https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Jackshi2006&feedformat=atom AoPS Wiki - User contributions [en] 2020-12-04T20:49:48Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_II_Problems/Problem_6&diff=137711 2004 AIME II Problems/Problem 6 2020-11-18T23:55:24Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a [[whole number]] of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the [[ratio]] &lt;math&gt; 3: 2: 1, &lt;/math&gt;what is the least possible total for the number of bananas?<br /> <br /> == Solution ==<br /> Denote the number of bananas the first monkey took from the pile as &lt;math&gt;b_1&lt;/math&gt;, the second &lt;math&gt;b_2&lt;/math&gt;, and the third &lt;math&gt;b_3&lt;/math&gt;; the total is &lt;math&gt;b_1 + b_2 + b_3&lt;/math&gt;. Thus, the first monkey got &lt;math&gt;\frac{3}{4}b_1 + \frac{3}{8}b_2 + \frac{11}{24}b_3&lt;/math&gt;, the second monkey got &lt;math&gt;\frac{1}{8}b_1 + \frac{1}{4}b_2 + \frac{11}{24}b_3&lt;/math&gt;, and the third monkey got &lt;math&gt;\frac{1}{8}b_1 + \frac{3}{8}b_2 + \frac{1}{12}b_3&lt;/math&gt;. <br /> <br /> Taking into account the ratio aspect, say that the third monkey took &lt;math&gt;x&lt;/math&gt; bananas in total. Then,<br /> <br /> &lt;math&gt;x = \frac{1}{4}b_1 + \frac{1}{8}b_2 + \frac{11}{72}b_3 = \frac{1}{16}b_1 + \frac{1}{8}b_2 + \frac{11}{48}b_3 = \frac{1}{8}b_1 + \frac{3}{8}b_2 + \frac{1}{12}b_3&lt;/math&gt;<br /> <br /> Solve this to find that &lt;math&gt;\frac{b_1}{11} = \frac{b_2}{13} = \frac{b_3}{27}&lt;/math&gt;. All three fractions must be integral. Also note some other conditions we have picked up in the course of the problem, namely that &lt;math&gt;b_1&lt;/math&gt; is divisible by &lt;math&gt;8&lt;/math&gt;, &lt;math&gt;b_2&lt;/math&gt; is divisible by &lt;math&gt;8&lt;/math&gt;, and &lt;math&gt;b_3&lt;/math&gt; is divisible by &lt;math&gt;72&lt;/math&gt; (however, since the denominator contains a &lt;math&gt;27&lt;/math&gt;, the factors of &lt;math&gt;3&lt;/math&gt; cancel, and it only really needs to be divisible by &lt;math&gt;8&lt;/math&gt;). Thus, the minimal value is when each fraction is equal to &lt;math&gt;8&lt;/math&gt;, and the solution is &lt;math&gt;8(11 + 13 + 27) = \boxed{408}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> Let the first monkey take &lt;math&gt;8x&lt;/math&gt; bananas, the second monkey take &lt;math&gt;8y&lt;/math&gt;, and the third monkey take &lt;math&gt;24z&lt;/math&gt;. I chose these numbers to make it so, when each monkey splits his bananas, they will get an integer amount of each variable. So, when the first monkey distributes his bananas, he gets &lt;math&gt;6x&lt;/math&gt;, and the other monkeys get &lt;math&gt;1x&lt;/math&gt;. So, we can make expressions for how many bananas each monkey gets. Also, the variables have to be integers too.<br /> <br /> Monkey 1: &lt;math&gt;6x+3y+11z&lt;/math&gt; <br /> <br /> Monkey 2: &lt;math&gt;1x+2y+11z&lt;/math&gt; <br /> <br /> Monkey 3: &lt;math&gt;1x+3y+2z&lt;/math&gt; <br /> <br /> So, they are in a ratio &lt;math&gt;3:2:1&lt;/math&gt;. But, we can turn it into an equation by multiplying the amount of bananas each monkey has by 2, 3, 6. Now, the ratio is &lt;math&gt;6:6:6&lt;/math&gt;, so, &lt;math&gt;12x+6y+22z = 3x+6y+33z = 6x+18y+12z&lt;/math&gt;. Subtracting &lt;math&gt;3x+6y+12z&lt;/math&gt; from all, we get &lt;math&gt;9x+10z = 21z = 3x+12y&lt;/math&gt;. Let's split this into 3 equations.<br /> <br /> &lt;cmath&gt;9x+10z = 21z&lt;/cmath&gt;<br /> &lt;cmath&gt;21z = 3x+12y&lt;/cmath&gt;<br /> &lt;cmath&gt;9x+10z = 3x+12y&lt;/cmath&gt;<br /> <br /> Let's look at the first equation. Rearranging, it gets us &lt;math&gt;9x = 11z&lt;/math&gt;<br /> <br /> We can rearrange the third equation, then divide by 2, then subtract the second equation.<br /> &lt;cmath&gt;21z-6y = 12y-5z&lt;/cmath&gt;<br /> &lt;cmath&gt;26z = 18y&lt;/cmath&gt;<br /> &lt;cmath&gt;13z = 9y&lt;/cmath&gt;.<br /> <br /> It is clear &lt;math&gt;z&lt;/math&gt; is a multiple of 9, so let &lt;math&gt;z = 9&lt;/math&gt;. Then we get the &lt;math&gt;x = 11&lt;/math&gt;, and &lt;math&gt;y = 13&lt;/math&gt;. Testing, we confirm this will get the first monkey 204 bananas, the second 136, and the third, 68. Adding them up, we get that there were &lt;math&gt;\boxed{408}&lt;/math&gt; bananas originally in the pile.<br /> <br /> -AlexLikeMath<br /> <br /> ==Solution 3==<br /> In this solution, you start out the same as solution one. Convert everything into the fractions of largest denominator terms (this is necessary) until you get <br /> <br /> &lt;math&gt;27x=11z&lt;/math&gt;<br /> &lt;math&gt;27y=13z&lt;/math&gt;<br /> <br /> While solving, make sure to leave a list of numbers that must divide &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt;, and &lt;math&gt;z&lt;/math&gt;. For example, just by looking at the basic fractions you receive from writing the starting equations, 24 divides &lt;math&gt;z&lt;/math&gt;, 8 divides &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt;. In the expression above, it's also clear that 27 divides &lt;math&gt;z&lt;/math&gt;, 13 divides &lt;math&gt;y&lt;/math&gt;, and 11 divides &lt;math&gt;z&lt;/math&gt;. You might be wondering why I wrote the expression in terms of z. That's because &lt;math&gt;z&lt;/math&gt; has the largest divisor. The LCM of all it's divisors shows that &lt;math&gt;z&lt;/math&gt; must be divisible by 216. The total amount of peanuts can be found to equal to &lt;math&gt;17z/9&lt;/math&gt;. This means there are two possible solutions under 1000: 408 and 816. Trial and error can be done quickly to find the smallest possible solution to be &lt;math&gt;\boxed{408}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=II|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_4&diff=135979 2006 AIME II Problems/Problem 4 2020-10-28T02:53:57Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let &lt;math&gt; (a_1,a_2,a_3,\ldots,a_{12}) &lt;/math&gt; be a permutation of &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt; for which<br /> <br /> &lt;center&gt;&lt;math&gt; a_1&gt;a_2&gt;a_3&gt;a_4&gt;a_5&gt;a_6 \mathrm{\ and \ } a_6&lt;a_7&lt;a_8&lt;a_9&lt;a_{10}&lt;a_{11}&lt;a_{12}. &lt;/math&gt;&lt;/center&gt;<br /> <br /> An example of such a permutation is &lt;math&gt; (6,5,4,3,2,1,7,8,9,10,11,12). &lt;/math&gt; Find the number of such permutations.<br /> <br /> == Solution ==<br /> <br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;. Now, consider selecting &lt;math&gt;5&lt;/math&gt; of the remaining &lt;math&gt;11&lt;/math&gt; values. Sort these values in descending order, and sort the other &lt;math&gt;6&lt;/math&gt; values in ascending order. Now, let the &lt;math&gt;5&lt;/math&gt; selected values be &lt;math&gt;a_1&lt;/math&gt; through &lt;math&gt;a_5&lt;/math&gt;, and let the remaining &lt;math&gt;6&lt;/math&gt; be &lt;math&gt;a_7&lt;/math&gt; through &lt;math&gt;{a_{12}}&lt;/math&gt;. It is now clear that there is a [[bijection]] between the number of ways to select &lt;math&gt;5&lt;/math&gt; values from &lt;math&gt;11&lt;/math&gt; and ordered 12-tuples &lt;math&gt;(a_1,\ldots,a_{12})&lt;/math&gt;. Thus, there will be &lt;math&gt;{11 \choose 5}=\boxed{462}&lt;/math&gt; such ordered 12-tuples.<br /> <br /> == Solution 2 ==<br /> There are &lt;math&gt;\binom{12}{6}&lt;/math&gt; ways to choose 6 numbers from &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt;, and then there will only be one way to order them. And since that &lt;math&gt;a_6&lt;a_7&lt;/math&gt;, only half of the choices will work, so the answer is &lt;math&gt;\frac{\binom{12}{6}}{2}=462&lt;/math&gt; 12-tuples - mathleticguyyy<br /> <br /> ==Solution 3==<br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;, and either &lt;math&gt;a_1&lt;/math&gt; or &lt;math&gt;a_{12}&lt;/math&gt; is 12.<br /> <br /> <br /> Case 1: &lt;math&gt;a_1 = 12&lt;/math&gt;<br /> <br /> In this case, there are 4 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 6 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_{12}&lt;/math&gt;. &lt;math&gt;\binom{10}{4}&lt;/math&gt; is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.