /> ==Video Solution 1==<br /> https://youtu.be/M22S82Am2zM<br /> <br /> ~IceMatrix<br /> <br /> == Video Solution 2 ==<br /> https://youtu.be/4_x1sgcQCp4?t=3790<br /> <br /> ~ pi_is_3.14<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_11&diff=151613 2017 AMC 10A Problems/Problem 11 2021-04-15T22:36:41Z <p>Bronzetruck2016: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The region consisting of all points in three-dimensional space within &lt;math&gt;3&lt;/math&gt; units of line segment &lt;math&gt;\overline{AB}&lt;/math&gt; has volume &lt;math&gt;216\pi&lt;/math&gt;. What is the length &lt;math&gt;\textit{AB}&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 6\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 20\qquad\textbf{(E)}\ 24&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> In order to solve this problem, we must first visualize what the region contained looks like. We know that, in a three dimensional plane, the region consisting of all points within &lt;math&gt;3&lt;/math&gt; units of a point would be a sphere with radius &lt;math&gt;3&lt;/math&gt;. However, we need to find the region containing all points within &lt;math&gt;3&lt;/math&gt; units of a segment. It can be seen that our region is a cylinder with two hemispheres on either end. We know the volume of our region, so we set up the following equation (the volume of our cylinder + the volume of our two hemispheres will equal &lt;math&gt;216 \pi&lt;/math&gt;):<br /> <br /> &lt;math&gt;\frac{4 \pi }{3} \cdot 3^3+9 \pi x=216 \pi&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; is equal to the length of our line segment.<br /> <br /> Solving, we find that &lt;math&gt;x = \boxed{\textbf{(D)}\ 20}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> Because this is just a cylinder and &lt;math&gt;2&lt;/math&gt; &quot;half spheres&quot;, and the radius is &lt;math&gt;3&lt;/math&gt;, the volume of the &lt;math&gt;2&lt;/math&gt; half spheres is &lt;math&gt;\frac{4(3^3)\pi}{3} = 36 \pi&lt;/math&gt;. Since we also know that the volume of this whole thing is &lt;math&gt;216 \pi&lt;/math&gt;, we do &lt;math&gt;216-36&lt;/math&gt; to get &lt;math&gt;180 \pi&lt;/math&gt; as the volume of the cylinder. Thus the height is &lt;math&gt;180 \pi&lt;/math&gt; over the base, or &lt;math&gt;\frac{180 \pi}{9\pi}=20&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(D)}\ 20}.&lt;/math&gt;<br /> <br /> ~Minor edit by virjoy2001<br /> <br /> ==Diagram for Solution==<br /> http://i.imgur.com/cwNt293.png<br /> <br /> ==Video Solution==<br /> https://youtu.be/s4vnGlwwHHw<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12A_Problems/Problem_16&diff=149457 2002 AMC 12A Problems/Problem 16 2021-03-14T04:00:09Z <p>Bronzetruck2016: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #16]] and [[2002 AMC 10A Problems|2002 AMC 10A #24]]}}<br /> <br /> <br /> ==Problem==<br /> Tina randomly selects two distinct numbers from the set &lt;math&gt;\{ 1, 2, 3, 4, 5 \}&lt;/math&gt;, and Sergio randomly selects a number from the set &lt;math&gt;\{ 1, 2, ..., 10 \}&lt;/math&gt;. What is the probability that Sergio's number is larger than the sum of the two numbers chosen by Tina? <br /> <br /> &lt;math&gt;\text{(A)}\ 2/5 \qquad \text{(B)}\ 9/20 \qquad \text{(C)}\ 1/2 \qquad \text{(D)}\ 11/20 \qquad \text{(E)}\ 24/25&lt;/math&gt;<br /> <br /> == Video Solution ==<br /> https://youtu.be/8WrdYLw9_ns?t=381<br /> <br /> ~ pi_is_3.14<br /> <br /> ==Solution==<br /> <br /> === Solution 1 ===<br /> <br /> This is not too bad using casework. <br /> <br /> Tina gets a sum of 3: This happens in only one way &lt;math&gt;(1,2)&lt;/math&gt; and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.