<br /> <br /> <br /> Case 2: &lt;math&gt;a_{12} = 12&lt;/math&gt;<br /> <br /> In this case, there are 5 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 5 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_{12}&lt;/math&gt;. &lt;math&gt;\binom{10}{5}&lt;/math&gt; is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.<br /> <br /> &lt;math&gt;210 + 252 = \boxed{462}&lt;/math&gt; ordered 12-tuples.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_4&diff=135978 2006 AIME II Problems/Problem 4 2020-10-28T02:51:48Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let &lt;math&gt; (a_1,a_2,a_3,\ldots,a_{12}) &lt;/math&gt; be a permutation of &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt; for which<br /> <br /> &lt;center&gt;&lt;math&gt; a_1&gt;a_2&gt;a_3&gt;a_4&gt;a_5&gt;a_6 \mathrm{\ and \ } a_6&lt;a_7&lt;a_8&lt;a_9&lt;a_{10}&lt;a_{11}&lt;a_{12}. &lt;/math&gt;&lt;/center&gt;<br /> <br /> An example of such a permutation is &lt;math&gt; (6,5,4,3,2,1,7,8,9,10,11,12). &lt;/math&gt; Find the number of such permutations.<br /> <br /> == Solution ==<br /> <br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;. Now, consider selecting &lt;math&gt;5&lt;/math&gt; of the remaining &lt;math&gt;11&lt;/math&gt; values. Sort these values in descending order, and sort the other &lt;math&gt;6&lt;/math&gt; values in ascending order. Now, let the &lt;math&gt;5&lt;/math&gt; selected values be &lt;math&gt;a_1&lt;/math&gt; through &lt;math&gt;a_5&lt;/math&gt;, and let the remaining &lt;math&gt;6&lt;/math&gt; be &lt;math&gt;a_7&lt;/math&gt; through &lt;math&gt;{a_{12}}&lt;/math&gt;. It is now clear that there is a [[bijection]] between the number of ways to select &lt;math&gt;5&lt;/math&gt; values from &lt;math&gt;11&lt;/math&gt; and ordered 12-tuples &lt;math&gt;(a_1,\ldots,a_{12})&lt;/math&gt;. Thus, there will be &lt;math&gt;{11 \choose 5}=\boxed{462}&lt;/math&gt; such ordered 12-tuples.<br /> <br /> == Solution 2 ==<br /> There are &lt;math&gt;\binom{12}{6}&lt;/math&gt; ways to choose 6 numbers from &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt;, and then there will only be one way to order them. And since that &lt;math&gt;a_6&lt;a_7&lt;/math&gt;, only half of the choices will work, so the answer is &lt;math&gt;\frac{\binom{12}{6}}{2}=462&lt;/math&gt; 12-tuples - mathleticguyyy<br /> <br /> ==Solution 3==<br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;, and either &lt;math&gt;a_1&lt;/math&gt; or &lt;math&gt;a_{12}&lt;/math&gt; is 12.<br /> <br /> <br /> Case 1: &lt;math&gt;a_1 = 12&lt;/math&gt;<br /> <br /> In this case, there are 4 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 6 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_{12}&lt;/math&gt;. &lt;math&gt;\binom{10}{4}&lt;/math&gt; is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.<br /> <br /> <br /> Case 2: &lt;math&gt;a_{12} = 12&lt;/math&gt;<br /> <br /> In this case, there are 5 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 5 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_{12}&lt;/math&gt;. &lt;math&gt;\binom{10}{5}&lt;/math&gt; is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.<br /> <br /> &lt;math&gt;210 + 252 = \boxed{462}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_4&diff=135977 2006 AIME II Problems/Problem 4 2020-10-28T02:51:31Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let &lt;math&gt; (a_1,a_2,a_3,\ldots,a_{12}) &lt;/math&gt; be a permutation of &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt; for which<br /> <br /> &lt;center&gt;&lt;math&gt; a_1&gt;a_2&gt;a_3&gt;a_4&gt;a_5&gt;a_6 \mathrm{\ and \ } a_6&lt;a_7&lt;a_8&lt;a_9&lt;a_{10}&lt;a_{11}&lt;a_{12}. &lt;/math&gt;&lt;/center&gt;<br /> <br /> An example of such a permutation is &lt;math&gt; (6,5,4,3,2,1,7,8,9,10,11,12). &lt;/math&gt; Find the number of such permutations.<br /> <br /> == Solution ==<br /> <br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;. Now, consider selecting &lt;math&gt;5&lt;/math&gt; of the remaining &lt;math&gt;11&lt;/math&gt; values. Sort these values in descending order, and sort the other &lt;math&gt;6&lt;/math&gt; values in ascending order. Now, let the &lt;math&gt;5&lt;/math&gt; selected values be &lt;math&gt;a_1&lt;/math&gt; through &lt;math&gt;a_5&lt;/math&gt;, and let the remaining &lt;math&gt;6&lt;/math&gt; be &lt;math&gt;a_7&lt;/math&gt; through &lt;math&gt;{a_{12}}&lt;/math&gt;. It is now clear that there is a [[bijection]] between the number of ways to select &lt;math&gt;5&lt;/math&gt; values from &lt;math&gt;11&lt;/math&gt; and ordered 12-tuples &lt;math&gt;(a_1,\ldots,a_{12})&lt;/math&gt;. Thus, there will be &lt;math&gt;{11 \choose 5}=\boxed{462}&lt;/math&gt; such ordered 12-tuples.<br /> <br /> == Solution 2 ==<br /> There are &lt;math&gt;\binom{12}{6}&lt;/math&gt; ways to choose 6 numbers from &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt;, and then there will only be one way to order them. And since that &lt;math&gt;a_6&lt;a_7&lt;/math&gt;, only half of the choices will work, so the answer is &lt;math&gt;\frac{\binom{12}{6}}{2}=462&lt;/math&gt; 12-tuples - mathleticguyyy<br /> <br /> ==Solution 3==<br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;, and either &lt;math&gt;a_1&lt;/math&gt; or &lt;math&gt;a_{12}&lt;/math&gt; is 12.<br /> <br /> <br /> Case 1: &lt;math&gt;a_1 = 12&lt;/math&gt;<br /> <br /> In this case, there are 4 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 6 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_12&lt;/math&gt;. &lt;math&gt;\binom{10}{4}&lt;/math&gt; is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.<br /> <br /> <br /> Case 2: &lt;math&gt;a_12 = 12&lt;/math&gt;<br /> <br /> In this case, there are 5 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 5 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_12&lt;/math&gt;. &lt;math&gt;\binom{10}{5}&lt;/math&gt; is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.<br /> <br /> &lt;math&gt;210 + 252 = \boxed{462}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_4&diff=135976 2006 AIME II Problems/Problem 4 2020-10-28T02:49:36Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let &lt;math&gt; (a_1,a_2,a_3,\ldots,a_{12}) &lt;/math&gt; be a permutation of &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt; for which<br /> <br /> &lt;center&gt;&lt;math&gt; a_1&gt;a_2&gt;a_3&gt;a_4&gt;a_5&gt;a_6 \mathrm{\ and \ } a_6&lt;a_7&lt;a_8&lt;a_9&lt;a_{10}&lt;a_{11}&lt;a_{12}. &lt;/math&gt;&lt;/center&gt;<br /> <br /> An example of such a permutation is &lt;math&gt; (6,5,4,3,2,1,7,8,9,10,11,12). &lt;/math&gt; Find the number of such permutations.<br /> <br /> == Solution ==<br /> <br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;. Now, consider selecting &lt;math&gt;5&lt;/math&gt; of the remaining &lt;math&gt;11&lt;/math&gt; values. Sort these values in descending order, and sort the other &lt;math&gt;6&lt;/math&gt; values in ascending order. Now, let the &lt;math&gt;5&lt;/math&gt; selected values be &lt;math&gt;a_1&lt;/math&gt; through &lt;math&gt;a_5&lt;/math&gt;, and let the remaining &lt;math&gt;6&lt;/math&gt; be &lt;math&gt;a_7&lt;/math&gt; through &lt;math&gt;{a_{12}}&lt;/math&gt;. It is now clear that there is a [[bijection]] between the number of ways to select &lt;math&gt;5&lt;/math&gt; values from &lt;math&gt;11&lt;/math&gt; and ordered 12-tuples &lt;math&gt;(a_1,\ldots,a_{12})&lt;/math&gt;. Thus, there will be &lt;math&gt;{11 \choose 5}=\boxed{462}&lt;/math&gt; such ordered 12-tuples.<br /> <br /> == Solution 2 ==<br /> There are &lt;math&gt;\binom{12}{6}&lt;/math&gt; ways to choose 6 numbers from &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt;, and then there will only be one way to order them. And since that &lt;math&gt;a_6&lt;a_7&lt;/math&gt;, only half of the choices will work, so the answer is &lt;math&gt;\frac{\binom{12}{6}}{2}=462&lt;/math&gt; 12-tuples - mathleticguyyy<br /> <br /> ==Solution 3==<br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;, and either &lt;math&gt;a_1&lt;/math&gt; or &lt;math&gt;a_12&lt;/math&gt; is 12.<br /> <br /> <br /> Case 1: &lt;math&gt;a_1 = 12&lt;/math&gt;<br /> <br /> In this case, there are 4 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 6 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_12&lt;/math&gt;. &lt;math&gt;\binom{10}{4}&lt;/math&gt; is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.