<br /> <br /> Tina gets a sum of 4: This once again happens in only one way &lt;math&gt;(1,3)&lt;/math&gt;. Sergio can choose a number from 5 to 10, so 6 ways here.<br /> <br /> Tina gets a sum of 5: This can happen in two ways &lt;math&gt;(1,4)&lt;/math&gt; and &lt;math&gt;(2,3)&lt;/math&gt;. Sergio can choose a number from 6 to 10, so &lt;math&gt;2\cdot5=10&lt;/math&gt; ways here.<br /> <br /> Tina gets a sum of 6: Two ways here &lt;math&gt;(1,5)&lt;/math&gt; and &lt;math&gt;(2,4)&lt;/math&gt;. Sergio can choose a number from 7 to 10, so &lt;math&gt;2\cdot4=8&lt;/math&gt; here.<br /> <br /> Tina gets a sum of 7: Two ways here &lt;math&gt;(2,5)&lt;/math&gt; and &lt;math&gt;(3,4)&lt;/math&gt;. Sergio can choose from 8 to 10, so &lt;math&gt;2\cdot3=6&lt;/math&gt; ways here.<br /> <br /> Tina gets a sum of 8: Only one way possible &lt;math&gt;(3,5&lt;/math&gt;). Sergio chooses 9 or 10, so 2 ways here.<br /> <br /> Tina gets a sum of 9: Only one way &lt;math&gt;(4,5)&lt;/math&gt;. Sergio must choose 10, so 1 way.<br /> <br /> In all, there are &lt;math&gt;7+6+10+8+6+2+1=40&lt;/math&gt; ways. Tina chooses two distinct numbers in &lt;math&gt;\binom{5}{2}=10&lt;/math&gt; ways while Sergio chooses a number in &lt;math&gt;10&lt;/math&gt; ways, so there are &lt;math&gt;10\cdot 10=100&lt;/math&gt; ways in all. Since &lt;math&gt;\frac{40}{100}=\frac{2}{5}&lt;/math&gt;, our answer is &lt;math&gt;\boxed{\text{(A)}\frac{2}{5}}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> We want to find the average of the smallest possible chance of Sergio winning and the largest possible chance of Sergio winning. This is because the probability decreases linearly. The largest possibility of Sergio winning if Tina chooses a 1 and a 1. The chances of Sergio winning is then &lt;math&gt; \frac{8}{10} = \frac{4}{5} &lt;/math&gt; . The smallest possibility of Sergio winning is if Tina chooses a 5 and a 5. The chances of Sergio winning then is &lt;math&gt; \frac{0}{10} = 0 &lt;/math&gt;. The average of &lt;math&gt; \frac{4}{5} &lt;/math&gt; and &lt;math&gt; 0 &lt;/math&gt; is &lt;math&gt;\boxed{\text{(A)}\frac{2}{5}}&lt;/math&gt;.<br /> <br /> -Minor edits by bronzetruck2016<br /> <br /> === Solution 3 ===<br /> We invoke some symmetry. Let &lt;math&gt;T&lt;/math&gt; denote Tina's sum, and let &lt;math&gt;S&lt;/math&gt; denote Sergio's number. Observe that, for &lt;math&gt;i = 2, 3, \ldots, 10&lt;/math&gt;, &lt;math&gt;\text{Pr}(T=i) = \text{Pr}(T=12-i)&lt;/math&gt;.