<br /> <br /> <br /> Case 2: &lt;math&gt;a_12 = 12&lt;/math&gt;<br /> <br /> In this case, there are 5 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 5 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_12&lt;/math&gt;. &lt;math&gt;\binom{10}{5}&lt;/math&gt; is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.<br /> <br /> &lt;math&gt;210 + 252 = \boxed{462}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_4&diff=135975 2006 AIME II Problems/Problem 4 2020-10-28T02:49:28Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let &lt;math&gt; (a_1,a_2,a_3,\ldots,a_{12}) &lt;/math&gt; be a permutation of &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt; for which<br /> <br /> &lt;center&gt;&lt;math&gt; a_1&gt;a_2&gt;a_3&gt;a_4&gt;a_5&gt;a_6 \mathrm{\ and \ } a_6&lt;a_7&lt;a_8&lt;a_9&lt;a_{10}&lt;a_{11}&lt;a_{12}. &lt;/math&gt;&lt;/center&gt;<br /> <br /> An example of such a permutation is &lt;math&gt; (6,5,4,3,2,1,7,8,9,10,11,12). &lt;/math&gt; Find the number of such permutations.<br /> <br /> == Solution ==<br /> <br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;. Now, consider selecting &lt;math&gt;5&lt;/math&gt; of the remaining &lt;math&gt;11&lt;/math&gt; values. Sort these values in descending order, and sort the other &lt;math&gt;6&lt;/math&gt; values in ascending order. Now, let the &lt;math&gt;5&lt;/math&gt; selected values be &lt;math&gt;a_1&lt;/math&gt; through &lt;math&gt;a_5&lt;/math&gt;, and let the remaining &lt;math&gt;6&lt;/math&gt; be &lt;math&gt;a_7&lt;/math&gt; through &lt;math&gt;{a_{12}}&lt;/math&gt;. It is now clear that there is a [[bijection]] between the number of ways to select &lt;math&gt;5&lt;/math&gt; values from &lt;math&gt;11&lt;/math&gt; and ordered 12-tuples &lt;math&gt;(a_1,\ldots,a_{12})&lt;/math&gt;. Thus, there will be &lt;math&gt;{11 \choose 5}=\boxed{462}&lt;/math&gt; such ordered 12-tuples.<br /> <br /> == Solution 2 ==<br /> There are &lt;math&gt;\binom{12}{6}&lt;/math&gt; ways to choose 6 numbers from &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt;, and then there will only be one way to order them. And since that &lt;math&gt;a_6&lt;a_7&lt;/math&gt;, only half of the choices will work, so the answer is &lt;math&gt;\frac{\binom{12}{6}}{2}=462&lt;/math&gt; 12-tuples - mathleticguyyy<br /> <br /> ==Solution 3==<br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;, and either &lt;math&gt;a_1&lt;/math&gt; or &lt;math&gt;a_12&lt;/math&gt; is 12.<br /> <br /> Case 1: &lt;math&gt;a_1 = 12&lt;/math&gt;<br /> <br /> In this case, there are 4 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 6 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_12&lt;/math&gt;. &lt;math&gt;\binom{10}{4}&lt;/math&gt; is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.<br /> <br /> <br /> Case 2: &lt;math&gt;a_12 = 12&lt;/math&gt;<br /> <br /> In this case, there are 5 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 5 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_12&lt;/math&gt;. &lt;math&gt;\binom{10}{5}&lt;/math&gt; is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.<br /> <br /> &lt;math&gt;210 + 252 = \boxed{462}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_4&diff=135974 2006 AIME II Problems/Problem 4 2020-10-28T02:49:21Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let &lt;math&gt; (a_1,a_2,a_3,\ldots,a_{12}) &lt;/math&gt; be a permutation of &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt; for which<br /> <br /> &lt;center&gt;&lt;math&gt; a_1&gt;a_2&gt;a_3&gt;a_4&gt;a_5&gt;a_6 \mathrm{\ and \ } a_6&lt;a_7&lt;a_8&lt;a_9&lt;a_{10}&lt;a_{11}&lt;a_{12}. &lt;/math&gt;&lt;/center&gt;<br /> <br /> An example of such a permutation is &lt;math&gt; (6,5,4,3,2,1,7,8,9,10,11,12). &lt;/math&gt; Find the number of such permutations.<br /> <br /> == Solution ==<br /> <br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;. Now, consider selecting &lt;math&gt;5&lt;/math&gt; of the remaining &lt;math&gt;11&lt;/math&gt; values. Sort these values in descending order, and sort the other &lt;math&gt;6&lt;/math&gt; values in ascending order. Now, let the &lt;math&gt;5&lt;/math&gt; selected values be &lt;math&gt;a_1&lt;/math&gt; through &lt;math&gt;a_5&lt;/math&gt;, and let the remaining &lt;math&gt;6&lt;/math&gt; be &lt;math&gt;a_7&lt;/math&gt; through &lt;math&gt;{a_{12}}&lt;/math&gt;. It is now clear that there is a [[bijection]] between the number of ways to select &lt;math&gt;5&lt;/math&gt; values from &lt;math&gt;11&lt;/math&gt; and ordered 12-tuples &lt;math&gt;(a_1,\ldots,a_{12})&lt;/math&gt;. Thus, there will be &lt;math&gt;{11 \choose 5}=\boxed{462}&lt;/math&gt; such ordered 12-tuples.<br /> <br /> == Solution 2 ==<br /> There are &lt;math&gt;\binom{12}{6}&lt;/math&gt; ways to choose 6 numbers from &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt;, and then there will only be one way to order them. And since that &lt;math&gt;a_6&lt;a_7&lt;/math&gt;, only half of the choices will work, so the answer is &lt;math&gt;\frac{\binom{12}{6}}{2}=462&lt;/math&gt; 12-tuples - mathleticguyyy<br /> <br /> ==Solution 3==<br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;, and either &lt;math&gt;a_1&lt;/math&gt; or &lt;math&gt;a_12&lt;/math&gt; is 12.<br /> <br /> Case 1: &lt;math&gt;a_1 = 12&lt;/math&gt;<br /> <br /> In this case, there are 4 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 6 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_12&lt;/math&gt;. &lt;math&gt;\binom{10}{4}&lt;/math&gt; is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.<br /> <br /> Case 2: &lt;math&gt;a_12 = 12&lt;/math&gt;<br /> <br /> In this case, there are 5 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 5 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_12&lt;/math&gt;. &lt;math&gt;\binom{10}{5}&lt;/math&gt; is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.<br /> <br /> &lt;math&gt;210 + 252 = \boxed{462}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_4&diff=135973 2006 AIME II Problems/Problem 4 2020-10-28T02:49:12Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> <br /> Let &lt;math&gt; (a_1,a_2,a_3,\ldots,a_{12}) &lt;/math&gt; be a permutation of &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt; for which<br /> <br /> &lt;center&gt;&lt;math&gt; a_1&gt;a_2&gt;a_3&gt;a_4&gt;a_5&gt;a_6 \mathrm{\ and \ } a_6&lt;a_7&lt;a_8&lt;a_9&lt;a_{10}&lt;a_{11}&lt;a_{12}. &lt;/math&gt;&lt;/center&gt;<br /> <br /> An example of such a permutation is &lt;math&gt; (6,5,4,3,2,1,7,8,9,10,11,12). &lt;/math&gt; Find the number of such permutations.<br /> <br /> == Solution ==<br /> <br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;. Now, consider selecting &lt;math&gt;5&lt;/math&gt; of the remaining &lt;math&gt;11&lt;/math&gt; values. Sort these values in descending order, and sort the other &lt;math&gt;6&lt;/math&gt; values in ascending order. Now, let the &lt;math&gt;5&lt;/math&gt; selected values be &lt;math&gt;a_1&lt;/math&gt; through &lt;math&gt;a_5&lt;/math&gt;, and let the remaining &lt;math&gt;6&lt;/math&gt; be &lt;math&gt;a_7&lt;/math&gt; through &lt;math&gt;{a_{12}}&lt;/math&gt;. It is now clear that there is a [[bijection]] between the number of ways to select &lt;math&gt;5&lt;/math&gt; values from &lt;math&gt;11&lt;/math&gt; and ordered 12-tuples &lt;math&gt;(a_1,\ldots,a_{12})&lt;/math&gt;. Thus, there will be &lt;math&gt;{11 \choose 5}=\boxed{462}&lt;/math&gt; such ordered 12-tuples.<br /> <br /> == Solution 2 ==<br /> There are &lt;math&gt;\binom{12}{6}&lt;/math&gt; ways to choose 6 numbers from &lt;math&gt; (1,2,3,\ldots,12) &lt;/math&gt;, and then there will only be one way to order them. And since that &lt;math&gt;a_6&lt;a_7&lt;/math&gt;, only half of the choices will work, so the answer is &lt;math&gt;\frac{\binom{12}{6}}{2}=462&lt;/math&gt; 12-tuples - mathleticguyyy<br /> <br /> ==Solution 3==<br /> Clearly, &lt;math&gt;a_6=1&lt;/math&gt;, and either &lt;math&gt;a_1&lt;/math&gt; or &lt;math&gt;a_12&lt;/math&gt; is 12.