<br /> <br /> If Tina's sum is &lt;math&gt;i&lt;/math&gt;, then the probability that Sergio's number is larger than Tina's sum is &lt;math&gt;\frac{10-i}{10}&lt;/math&gt;. Thus, the probability &lt;math&gt;P&lt;/math&gt; is<br /> <br /> &lt;cmath&gt;P = \text{Pr}(S&gt;T) = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{10-i}{10}&lt;/cmath&gt;<br /> <br /> Using the symmetry observation, we can also write the above sum as<br /> &lt;cmath&gt; P = \sum_{i=2}^{10} \text{Pr}(T=12-i) \times \frac{10-i}{10} = \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{i-2}{10}&lt;/cmath&gt;<br /> where the last equality follows as we reversed the indices of the sum (by replacing &lt;math&gt;12-i&lt;/math&gt; with &lt;math&gt;i&lt;/math&gt;). Thus, adding the two equivalent expressions for &lt;math&gt;P&lt;/math&gt;, we have<br /> <br /> &lt;cmath&gt;<br /> \begin{align*}<br /> 2P &amp;= \sum_{i=2}^{10} \text{Pr}(T=i) \times \left(\frac{10-i}{10} + \frac{i-2}{10}\right) \\<br /> &amp;= \sum_{i=2}^{10} \text{Pr}(T=i) \times \frac{4}{5} \\<br /> &amp;= \frac{4}{5} \sum_{i=2}^{10} \text{Pr}(T=i) \\<br /> &amp;= \frac{4}{5}<br /> \end{align*}<br /> &lt;/cmath&gt;<br /> <br /> Since this represents twice the desired probability, the answer is &lt;math&gt;P = \boxed{\textbf{(A)} \frac{2}{5}}&lt;/math&gt;. -scrabbler94<br /> <br /> ==Solution 4==<br /> We have 5 cases, if Tina choose &lt;math&gt;1, 2, 3, 4,&lt;/math&gt; or &lt;math&gt;5.&lt;/math&gt;<br /> <br /> The number of ways of choosing 2 numbers from &lt;math&gt;5&lt;/math&gt; are &lt;math&gt;\binom{5}{2}&lt;/math&gt;.<br /> --------------------------<br /> Case 1: Tina chooses &lt;math&gt;1&lt;/math&gt;.<br /> <br /> In this case, since the numbers are distinct, Tina can choose &lt;math&gt;(1, 2), (1, 3), (1, 4),&lt;/math&gt; or &lt;math&gt;(1, 5).&lt;/math&gt;<br /> <br /> If Tina chooses &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; which some to &lt;math&gt;3&lt;/math&gt;, Sergio only has &lt;math&gt;10-3=7&lt;/math&gt; choices.<br /> <br /> Since the sum of the combined numbers increases by &lt;math&gt;1&lt;/math&gt; every time for this specific case, Sergio has &lt;math&gt;1&lt;/math&gt; less choice every time.<br /> <br /> Therefore, the probability of this is &lt;math&gt;\frac{7+6+5+4}{10 \cdot \binom{5}{2}}&lt;/math&gt;.<br /> --------------------------<br /> If you do this over and over again you will see that you have &lt;math&gt;\frac{(7+6+5+4)+(7+5+4+3)+(6+5+3+2)+(5+4+3+1)+(4+3+2+1)}{10 \cdot \binom{5}{2}} = \frac{80}{100} = \frac{4}{5}&lt;/math&gt; probability.<br /> <br /> But since we overcounted by 2 (e.g. &lt;math&gt;(1, 2)&lt;/math&gt; and &lt;math&gt;(2, 1)&lt;/math&gt;) we need to divide by &lt;math&gt;2.&lt;/math&gt;<br /> <br /> Thus our answer is &lt;math&gt;\frac{4}{5} \div 2 = \boxed{\textbf{(A)} \frac{2}{5}}.&lt;/math&gt;<br /> <br /> ~mathboy282<br /> <br /> Note: I will add in all the cases soon, kind of busy today so yea.<br /> <br /> === Solution 5 ===<br /> <br /> Assume Sergio chooses from &lt;math&gt;{2,3,\ldots,10}&lt;/math&gt;. The probably of Tina getting a sum of &lt;math&gt;6+x&lt;/math&gt; and &lt;math&gt;6-x&lt;/math&gt; (&lt;math&gt;x \leq 4&lt;/math&gt;) are equal due to symmetry. The probability of Sergio choosing numbers higher/lower than &lt;math&gt;6+x&lt;/math&gt; is equal to him choosing numbers lower/higher than &lt;math&gt;6-x&lt;/math&gt;. Therefore over all of Tina's sums, the probability of Sergio choosing a number higher is equal to the probability of choosing a number lower. <br /> <br /> The probability that they get the same value is &lt;math&gt;1/9&lt;/math&gt;, so the probability of Sergio getting a higher number is &lt;math&gt;\frac{(9-1)/2}{9} = \frac49&lt;/math&gt;. <br /> <br /> Sergio never wins when choosing &lt;math&gt;1&lt;/math&gt; so the probability is &lt;math&gt;\frac49 \frac{9}{10} + (0)\frac{1}{10} = \boxed{\textbf{(A)} \frac{2}{5}}.