<br /> <br /> Case 1: &lt;math&gt;a_1 = 12&lt;/math&gt;<br /> In this case, there are 4 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 6 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_12&lt;/math&gt;. &lt;math&gt;\binom{10}{4}&lt;/math&gt; is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.<br /> <br /> Case 2: &lt;math&gt;a_12 = 12&lt;/math&gt;<br /> In this case, there are 5 empty spaces between &lt;math&gt;a_1&lt;/math&gt; and &lt;math&gt;a_6&lt;/math&gt;, and 5 empty spaces between &lt;math&gt;a_6&lt;/math&gt; and &lt;math&gt;a_12&lt;/math&gt;. &lt;math&gt;\binom{10}{5}&lt;/math&gt; is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.<br /> <br /> &lt;math&gt;210 + 252 = \boxed{462}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=II|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_3&diff=135971 2006 AIME II Problems/Problem 3 2020-10-28T02:34:33Z <p>Jackshi2006: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P &lt;/math&gt; be the product of the first &lt;math&gt;100&lt;/math&gt; [[positive integer | positive]] [[odd integer]]s. Find the largest integer &lt;math&gt;k &lt;/math&gt; such that &lt;math&gt;P &lt;/math&gt; is divisible by &lt;math&gt;3^k .&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Note that the product of the first &lt;math&gt;100&lt;/math&gt; positive odd integers can be written as &lt;math&gt;1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}&lt;/math&gt;<br /> <br /> Hence, we seek the number of threes in &lt;math&gt;200!&lt;/math&gt; decreased by the number of threes in &lt;math&gt;100!.&lt;/math&gt;<br /> <br /> There are <br /> <br /> &lt;math&gt;\left\lfloor \frac{200}{3}\right\rfloor+\left\lfloor\frac{200}{9}\right\rfloor+\left\lfloor \frac{200}{27}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor =66+22+7+2=97&lt;/math&gt;<br /> <br /> threes in &lt;math&gt;200!&lt;/math&gt; and <br /> <br /> &lt;math&gt;\left\lfloor \frac{100}{3}\right\rfloor+\left\lfloor\frac{100}{9}\right\rfloor+\left\lfloor \frac{100}{27}\right\rfloor+\left\lfloor\frac{100}{81}\right\rfloor=33+11+3+1=48 &lt;/math&gt;<br /> <br /> threes in &lt;math&gt;100!&lt;/math&gt;<br /> <br /> Therefore, we have a total of &lt;math&gt;97-48=\boxed{049}&lt;/math&gt; threes.<br /> <br /> For more information, see also [[Factorial#Prime factorization| prime factorizations of a factorial]].<br /> <br /> <br /> == Solution 2 ==<br /> We count the multiples of &lt;math&gt;3^k&lt;/math&gt; below 200 and subtract the count of multiples of &lt;math&gt;2\cdot 3^k&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\left\lfloor \frac{200}{3}\right\rfloor - \left\lfloor \frac{200}{6}\right\rfloor +\left\lfloor\frac{200}{9}\right\rfloor - \left\lfloor \frac{200}{18}\right\rfloor +\left\lfloor \frac{200}{27}\right\rfloor - \left\lfloor \frac{200}{54}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor - \left\lfloor \frac{200}{162}\right\rfloor&lt;/cmath&gt;<br /> &lt;cmath&gt;= 66 - 33 + 22 - 11 + 7 - 3 + 2 - 1 = 49.&lt;/cmath&gt;<br /> <br /> == Solution 3 ==<br /> We can use a modified version of Legendre's Formula. First, we count the number of multiples of 3 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. <br /> <br /> This is the same as the number of multiples of 3 in the sequence 3, 9, 15, 21, ..., 192, 195. There are clearly 33 terms in this sequence. <br /> <br /> Next, we count the number of multiples of 9 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. This is the same as the number of multiples of 9 in the sequence 9, 18, 27, 36, ...., 189, 198 - but there's a catch. Note that every other member of this sequence isn't odd and thus is not part of the product of the first 100 odd integers, so our new sequence is actually 9, 27, 45...189. Divide every term by 9 to get a new sequence; 1, 3, 5...21, which clearly has 11 terms. <br /> <br /> Next, we similarly count the number of multiples of 27 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. This is just 27, 81, 135, 189, so 4 multiples here. <br /> <br /> Finally, we count the number of multiples of 81 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. There is only one such multiple, 81. <br /> <br /> Any power of 3 above 81 doesn't fit into our sequence. <br /> <br /> Finally, we have 33+11+4+1=49.<br /> <br /> Our final answer is 49.<br /> <br /> (Note that I oversimplified this a lot, in real life we wouldn't have to list out the sequences as tediously as I did).<br /> <br /> ==Solution 4==<br /> We are obviously searching for multiples of three set S of odd numbers 1-199. Starting with 3, every number &lt;math&gt;2mod3&lt;/math&gt; in set S will be divisible by 3. In other words, every number &lt;math&gt;3mod6&lt;/math&gt;. This is because the LCM must be divisible by 3, and 2, because the set is comprised of only odd numbers. Using some simple math, there are 33 numbers that fit this description. All you have to do is find the largest odd number that is divisible by 3 and below 200, then add 3 and divide by 6. Next, we find the number of odd digits below 200 that are divisible by 9. The same strategy works, and gives us 11. 27 gives 3, and 81 gives 1. &lt;math&gt;33 + 11 + 4 + 1 = \boxed{45}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> * [[Number Theory]]<br /> {{AIME box|year=2006|n=II|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_II_Problems/Problem_3&diff=135970 2006 AIME II Problems/Problem 3 2020-10-28T02:34:22Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P &lt;/math&gt; be the product of the first &lt;math&gt;100&lt;/math&gt; [[positive integer | positive]] [[odd integer]]s. Find the largest integer &lt;math&gt;k &lt;/math&gt; such that &lt;math&gt;P &lt;/math&gt; is divisible by &lt;math&gt;3^k .&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Note that the product of the first &lt;math&gt;100&lt;/math&gt; positive odd integers can be written as &lt;math&gt;1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}&lt;/math&gt;<br /> <br /> Hence, we seek the number of threes in &lt;math&gt;200!&lt;/math&gt; decreased by the number of threes in &lt;math&gt;100!.&lt;/math&gt;<br /> <br /> There are <br /> <br /> &lt;math&gt;\left\lfloor \frac{200}{3}\right\rfloor+\left\lfloor\frac{200}{9}\right\rfloor+\left\lfloor \frac{200}{27}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor =66+22+7+2=97&lt;/math&gt;<br /> <br /> threes in &lt;math&gt;200!&lt;/math&gt; and <br /> <br /> &lt;math&gt;\left\lfloor \frac{100}{3}\right\rfloor+\left\lfloor\frac{100}{9}\right\rfloor+\left\lfloor \frac{100}{27}\right\rfloor+\left\lfloor\frac{100}{81}\right\rfloor=33+11+3+1=48 &lt;/math&gt;<br /> <br /> threes in &lt;math&gt;100!&lt;/math&gt;<br /> <br /> Therefore, we have a total of &lt;math&gt;97-48=\boxed{049}&lt;/math&gt; threes.<br /> <br /> For more information, see also [[Factorial#Prime factorization| prime factorizations of a factorial]].<br /> <br /> <br /> == Solution 2 ==<br /> We count the multiples of &lt;math&gt;3^k&lt;/math&gt; below 200 and subtract the count of multiples of &lt;math&gt;2\cdot 3^k&lt;/math&gt;:<br /> <br /> &lt;cmath&gt;\left\lfloor \frac{200}{3}\right\rfloor - \left\lfloor \frac{200}{6}\right\rfloor +\left\lfloor\frac{200}{9}\right\rfloor - \left\lfloor \frac{200}{18}\right\rfloor +\left\lfloor \frac{200}{27}\right\rfloor - \left\lfloor \frac{200}{54}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor - \left\lfloor \frac{200}{162}\right\rfloor&lt;/cmath&gt;<br /> &lt;cmath&gt;= 66 - 33 + 22 - 11 + 7 - 3 + 2 - 1 = 49.&lt;/cmath&gt;<br /> <br /> == Solution 3 ==<br /> We can use a modified version of Legendre's Formula. First, we count the number of multiples of 3 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. <br /> <br /> This is the same as the number of multiples of 3 in the sequence 3, 9, 15, 21, ..., 192, 195. There are clearly 33 terms in this sequence. <br /> <br /> Next, we count the number of multiples of 9 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. This is the same as the number of multiples of 9 in the sequence 9, 18, 27, 36, ...., 189, 198 - but there's a catch. Note that every other member of this sequence isn't odd and thus is not part of the product of the first 100 odd integers, so our new sequence is actually 9, 27, 45...189. Divide every term by 9 to get a new sequence; 1, 3, 5...21, which clearly has 11 terms. <br /> <br /> Next, we similarly count the number of multiples of 27 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. This is just 27, 81, 135, 189, so 4 multiples here. <br /> <br /> Finally, we count the number of multiples of 81 in the sequence 1, 3, 5, 7, 9, ..., 195, 197, 199. There is only one such multiple, 81. <br /> <br /> Any power of 3 above 81 doesn't fit into our sequence. <br /> <br /> Finally, we have 33+11+4+1=49.<br /> <br /> Our final answer is 49.<br /> <br /> (Note that I oversimplified this a lot, in real life we wouldn't have to list out the sequences as tediously as I did).<br /> <br /> ==Solution 4==<br /> We are obviously searching for multiples of three set S of odd numbers 1-199. Starting with 3, every number &lt;math&gt;2mod3&lt;/math&gt; in set S will be divisible by 3. In other words, every number &lt;math&gt;3mod6&lt;/math&gt;. This is because the LCM must be divisible by 3, and 2, because the set is comprised of only odd numbers. Using some simple math, there are 33 numbers that fit this description. All you have to do is find the largest odd number that is divisible by 3 and below 200, then add 3 and divide by 6. Next, we find the number of odd digits below 200 that are divisible by 9. The same strategy works, and gives us 11. 27 gives 3, and 81 gives 1. &lt;math&gt;33 + 11 + 4 + 1 = \boxed{45}&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[Number Theory]]<br /> {{AIME box|year=2006|n=II|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_10&diff=135843 2006 AIME I Problems/Problem 10 2020-10-26T03:27:24Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Eight [[circle]]s of [[diameter]] 1 are packed in the first [[quadrant]] of the [[coordinate plane]] as shown. Let region &lt;math&gt; \mathcal{R} &lt;/math&gt; be the union of the eight circular regions. Line &lt;math&gt; l, &lt;/math&gt; with slope 3, divides &lt;math&gt; \mathcal{R} &lt;/math&gt; into two regions of equal area. Line &lt;math&gt; l &lt;/math&gt;'s equation can be expressed in the form &lt;math&gt; ax=by+c, &lt;/math&gt; where &lt;math&gt; a, b, &lt;/math&gt; and &lt;math&gt; c &lt;/math&gt; are positive integers whose [[greatest common divisor]] is 1. Find &lt;math&gt; a^2+b^2+c^2. &lt;/math&gt; <br /> &lt;asy&gt;<br /> size(150);defaultpen(linewidth(0.7));<br /> draw((6.5,0)--origin--(0,6.5), Arrows(5));<br /> int[] array={3,3,2};<br /> int i,j;<br /> for(i=0; i&lt;3; i=i+1) {<br /> for(j=0; j&lt;array[i]; j=j+1) {<br /> draw(Circle((1+2*i,1+2*j),1));<br /> }}<br /> label(&quot;x&quot;, (7,0));<br /> label(&quot;y&quot;, (0,7));&lt;/asy&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> The line passing through the [[tangent (geometry)|tangency point]] of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through &lt;math&gt;\left(1,\frac 12\right)&lt;/math&gt; and &lt;math&gt;\left(\frac 32,2\right)&lt;/math&gt;, which can be easily solved to be &lt;math&gt;6x = 2y + 5&lt;/math&gt;. Thus, &lt;math&gt;a^2 + b^2 + c^2 = \boxed{065}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Assume that if unit [[square]]s are drawn circumscribing the circles, then the line will divide the area of the [[concave]] hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the [[x-axis]] as &lt;math&gt;(x, 0)&lt;/math&gt;. <br /> <br /> The line divides the region into 2 sections. The left piece is a [[trapezoid]], with its area &lt;math&gt;\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}&lt;/math&gt;. The right piece is the addition of a [[trapezoid]] and a [[rectangle]], and the areas are &lt;math&gt;\frac{1}{2}((1-x) + (2-x))(3)&lt;/math&gt; and &lt;math&gt;2 \cdot 1 = 2&lt;/math&gt;, totaling &lt;math&gt;\frac{13}{2} - 3x&lt;/math&gt;. Since we want the two regions to be equal, we find that &lt;math&gt;3x + \frac 32 = \frac {13}2 - 3x&lt;/math&gt;, so &lt;math&gt;x = \frac{5}{6}&lt;/math&gt;. <br /> <br /> We have that &lt;math&gt;\left(\frac 56, 0\right)&lt;/math&gt; is a point on the line of slope 3, so &lt;math&gt;y - 0 = 3\left(x - \frac 56\right) \Longrightarrow 6x = 2y + 5&lt;/math&gt;. Our answer is &lt;math&gt;2^2 + 5^2 + 6^2 = 65&lt;/math&gt;.<br /> <br /> We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.<br /> <br /> <br /> === Solution 3===<br /> This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: &lt;math&gt;6x = 2y + 5&lt;/math&gt;. The answer is then &lt;math&gt;6^2 + 2^2 + 5^2 = \boxed{065}&lt;/math&gt;. This solution is an alternate explanation to solution 1.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_10&diff=135842 2006 AIME I Problems/Problem 10 2020-10-26T03:26:54Z <p>Jackshi2006: /* Solution 3 Alternate explanation to Solution 1 */</p> <hr /> <div>== Problem ==<br /> Eight [[circle]]s of [[diameter]] 1 are packed in the first [[quadrant]] of the [[coordinate plane]] as shown. Let region &lt;math&gt; \mathcal{R} &lt;/math&gt; be the union of the eight circular regions. Line &lt;math&gt; l, &lt;/math&gt; with slope 3, divides &lt;math&gt; \mathcal{R} &lt;/math&gt; into two regions of equal area. Line &lt;math&gt; l &lt;/math&gt;'s equation can be expressed in the form &lt;math&gt; ax=by+c, &lt;/math&gt; where &lt;math&gt; a, b, &lt;/math&gt; and &lt;math&gt; c &lt;/math&gt; are positive integers whose [[greatest common divisor]] is 1. Find &lt;math&gt; a^2+b^2+c^2. &lt;/math&gt; <br /> &lt;asy&gt;<br /> size(150);defaultpen(linewidth(0.7));<br /> draw((6.5,0)--origin--(0,6.5), Arrows(5));<br /> int[] array={3,3,2};<br /> int i,j;<br /> for(i=0; i&lt;3; i=i+1) {<br /> for(j=0; j&lt;array[i]; j=j+1) {<br /> draw(Circle((1+2*i,1+2*j),1));<br /> }}<br /> label(&quot;x&quot;, (7,0));<br /> label(&quot;y&quot;, (0,7));&lt;/asy&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> The line passing through the [[tangent (geometry)|tangency point]] of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through &lt;math&gt;\left(1,\frac 12\right)&lt;/math&gt; and &lt;math&gt;\left(\frac 32,2\right)&lt;/math&gt;, which can be easily solved to be &lt;math&gt;6x = 2y + 5&lt;/math&gt;. Thus, &lt;math&gt;a^2 + b^2 + c^2 = \boxed{065}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Assume that if unit [[square]]s are drawn circumscribing the circles, then the line will divide the area of the [[concave]] hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the [[x-axis]] as &lt;math&gt;(x, 0)&lt;/math&gt;. <br /> <br /> The line divides the region into 2 sections. The left piece is a [[trapezoid]], with its area &lt;math&gt;\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}&lt;/math&gt;. The right piece is the addition of a [[trapezoid]] and a [[rectangle]], and the areas are &lt;math&gt;\frac{1}{2}((1-x) + (2-x))(3)&lt;/math&gt; and &lt;math&gt;2 \cdot 1 = 2&lt;/math&gt;, totaling &lt;math&gt;\frac{13}{2} - 3x&lt;/math&gt;. Since we want the two regions to be equal, we find that &lt;math&gt;3x + \frac 32 = \frac {13}2 - 3x&lt;/math&gt;, so &lt;math&gt;x = \frac{5}{6}&lt;/math&gt;. <br /> <br /> We have that &lt;math&gt;\left(\frac 56, 0\right)&lt;/math&gt; is a point on the line of slope 3, so &lt;math&gt;y - 0 = 3\left(x - \frac 56\right) \Longrightarrow 6x = 2y + 5&lt;/math&gt;. Our answer is &lt;math&gt;2^2 + 5^2 + 6^2 = 65&lt;/math&gt;.<br /> <br /> We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.