&lt;/math&gt;<br /> <br /> ~zeric<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2002|ab=A|num-b=15|num-a=17}}<br /> {{AMC10 box|year=2002|ab=A|num-b=23|num-a=25}}<br /> <br /> [[Category:Introductory Probability Problems]]<br /> [[Category:Introductory Combinatorics Problems]]<br /> {{MAA Notice}}</div> Bronzetruck2016 https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10A_Problems/Problem_25&diff=141686 2017 AMC 10A Problems/Problem 25 2021-01-07T20:33:47Z <p>Bronzetruck2016: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> How many integers between &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt;, inclusive, have the property that some permutation of its digits is a multiple of &lt;math&gt;11&lt;/math&gt; between &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;999?&lt;/math&gt; For example, both &lt;math&gt;121&lt;/math&gt; and &lt;math&gt;211&lt;/math&gt; have this property.<br /> <br /> &lt;math&gt; \mathrm{\textbf{(A)} \ }226\qquad \mathrm{\textbf{(B)} \ } 243 \qquad \mathrm{\textbf{(C)} \ } 270 \qquad \mathrm{\textbf{(D)} \ }469\qquad \mathrm{\textbf{(E)} \ } 486&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> There are 81 multiples of 11. Some have digits repeated twice, making 3 permutations.<br /> <br /> Others that have no repeated digits have 6 permutations, but switching the hundreds and units digits also yield a multiple of 11. Therefore, assign 3 permutations to each multiple.<br /> <br /> There are now 81*3 = 243 permutations, but we have overcounted*. Some multiples of 11 have a zero, and we must subtract a permutation for each. <br /> <br /> There are 110, 220, 330 ... 990, yielding 9 extra permutations<br /> <br /> Also, there are 209, 308, 407...902, yielding 8 more permutations.<br /> <br /> Now, just subtract these 17 from the total (243), getting 226. &lt;math&gt;\boxed{\textbf{(A) } 226}&lt;/math&gt;<br /> <br /> *Note: If short on time, note that 226 is the only answer choice less than 243, and therefore is the only feasible answer. <br /> <br /> ===Solution 2===<br /> <br /> Let the three-digit number be &lt;math&gt;ACB&lt;/math&gt;:<br /> <br /> If a number is divisible by &lt;math&gt;11&lt;/math&gt;, then the difference between the sums of alternating digits is a multiple of &lt;math&gt;11&lt;/math&gt;.<br /> <br /> There are two cases:<br /> &lt;math&gt;A+B=C&lt;/math&gt; and &lt;math&gt;A+B=C+11&lt;/math&gt;<br /> <br /> We now proceed to break down the cases. Note: let &lt;math&gt;A \geq B&lt;/math&gt; so that we avoid counting the same permutations and having to subtract them later.<br /> <br /> <br /> &lt;math&gt;\textbf{Case 1}&lt;/math&gt;: &lt;math&gt;A+B=C&lt;/math&gt;. <br /> <br /> <br /> <br /> &lt;math&gt;\textbf{Part 1}&lt;/math&gt;: &lt;math&gt;B=0, A=C&lt;/math&gt;, this case results in &lt;math&gt;110, 220, 330...990&lt;/math&gt;. There are two ways to arrange the digits in each of those numbers.