<br /> <br /> <br /> === Solution 3===<br /> This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: &lt;math&gt;6x = 2y + 5&lt;/math&gt;. The answer is then &lt;math&gt;6^2 + 2^2 + 5^2 = \boxed{65}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_10&diff=135841 2006 AIME I Problems/Problem 10 2020-10-26T03:26:43Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Eight [[circle]]s of [[diameter]] 1 are packed in the first [[quadrant]] of the [[coordinate plane]] as shown. Let region &lt;math&gt; \mathcal{R} &lt;/math&gt; be the union of the eight circular regions. Line &lt;math&gt; l, &lt;/math&gt; with slope 3, divides &lt;math&gt; \mathcal{R} &lt;/math&gt; into two regions of equal area. Line &lt;math&gt; l &lt;/math&gt;'s equation can be expressed in the form &lt;math&gt; ax=by+c, &lt;/math&gt; where &lt;math&gt; a, b, &lt;/math&gt; and &lt;math&gt; c &lt;/math&gt; are positive integers whose [[greatest common divisor]] is 1. Find &lt;math&gt; a^2+b^2+c^2. &lt;/math&gt; <br /> &lt;asy&gt;<br /> size(150);defaultpen(linewidth(0.7));<br /> draw((6.5,0)--origin--(0,6.5), Arrows(5));<br /> int[] array={3,3,2};<br /> int i,j;<br /> for(i=0; i&lt;3; i=i+1) {<br /> for(j=0; j&lt;array[i]; j=j+1) {<br /> draw(Circle((1+2*i,1+2*j),1));<br /> }}<br /> label(&quot;x&quot;, (7,0));<br /> label(&quot;y&quot;, (0,7));&lt;/asy&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> The line passing through the [[tangent (geometry)|tangency point]] of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through &lt;math&gt;\left(1,\frac 12\right)&lt;/math&gt; and &lt;math&gt;\left(\frac 32,2\right)&lt;/math&gt;, which can be easily solved to be &lt;math&gt;6x = 2y + 5&lt;/math&gt;. Thus, &lt;math&gt;a^2 + b^2 + c^2 = \boxed{065}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Assume that if unit [[square]]s are drawn circumscribing the circles, then the line will divide the area of the [[concave]] hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the [[x-axis]] as &lt;math&gt;(x, 0)&lt;/math&gt;. <br /> <br /> The line divides the region into 2 sections. The left piece is a [[trapezoid]], with its area &lt;math&gt;\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}&lt;/math&gt;. The right piece is the addition of a [[trapezoid]] and a [[rectangle]], and the areas are &lt;math&gt;\frac{1}{2}((1-x) + (2-x))(3)&lt;/math&gt; and &lt;math&gt;2 \cdot 1 = 2&lt;/math&gt;, totaling &lt;math&gt;\frac{13}{2} - 3x&lt;/math&gt;. Since we want the two regions to be equal, we find that &lt;math&gt;3x + \frac 32 = \frac {13}2 - 3x&lt;/math&gt;, so &lt;math&gt;x = \frac{5}{6}&lt;/math&gt;. <br /> <br /> We have that &lt;math&gt;\left(\frac 56, 0\right)&lt;/math&gt; is a point on the line of slope 3, so &lt;math&gt;y - 0 = 3\left(x - \frac 56\right) \Longrightarrow 6x = 2y + 5&lt;/math&gt;. Our answer is &lt;math&gt;2^2 + 5^2 + 6^2 = 65&lt;/math&gt;.<br /> <br /> We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.<br /> <br /> <br /> === Solution 3 Alternate explanation to Solution 1===<br /> This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: &lt;math&gt;6x = 2y + 5&lt;/math&gt;. The answer is then &lt;math&gt;6^2 + 2^2 + 5^2 = \boxed{65}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_10&diff=135840 2006 AIME I Problems/Problem 10 2020-10-26T03:26:20Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Eight [[circle]]s of [[diameter]] 1 are packed in the first [[quadrant]] of the [[coordinate plane]] as shown. Let region &lt;math&gt; \mathcal{R} &lt;/math&gt; be the union of the eight circular regions. Line &lt;math&gt; l, &lt;/math&gt; with slope 3, divides &lt;math&gt; \mathcal{R} &lt;/math&gt; into two regions of equal area. Line &lt;math&gt; l &lt;/math&gt;'s equation can be expressed in the form &lt;math&gt; ax=by+c, &lt;/math&gt; where &lt;math&gt; a, b, &lt;/math&gt; and &lt;math&gt; c &lt;/math&gt; are positive integers whose [[greatest common divisor]] is 1. Find &lt;math&gt; a^2+b^2+c^2. &lt;/math&gt; <br /> &lt;asy&gt;<br /> size(150);defaultpen(linewidth(0.7));<br /> draw((6.5,0)--origin--(0,6.5), Arrows(5));<br /> int[] array={3,3,2};<br /> int i,j;<br /> for(i=0; i&lt;3; i=i+1) {<br /> for(j=0; j&lt;array[i]; j=j+1) {<br /> draw(Circle((1+2*i,1+2*j),1));<br /> }}<br /> label(&quot;x&quot;, (7,0));<br /> label(&quot;y&quot;, (0,7));&lt;/asy&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> The line passing through the [[tangent (geometry)|tangency point]] of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through &lt;math&gt;\left(1,\frac 12\right)&lt;/math&gt; and &lt;math&gt;\left(\frac 32,2\right)&lt;/math&gt;, which can be easily solved to be &lt;math&gt;6x = 2y + 5&lt;/math&gt;. Thus, &lt;math&gt;a^2 + b^2 + c^2 = \boxed{065}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Assume that if unit [[square]]s are drawn circumscribing the circles, then the line will divide the area of the [[concave]] hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the [[x-axis]] as &lt;math&gt;(x, 0)&lt;/math&gt;. <br /> <br /> The line divides the region into 2 sections. The left piece is a [[trapezoid]], with its area &lt;math&gt;\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}&lt;/math&gt;. The right piece is the addition of a [[trapezoid]] and a [[rectangle]], and the areas are &lt;math&gt;\frac{1}{2}((1-x) + (2-x))(3)&lt;/math&gt; and &lt;math&gt;2 \cdot 1 = 2&lt;/math&gt;, totaling &lt;math&gt;\frac{13}{2} - 3x&lt;/math&gt;. Since we want the two regions to be equal, we find that &lt;math&gt;3x + \frac 32 = \frac {13}2 - 3x&lt;/math&gt;, so &lt;math&gt;x = \frac{5}{6}&lt;/math&gt;. <br /> <br /> We have that &lt;math&gt;\left(\frac 56, 0\right)&lt;/math&gt; is a point on the line of slope 3, so &lt;math&gt;y - 0 = 3\left(x - \frac 56\right) \Longrightarrow 6x = 2y + 5&lt;/math&gt;. Our answer is &lt;math&gt;2^2 + 5^2 + 6^2 = 65&lt;/math&gt;.<br /> <br /> We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.<br /> <br /> <br /> === Solution 3 ===<br /> A better explanation for solution 1:<br /> This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: &lt;math&gt;6x = 2y + 5&lt;/math&gt;. The answer is then &lt;math&gt;6^2 + 2^2 + 5^2 = \boxed{65}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_10&diff=135839 2006 AIME I Problems/Problem 10 2020-10-26T03:25:41Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Eight [[circle]]s of [[diameter]] 1 are packed in the first [[quadrant]] of the [[coordinate plane]] as shown. Let region &lt;math&gt; \mathcal{R} &lt;/math&gt; be the union of the eight circular regions. Line &lt;math&gt; l, &lt;/math&gt; with slope 3, divides &lt;math&gt; \mathcal{R} &lt;/math&gt; into two regions of equal area. Line &lt;math&gt; l &lt;/math&gt;'s equation can be expressed in the form &lt;math&gt; ax=by+c, &lt;/math&gt; where &lt;math&gt; a, b, &lt;/math&gt; and &lt;math&gt; c &lt;/math&gt; are positive integers whose [[greatest common divisor]] is 1. Find &lt;math&gt; a^2+b^2+c^2. &lt;/math&gt; <br /> &lt;asy&gt;<br /> size(150);defaultpen(linewidth(0.7));<br /> draw((6.5,0)--origin--(0,6.5), Arrows(5));<br /> int[] array={3,3,2};<br /> int i,j;<br /> for(i=0; i&lt;3; i=i+1) {<br /> for(j=0; j&lt;array[i]; j=j+1) {<br /> draw(Circle((1+2*i,1+2*j),1));<br /> }}<br /> label(&quot;x&quot;, (7,0));<br /> label(&quot;y&quot;, (0,7));&lt;/asy&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> The line passing through the [[tangent (geometry)|tangency point]] of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through &lt;math&gt;\left(1,\frac 12\right)&lt;/math&gt; and &lt;math&gt;\left(\frac 32,2\right)&lt;/math&gt;, which can be easily solved to be &lt;math&gt;6x = 2y + 5&lt;/math&gt;. Thus, &lt;math&gt;a^2 + b^2 + c^2 = \boxed{065}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Assume that if unit [[square]]s are drawn circumscribing the circles, then the line will divide the area of the [[concave]] hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the [[x-axis]] as &lt;math&gt;(x, 0)&lt;/math&gt;. <br /> <br /> The line divides the region into 2 sections. The left piece is a [[trapezoid]], with its area &lt;math&gt;\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}&lt;/math&gt;. The right piece is the addition of a [[trapezoid]] and a [[rectangle]], and the areas are &lt;math&gt;\frac{1}{2}((1-x) + (2-x))(3)&lt;/math&gt; and &lt;math&gt;2 \cdot 1 = 2&lt;/math&gt;, totaling &lt;math&gt;\frac{13}{2} - 3x&lt;/math&gt;. Since we want the two regions to be equal, we find that &lt;math&gt;3x + \frac 32 = \frac {13}2 - 3x&lt;/math&gt;, so &lt;math&gt;x = \frac{5}{6}&lt;/math&gt;. <br /> <br /> We have that &lt;math&gt;\left(\frac 56, 0\right)&lt;/math&gt; is a point on the line of slope 3, so &lt;math&gt;y - 0 = 3\left(x - \frac 56\right) \Longrightarrow 6x = 2y + 5&lt;/math&gt;. Our answer is &lt;math&gt;2^2 + 5^2 + 6^2 = 65&lt;/math&gt;.<br /> <br /> We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.