<br /> &lt;math&gt;2 \cdot 9 = 18&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Part 2}&lt;/math&gt;:<br /> &lt;math&gt;B=1, A+1=C&lt;/math&gt;, this case results in &lt;math&gt;121, 231,... 891&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange the digits in all of those number except the first, and &lt;math&gt;3&lt;/math&gt; ways for the first. This leads to &lt;math&gt;45&lt;/math&gt; cases.<br /> <br /> &lt;math&gt;\textbf{Part 3}&lt;/math&gt;: &lt;math&gt;B=2, A+2=C&lt;/math&gt;, this case results in &lt;math&gt;242, 352,... 792&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to &lt;math&gt;33&lt;/math&gt; cases.<br /> <br /> &lt;math&gt;\textbf{Part 4}&lt;/math&gt;: &lt;math&gt;B=3, A+3=C&lt;/math&gt;, this case results in &lt;math&gt;363, 473,...693&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange the digits in all of those number except the first, and 3 ways for the first. This leads to &lt;math&gt;21&lt;/math&gt; cases.<br /> <br /> &lt;math&gt;\textbf{Part 5}&lt;/math&gt;: &lt;math&gt;B=4, A+4=C&lt;/math&gt;, this case results in &lt;math&gt;484&lt;/math&gt; and &lt;math&gt;594&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange the digits in 594 and 3 ways for 484. This leads to &lt;math&gt;9&lt;/math&gt; cases.<br /> <br /> This case has &lt;math&gt;18+45+33+21+9=126&lt;/math&gt; subcases.<br /> <br /> <br /> <br /> &lt;math&gt;\textbf{Case 2}&lt;/math&gt;: &lt;math&gt;A+B=C+11&lt;/math&gt;. <br /> <br /> <br /> &lt;math&gt;\textbf{Part 1}&lt;/math&gt;: &lt;math&gt;C=0, A+B=11&lt;/math&gt;, this cases results in &lt;math&gt;209, 308, 407, 506&lt;/math&gt;. There are &lt;math&gt;4&lt;/math&gt; ways to arrange each of those cases. This leads to &lt;math&gt;16&lt;/math&gt; cases.<br /> <br /> &lt;math&gt;\textbf{Part 2}&lt;/math&gt;: &lt;math&gt;C=1, A+B=12&lt;/math&gt;, this cases results in &lt;math&gt;319, 418,517,616&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange each of those cases, except the last. This leads to &lt;math&gt;21&lt;/math&gt; cases.<br /> <br /> &lt;math&gt;\textbf{Part 3}&lt;/math&gt;: &lt;math&gt;C=2, A+B=13&lt;/math&gt;, this cases results in &lt;math&gt;429, 528, 627&lt;/math&gt;. There are &lt;math&gt;6&lt;/math&gt; ways to arrange each of those cases. This leads to &lt;math&gt;18&lt;/math&gt; cases.<br /> <br /> ...<br /> If we continue this counting, we receive &lt;math&gt;16+21+18+15+12+9+6+3=100&lt;/math&gt; subcases.<br /> <br /> &lt;math&gt;100+126=\boxed{\textbf{(A) } 226}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> We note that we only have to consider multiples of &lt;math&gt;11&lt;/math&gt; and see how many valid permutations each has. We can do casework on the number of repeating digits that the multiple of &lt;math&gt;11&lt;/math&gt; has:<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; All three digits are the same. <br /> By inspection, we find that there are no multiples of &lt;math&gt;11&lt;/math&gt; here.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; Two of the digits are the same, and the third is different.<br /> <br /> &lt;math&gt;\textbf{Case 2a:}&lt;/math&gt;<br /> There are &lt;math&gt;8&lt;/math&gt; multiples of &lt;math&gt;11&lt;/math&gt; without a zero that have this property:<br /> &lt;math&gt;121&lt;/math&gt;, &lt;math&gt;242&lt;/math&gt;, &lt;math&gt;363&lt;/math&gt;, &lt;math&gt;484&lt;/math&gt;, &lt;math&gt;616&lt;/math&gt;, &lt;math&gt;737&lt;/math&gt;, &lt;math&gt;858&lt;/math&gt;, &lt;math&gt;979&lt;/math&gt;.