<br /> <br /> <br /> === Solution 3 ===<br /> This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slope 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: &lt;math&gt;6x = 2y + 5&lt;/math&gt;. The answer is then &lt;math&gt;6^2 + 2^2 + 5^2 = \boxed{65}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_10&diff=135838 2006 AIME I Problems/Problem 10 2020-10-26T03:25:26Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Eight [[circle]]s of [[diameter]] 1 are packed in the first [[quadrant]] of the [[coordinate plane]] as shown. Let region &lt;math&gt; \mathcal{R} &lt;/math&gt; be the union of the eight circular regions. Line &lt;math&gt; l, &lt;/math&gt; with slope 3, divides &lt;math&gt; \mathcal{R} &lt;/math&gt; into two regions of equal area. Line &lt;math&gt; l &lt;/math&gt;'s equation can be expressed in the form &lt;math&gt; ax=by+c, &lt;/math&gt; where &lt;math&gt; a, b, &lt;/math&gt; and &lt;math&gt; c &lt;/math&gt; are positive integers whose [[greatest common divisor]] is 1. Find &lt;math&gt; a^2+b^2+c^2. &lt;/math&gt; <br /> &lt;asy&gt;<br /> size(150);defaultpen(linewidth(0.7));<br /> draw((6.5,0)--origin--(0,6.5), Arrows(5));<br /> int[] array={3,3,2};<br /> int i,j;<br /> for(i=0; i&lt;3; i=i+1) {<br /> for(j=0; j&lt;array[i]; j=j+1) {<br /> draw(Circle((1+2*i,1+2*j),1));<br /> }}<br /> label(&quot;x&quot;, (7,0));<br /> label(&quot;y&quot;, (0,7));&lt;/asy&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> The line passing through the [[tangent (geometry)|tangency point]] of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through &lt;math&gt;\left(1,\frac 12\right)&lt;/math&gt; and &lt;math&gt;\left(\frac 32,2\right)&lt;/math&gt;, which can be easily solved to be &lt;math&gt;6x = 2y + 5&lt;/math&gt;. Thus, &lt;math&gt;a^2 + b^2 + c^2 = \boxed{065}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Assume that if unit [[square]]s are drawn circumscribing the circles, then the line will divide the area of the [[concave]] hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the [[x-axis]] as &lt;math&gt;(x, 0)&lt;/math&gt;. <br /> <br /> The line divides the region into 2 sections. The left piece is a [[trapezoid]], with its area &lt;math&gt;\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}&lt;/math&gt;. The right piece is the addition of a [[trapezoid]] and a [[rectangle]], and the areas are &lt;math&gt;\frac{1}{2}((1-x) + (2-x))(3)&lt;/math&gt; and &lt;math&gt;2 \cdot 1 = 2&lt;/math&gt;, totaling &lt;math&gt;\frac{13}{2} - 3x&lt;/math&gt;. Since we want the two regions to be equal, we find that &lt;math&gt;3x + \frac 32 = \frac {13}2 - 3x&lt;/math&gt;, so &lt;math&gt;x = \frac{5}{6}&lt;/math&gt;. <br /> <br /> We have that &lt;math&gt;\left(\frac 56, 0\right)&lt;/math&gt; is a point on the line of slope 3, so &lt;math&gt;y - 0 = 3\left(x - \frac 56\right) \Longrightarrow 6x = 2y + 5&lt;/math&gt;. Our answer is &lt;math&gt;2^2 + 5^2 + 6^2 = 65&lt;/math&gt;.<br /> <br /> We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.<br /> <br /> <br /> === Solution 3 ===<br /> This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines that simplify the problem by removing sections of the total area. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slop 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: &lt;math&gt;6x = 2y + 5&lt;/math&gt;. The answer is then &lt;math&gt;6^2 + 2^2 + 5^2 = \boxed{65}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_10&diff=135837 2006 AIME I Problems/Problem 10 2020-10-26T03:25:02Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Eight [[circle]]s of [[diameter]] 1 are packed in the first [[quadrant]] of the [[coordinate plane]] as shown. Let region &lt;math&gt; \mathcal{R} &lt;/math&gt; be the union of the eight circular regions. Line &lt;math&gt; l, &lt;/math&gt; with slope 3, divides &lt;math&gt; \mathcal{R} &lt;/math&gt; into two regions of equal area. Line &lt;math&gt; l &lt;/math&gt;'s equation can be expressed in the form &lt;math&gt; ax=by+c, &lt;/math&gt; where &lt;math&gt; a, b, &lt;/math&gt; and &lt;math&gt; c &lt;/math&gt; are positive integers whose [[greatest common divisor]] is 1. Find &lt;math&gt; a^2+b^2+c^2. &lt;/math&gt; <br /> &lt;asy&gt;<br /> size(150);defaultpen(linewidth(0.7));<br /> draw((6.5,0)--origin--(0,6.5), Arrows(5));<br /> int[] array={3,3,2};<br /> int i,j;<br /> for(i=0; i&lt;3; i=i+1) {<br /> for(j=0; j&lt;array[i]; j=j+1) {<br /> draw(Circle((1+2*i,1+2*j),1));<br /> }}<br /> label(&quot;x&quot;, (7,0));<br /> label(&quot;y&quot;, (0,7));&lt;/asy&gt;<br /> <br /> == Solutions ==<br /> === Solution 1 ===<br /> The line passing through the [[tangent (geometry)|tangency point]] of the bottom left circle and the one to its right and through the tangency of the top circle in the middle column and the one beneath it is the line we are looking for: a line passing through the tangency of two circles cuts congruent areas, so our line cuts through the four aforementioned circles splitting into congruent areas, and there are an additional two circles on each side. The line passes through &lt;math&gt;\left(1,\frac 12\right)&lt;/math&gt; and &lt;math&gt;\left(\frac 32,2\right)&lt;/math&gt;, which can be easily solved to be &lt;math&gt;6x = 2y + 5&lt;/math&gt;. Thus, &lt;math&gt;a^2 + b^2 + c^2 = \boxed{065}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Assume that if unit [[square]]s are drawn circumscribing the circles, then the line will divide the area of the [[concave]] hexagonal region of the squares equally (as of yet, there is no substantiation that such would work, and definitely will not work in general). Denote the intersection of the line and the [[x-axis]] as &lt;math&gt;(x, 0)&lt;/math&gt;. <br /> <br /> The line divides the region into 2 sections. The left piece is a [[trapezoid]], with its area &lt;math&gt;\frac{1}{2}((x) + (x+1))(3) = 3x + \frac{3}{2}&lt;/math&gt;. The right piece is the addition of a [[trapezoid]] and a [[rectangle]], and the areas are &lt;math&gt;\frac{1}{2}((1-x) + (2-x))(3)&lt;/math&gt; and &lt;math&gt;2 \cdot 1 = 2&lt;/math&gt;, totaling &lt;math&gt;\frac{13}{2} - 3x&lt;/math&gt;. Since we want the two regions to be equal, we find that &lt;math&gt;3x + \frac 32 = \frac {13}2 - 3x&lt;/math&gt;, so &lt;math&gt;x = \frac{5}{6}&lt;/math&gt;. <br /> <br /> We have that &lt;math&gt;\left(\frac 56, 0\right)&lt;/math&gt; is a point on the line of slope 3, so &lt;math&gt;y - 0 = 3\left(x - \frac 56\right) \Longrightarrow 6x = 2y + 5&lt;/math&gt;. Our answer is &lt;math&gt;2^2 + 5^2 + 6^2 = 65&lt;/math&gt;.<br /> <br /> We now assess the validity of our starting assumption. We can do that by seeing that our answer passes through the tangency of the two circles, cutting congruent areas, a result explored in solution 1.<br /> <br /> <br /> === Solution 3 ===<br /> This problem looks daunting at a first glance, but we can make geometric inequality inferences by drawing lines. These inferences simplify the problem by removing sections of the total area, and reducing the sample size we work with. To begin, we can eliminate the possibility of the line intersecting the circle on the top left (call it circle A), or the circle on the bottom right (call it circle B). This is can be seen visually by drawing a line with slop 3 that is tangent to either of these circles. The area is clearly larger on one side; this can be proven by counting full circles. We can go on with the same mindset and eliminate the circle below circle A and the circle above circle B. By removing pairs of circles and proving the line will never intersect with them, we can safely work with whatever is remaining. By now you should have 4 circles making an L shape (waluigi style). Now the two biggest contenders for this method are the two circles on the bottom row. Using the same strategy, we can see that a line that goes through the tangent point of these two circles also goes through the tangent point of the other two circles. This clearly will cut the 4 circles into two regions of equal area. Using super advanced linear algebra we get: &lt;math&gt;6x = 2y + 5&lt;/math&gt;. The answer is then &lt;math&gt;6^2 + 2^2 + 5^2 = /boxed{65}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=9|num-a=11}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_8&diff=135557 2006 AIME I Problems/Problem 8 2020-10-22T03:37:51Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> [[Hexagon]] &lt;math&gt; ABCDEF &lt;/math&gt; is divided into five [[rhombus]]es, &lt;math&gt; \mathcal{P, Q, R, S,} &lt;/math&gt; and &lt;math&gt; \mathcal{T,} &lt;/math&gt; as shown. Rhombuses &lt;math&gt; \mathcal{P, Q, R,} &lt;/math&gt; and &lt;math&gt; \mathcal{S} &lt;/math&gt; are [[congruent (geometry) | congruent]], and each has [[area]] &lt;math&gt; \sqrt{2006}. &lt;/math&gt; Let &lt;math&gt; K &lt;/math&gt; be the area of rhombus &lt;math&gt; \mathcal{T}&lt;/math&gt;. Given that &lt;math&gt; K &lt;/math&gt; is a [[positive integer]], find the number of possible values for &lt;math&gt; K&lt;/math&gt;.