<br /> Each contributes &lt;math&gt;3&lt;/math&gt; valid permutations, so there are &lt;math&gt;8 \cdot 3 = 24&lt;/math&gt; permutations in this subcase.<br /> <br /> &lt;math&gt;\textbf{Case 2b:}&lt;/math&gt;<br /> There are &lt;math&gt;9&lt;/math&gt; multiples of &lt;math&gt;11&lt;/math&gt; with a zero that have this property:<br /> &lt;math&gt;110&lt;/math&gt;, &lt;math&gt;220&lt;/math&gt;, &lt;math&gt;330&lt;/math&gt;, &lt;math&gt;440&lt;/math&gt;, &lt;math&gt;550&lt;/math&gt;, &lt;math&gt;660&lt;/math&gt;, &lt;math&gt;770&lt;/math&gt;, &lt;math&gt;880&lt;/math&gt;, &lt;math&gt;990&lt;/math&gt;.<br /> Each one contributes &lt;math&gt;2&lt;/math&gt; valid permutations (the first digit can't be zero), so there are &lt;math&gt;9 \cdot 2 = 18&lt;/math&gt; permutations in this subcase.<br /> <br /> &lt;math&gt;\textbf{Case 3:}&lt;/math&gt; All the digits are different.<br /> Since there are &lt;math&gt;\frac{990-110}{11}+1 = 81&lt;/math&gt; multiples of &lt;math&gt;11&lt;/math&gt; between &lt;math&gt;100&lt;/math&gt; and &lt;math&gt;999&lt;/math&gt;, there are &lt;math&gt;81-8-9 = 64&lt;/math&gt; multiples of &lt;math&gt;11&lt;/math&gt; remaining in this case. However, &lt;math&gt;8&lt;/math&gt; of them contain a zero, namely &lt;math&gt;209&lt;/math&gt;, &lt;math&gt;308&lt;/math&gt;, &lt;math&gt;407&lt;/math&gt;, &lt;math&gt;506&lt;/math&gt;, &lt;math&gt;605&lt;/math&gt;, &lt;math&gt;704&lt;/math&gt;, &lt;math&gt;803&lt;/math&gt;, and &lt;math&gt;902&lt;/math&gt;. Each of those multiples of &lt;math&gt;11&lt;/math&gt; contributes &lt;math&gt;2 \cdot 2=4&lt;/math&gt; valid permutations, but we overcounted by a factor of &lt;math&gt;2&lt;/math&gt;; every permutation of &lt;math&gt;209&lt;/math&gt;, for example, is also a permutation of &lt;math&gt;902&lt;/math&gt;. Therefore, there are &lt;math&gt;8 \cdot 4 / 2 = 16&lt;/math&gt;. Therefore, there are &lt;math&gt;64-8=56&lt;/math&gt; remaining multiples of &lt;math&gt;11&lt;/math&gt; without a &lt;math&gt;0&lt;/math&gt; in this case. Each one contributes &lt;math&gt;3! = 6&lt;/math&gt; valid permutations, but once again, we overcounted by a factor of &lt;math&gt;2&lt;/math&gt; (note that if a number ABC is a multiple of &lt;math&gt;11&lt;/math&gt;, then so is CBA). Therefore, there are &lt;math&gt;56 \cdot 6 / 2 = 168&lt;/math&gt; valid permutations in this subcase.<br /> <br /> Adding up all the permutations from all the cases, we have &lt;math&gt;24+18+16+168 = \boxed{\textbf{(A) } 226}&lt;/math&gt;.<br /> <br /> ==Solution 4 ==<br /> <br /> We can first overcount and then subtract.<br /> We know that there are &lt;math&gt;81&lt;/math&gt; multiples of &lt;math&gt;11&lt;/math&gt;.<br /> <br /> We can then multiply by &lt;math&gt;6&lt;/math&gt; for each permutation of these multiples. (Yet some multiples do not have six distinct permutations.)<br /> <br /> Now divide by &lt;math&gt;2&lt;/math&gt;, because if a number &lt;math&gt;abc&lt;/math&gt; with digits &lt;math&gt;a&lt;/math&gt;, &lt;math&gt;b&lt;/math&gt;, and &lt;math&gt;c&lt;/math&gt; is a multiple of &lt;math&gt;11&lt;/math&gt;, then &lt;math&gt;cba&lt;/math&gt; is also a multiple of &lt;math&gt;11&lt;/math&gt; so we have counted the same permutations twice. <br /> <br /> Basically, each multiple of &lt;math&gt;11&lt;/math&gt; has its own &lt;math&gt;3&lt;/math&gt; permutations (say &lt;math&gt;abc&lt;/math&gt; has &lt;math&gt;abc&lt;/math&gt; &lt;math&gt;acb&lt;/math&gt; and &lt;math&gt;bac&lt;/math&gt; whereas &lt;math&gt;cba&lt;/math&gt; has &lt;math&gt;cba&lt;/math&gt; &lt;math&gt;cab&lt;/math&gt; and &lt;math&gt;bca&lt;/math&gt;). We know that each multiple of &lt;math&gt;11&lt;/math&gt; has at least &lt;math&gt;3&lt;/math&gt; permutations because it cannot have &lt;math&gt;3&lt;/math&gt; repeating digits.<br /> <br /> Hence we have &lt;math&gt;243&lt;/math&gt; permutations without subtracting for overcounting.<br /> Now note that we overcounted cases in which we have &lt;math&gt;0&lt;/math&gt;'s at the start of each number. So, in theory, we could just answer &lt;math&gt;A&lt;/math&gt; and then move on.<br /> <br /> If we want to solve it, then we continue.<br /> <br /> We overcounted cases where the middle digit of the number is &lt;math&gt;0&lt;/math&gt; and the last digit is &lt;math&gt;0&lt;/math&gt;.<br /> <br /> Note that we assigned each multiple of &lt;math&gt;11&lt;/math&gt; three permutations.<br /> <br /> The last digit is &lt;math&gt;0&lt;/math&gt; gives &lt;math&gt;9&lt;/math&gt; possibilities where we overcounted by &lt;math&gt;1&lt;/math&gt; permutation for each of &lt;math&gt;110, 220, ... , 990&lt;/math&gt;.<br /> <br /> The middle digit is &lt;math&gt;0&lt;/math&gt; gives &lt;math&gt;8&lt;/math&gt; possibilities where we overcount by &lt;math&gt;1&lt;/math&gt;.<br /> &lt;math&gt;605, 704, 803, 902&lt;/math&gt; and &lt;math&gt;506, 407, 308, 209&lt;/math&gt;<br /> <br /> Subtracting &lt;math&gt;17&lt;/math&gt; gives &lt;math&gt;\boxed{\textbf{(A) } 226}&lt;/math&gt;.<br /> <br /> Now, we may ask if there is further overlap (i.e if two of &lt;math&gt;abc&lt;/math&gt; and &lt;math&gt;bac&lt;/math&gt; and &lt;math&gt;acb&lt;/math&gt; were multiples of &lt;math&gt;11&lt;/math&gt;). Thankfully, using divisibility rules, this can never happen, as taking the divisibility rule mod &lt;math&gt;11&lt;/math&gt; and adding, we get that &lt;math&gt;2a&lt;/math&gt;, &lt;math&gt;2b&lt;/math&gt;, or &lt;math&gt;2c&lt;/math&gt; is congruent to &lt;math&gt;0\ (mod\ 11)&lt;/math&gt;. Since &lt;math&gt;a, b, c&lt;/math&gt; are digits, this can never happen as none of them can equal &lt;math&gt;11&lt;/math&gt; and they can't equal &lt;math&gt;0&lt;/math&gt; as they are the leading digit of a three-digit number in each of the cases.<br /> <br /> == Solution 5: A Slightly Adjusted Version of Solution 2 ==<br /> <br /> <br /> &lt;math&gt;\textbf{WARNING:}&lt;/math&gt; If you do not feel comfortable looking at a massive amount of casework, please skip the solution.<br /> <br /> <br /> Recalling the divisibility rule for &lt;math&gt;11&lt;/math&gt;, if we have a number &lt;math&gt;ABC&lt;/math&gt; where &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt; are digits, then &lt;math&gt;11\mid -A+B-C&lt;/math&gt;.<br /> <br /> Notice that for any three-digit positive integer &lt;math&gt;ABC&lt;/math&gt;, &lt;math&gt;-A+B-C&lt;11&lt;/math&gt;, thus we have 2 possibilities: &lt;math&gt;-A+B-C=0&lt;/math&gt; and &lt;math&gt;-A+B-C=-11&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; &lt;math&gt;-A+B-C=0\Longrightarrow A+C=B&lt;/math&gt;<br /> <br /> Subcase &lt;math&gt;a&lt;/math&gt;: &lt;math&gt;A\neq B\neq C\neq0&lt;/math&gt;<br /> <br /> We have these values for &lt;math&gt;A+C=B&lt;/math&gt;<br /> &lt;cmath&gt;1+2=3,1+3=4,1+4=5,...,1+8=9&lt;/cmath&gt;<br /> &lt;cmath&gt;2+3=5,2+4=6,...,2+7=9&lt;/cmath&gt;<br /> &lt;cmath&gt;3+4=7,3+5=8,3+6=9&lt;/cmath&gt;<br /> &lt;cmath&gt;4+5=9&lt;/cmath&gt;<br /> From which we get &lt;math&gt;7+5+3+1=16&lt;/math&gt; triples &lt;math&gt;(A,B,C)&lt;/math&gt;. Counting every permutation, we have &lt;math&gt;16\cdot3!=96&lt;/math&gt; possibilities.<br /> <br /> Subcase &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;A=C&lt;/math&gt;, &lt;math&gt;A,B,C\neq0&lt;/math&gt;<br /> <br /> We have<br /> &lt;cmath&gt;1+1=2,2+2=4,3+3=6,4+4=8&lt;/cmath&gt;<br /> From which we get &lt;math&gt;4\cdot3=12&lt;/math&gt; possibilities.<br /> <br /> Subcase &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;A=B,C=0&lt;/math&gt;<br /> <br /> We have<br /> &lt;cmath&gt;1+0=1,2+0=2,...,9+0=9&lt;/cmath&gt;<br /> Since &lt;math&gt;0&lt;/math&gt; can't be the hundreds digit, from here we get &lt;math&gt;9\cdot2=18&lt;/math&gt; possibilities. Summing up case &lt;math&gt;1&lt;/math&gt;, we have &lt;math&gt;96+12+18=126&lt;/math&gt; possibilities.<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; &lt;math&gt;-A+B-C=-11\Longrightarrow A+C-B=11&lt;/math&gt;<br /> <br /> Subcase &lt;math&gt;a&lt;/math&gt;: &lt;math&gt;A\neq B\neq C\neq0&lt;/math&gt;<br /> <br /> We have these values for &lt;math&gt;A+C-B=11&lt;/math&gt;<br /> &lt;cmath&gt;9+8-6=11,9+7-5=11,...9+3-1=11&lt;/cmath&gt;<br /> &lt;cmath&gt;8+7-4=11,8+6-3=11,8+5-2=11,8+4-1&lt;/cmath&gt;<br /> &lt;cmath&gt;...&lt;/cmath&gt;<br /> From which we get &lt;math&gt;(6+4+2)\cdot3!=72&lt;/math&gt; possibilities.<br /> <br /> Subcase &lt;math&gt;b&lt;/math&gt;: &lt;math&gt;A=C&lt;/math&gt;, &lt;math&gt;A,B,C\neq0&lt;/math&gt;<br /> <br /> We have <br /> &lt;cmath&gt;9+9=7,8+8-5,7+7-3=11,6+6-1=11&lt;/cmath&gt;<br /> From which we get &lt;math&gt;4\cdot3=12&lt;/math&gt; possibilities.<br /> <br /> Subcase &lt;math&gt;c&lt;/math&gt;: &lt;math&gt;B=0\Longrightarrow A+C+11&lt;/math&gt;<br /> <br /> We have<br /> &lt;cmath&gt;9+2=8+3=7+4=6+5=11&lt;/cmath&gt;<br /> From which we get &lt;math&gt;2\cdot2\cdot4=16&lt;/math&gt; possibilities. Summing up case &lt;math&gt;2&lt;/math&gt;, we have &lt;math&gt;72+12+16=100&lt;/math&gt; possibilities.<br /> <br /> Adding the &lt;math&gt;2&lt;/math&gt; cases, we get a total of &lt;math&gt;126+100=226&lt;/math&gt; possibilities. &lt;math&gt;\boxed{\mathrm{(A)}}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> ==Video Solution==<br /> Two different variations on solving it.<br /> https://youtu.be/z5KNZEwmrWM<br /> <br /> https://youtu.be/MBcHwu30MX4<br /> -Video Solution by Richard Rusczyk<br /> <br /> https://youtu.be/Ly69GHOq9Yw<br /> <br /> ~savannahsolver<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2017|ab=A|num-b=24|after=Last Problem}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Bronzetruck2016