<br /> <br /> [[Image:2006AimeA8.PNG]]<br /> <br /> == Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; denote the common side length of the rhombi.<br /> Let &lt;math&gt;y&lt;/math&gt; denote one of the smaller interior [[angle]]s of rhombus &lt;math&gt; \mathcal{P} &lt;/math&gt;. Then &lt;math&gt;x^2\sin(y)=\sqrt{2006}&lt;/math&gt;. We also see that &lt;math&gt;K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y&lt;/math&gt;. Thus &lt;math&gt;K&lt;/math&gt; can be any positive integer in the [[interval]] &lt;math&gt;(0, 2\sqrt{2006})&lt;/math&gt;. <br /> &lt;math&gt;2\sqrt{2006} = \sqrt{8024}&lt;/math&gt; and &lt;math&gt;89^2 = 7921 &lt; 8024 &lt; 8100 = 90^2&lt;/math&gt;, so &lt;math&gt;K&lt;/math&gt; can be any [[integer]] between 1 and 89, inclusive. Thus the number of positive values for &lt;math&gt;K&lt;/math&gt; is &lt;math&gt;\boxed{089}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where &lt;math&gt;w*h=\sqrt{2006}&lt;/math&gt;. The height of rhombus T would be 2h, and the width would be &lt;math&gt;\sqrt{w^2-h^2}&lt;/math&gt;. Substitute the first equation to get &lt;math&gt;\sqrt{\frac{2006}{h^2}-h^2}&lt;/math&gt;. Then the area of the rhombus would be &lt;math&gt;2h * \sqrt{\frac{2006}{h^2}-h^2}&lt;/math&gt;. Combine like terms to get &lt;math&gt;2 * \sqrt{2006-h^4}&lt;/math&gt;. This expression equals an integer K. &lt;math&gt;2006-h^4&lt;/math&gt; specifically must be in the form &lt;math&gt;n^2/2&lt;/math&gt;. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of &lt;math&gt;n^2&lt;/math&gt; for &lt;math&gt;2006-h^4&lt;/math&gt;. Now, quick testing shows that &lt;math&gt;44^2 &lt; 2006&lt;/math&gt; and &lt;math&gt;45^2&gt;2006&lt;/math&gt;, but we must also test &lt;math&gt;44.5^2&lt;/math&gt;, because the product of two will make it an integer. &lt;math&gt;44.5^2&lt;/math&gt; is also less than &lt;math&gt;2006&lt;/math&gt;, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us &lt;math&gt;44*2+1=&lt;/math&gt; &lt;math&gt;\boxed{089}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_8&diff=135556 2006 AIME I Problems/Problem 8 2020-10-22T03:36:10Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> [[Hexagon]] &lt;math&gt; ABCDEF &lt;/math&gt; is divided into five [[rhombus]]es, &lt;math&gt; \mathcal{P, Q, R, S,} &lt;/math&gt; and &lt;math&gt; \mathcal{T,} &lt;/math&gt; as shown. Rhombuses &lt;math&gt; \mathcal{P, Q, R,} &lt;/math&gt; and &lt;math&gt; \mathcal{S} &lt;/math&gt; are [[congruent (geometry) | congruent]], and each has [[area]] &lt;math&gt; \sqrt{2006}. &lt;/math&gt; Let &lt;math&gt; K &lt;/math&gt; be the area of rhombus &lt;math&gt; \mathcal{T}&lt;/math&gt;. Given that &lt;math&gt; K &lt;/math&gt; is a [[positive integer]], find the number of possible values for &lt;math&gt; K&lt;/math&gt;.<br /> <br /> [[Image:2006AimeA8.PNG]]<br /> <br /> == Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; denote the common side length of the rhombi.<br /> Let &lt;math&gt;y&lt;/math&gt; denote one of the smaller interior [[angle]]s of rhombus &lt;math&gt; \mathcal{P} &lt;/math&gt;. Then &lt;math&gt;x^2\sin(y)=\sqrt{2006}&lt;/math&gt;. We also see that &lt;math&gt;K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y&lt;/math&gt;. Thus &lt;math&gt;K&lt;/math&gt; can be any positive integer in the [[interval]] &lt;math&gt;(0, 2\sqrt{2006})&lt;/math&gt;. <br /> &lt;math&gt;2\sqrt{2006} = \sqrt{8024}&lt;/math&gt; and &lt;math&gt;89^2 = 7921 &lt; 8024 &lt; 8100 = 90^2&lt;/math&gt;, so &lt;math&gt;K&lt;/math&gt; can be any [[integer]] between 1 and 89, inclusive. Thus the number of positive values for &lt;math&gt;K&lt;/math&gt; is &lt;math&gt;\boxed{089}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where &lt;math&gt;w*h=\sqrt{2006}&lt;/math&gt;. The height of rhombus T would be 2h, and the width would be &lt;math&gt;\sqrt{w^2-h^2}&lt;/math&gt;. Substitute the first equation to get &lt;math&gt;\sqrt{\frac{2006}{h^2}-h^2}&lt;/math&gt;. Then the area of the rhombus would be &lt;math&gt;2h * \sqrt{\frac{2006}{h^2}-h^2}&lt;/math&gt;. Combine like terms to get &lt;math&gt;2 * \sqrt{2006-h^4}&lt;/math&gt;. This expression equals an integer K. &lt;math&gt;2006-h^4&lt;/math&gt; specifically must be in the form &lt;math&gt;n^2/2&lt;/math&gt;. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of &lt;math&gt;n^2&lt;/math&gt; for &lt;math&gt;2006-h^4&lt;/math&gt;. Now, quick testing shows that &lt;math&gt;44^2 &lt; 2006&lt;/math&gt; and &lt;math&gt;45^2&gt;2006&lt;/math&gt;, but we must also test &lt;math&gt;44.5^2&lt;/math&gt;, because the product of two will make it an integer. &lt;math&gt;44.5^2&lt;/math&gt; is also less than &lt;math&gt;2006&lt;/math&gt;, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us &lt;math&gt;44*2+1=&lt;/math&gt; &lt;math&gt;\boxed{89}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_8&diff=135555 2006 AIME I Problems/Problem 8 2020-10-22T03:36:03Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> [[Hexagon]] &lt;math&gt; ABCDEF &lt;/math&gt; is divided into five [[rhombus]]es, &lt;math&gt; \mathcal{P, Q, R, S,} &lt;/math&gt; and &lt;math&gt; \mathcal{T,} &lt;/math&gt; as shown. Rhombuses &lt;math&gt; \mathcal{P, Q, R,} &lt;/math&gt; and &lt;math&gt; \mathcal{S} &lt;/math&gt; are [[congruent (geometry) | congruent]], and each has [[area]] &lt;math&gt; \sqrt{2006}. &lt;/math&gt; Let &lt;math&gt; K &lt;/math&gt; be the area of rhombus &lt;math&gt; \mathcal{T}&lt;/math&gt;. Given that &lt;math&gt; K &lt;/math&gt; is a [[positive integer]], find the number of possible values for &lt;math&gt; K&lt;/math&gt;.<br /> <br /> [[Image:2006AimeA8.PNG]]<br /> <br /> == Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; denote the common side length of the rhombi.<br /> Let &lt;math&gt;y&lt;/math&gt; denote one of the smaller interior [[angle]]s of rhombus &lt;math&gt; \mathcal{P} &lt;/math&gt;. Then &lt;math&gt;x^2\sin(y)=\sqrt{2006}&lt;/math&gt;. We also see that &lt;math&gt;K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y&lt;/math&gt;. Thus &lt;math&gt;K&lt;/math&gt; can be any positive integer in the [[interval]] &lt;math&gt;(0, 2\sqrt{2006})&lt;/math&gt;. <br /> &lt;math&gt;2\sqrt{2006} = \sqrt{8024}&lt;/math&gt; and &lt;math&gt;89^2 = 7921 &lt; 8024 &lt; 8100 = 90^2&lt;/math&gt;, so &lt;math&gt;K&lt;/math&gt; can be any [[integer]] between 1 and 89, inclusive. Thus the number of positive values for &lt;math&gt;K&lt;/math&gt; is &lt;math&gt;\boxed{089}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where &lt;math&gt;w*h=\sqrt{2006}&lt;/math&gt;. The height of rhombus T would be 2h, and the width would be &lt;math&gt;\sqrt{w^2-h^2}&lt;/math&gt;. Substitute the first equation to get &lt;math&gt;\sqrt{\frac{2006}{h^2}-h^2}&lt;/math&gt;. Then the area of the rhombus would be &lt;math&gt;2h * \sqrt{\frac{2006}{h^2}-h^2}&lt;/math&gt;. Combine like terms to get &lt;math&gt;2 * \sqrt{2006-h^4}&lt;/math&gt;. This expression equals an integer K. &lt;math&gt;2006-h^4&lt;/math&gt; specifically must be in the form &lt;math&gt;n^2/2&lt;/math&gt;. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of &lt;math&gt;n^2&lt;/math&gt; for &lt;math&gt;2006-h^4&lt;/math&gt;. Now, quick testing shows that &lt;math&gt;44^2 &lt; 2006&lt;/math&gt; and &lt;math&gt;45^2&gt;2006&lt;/math&gt;, but we must also test &lt;math&gt;44.5^2&lt;/math&gt;, because the product of two will make it an integer. &lt;math&gt;44.5^2&lt;/math&gt; is also less than &lt;math&gt;2006&lt;/math&gt;, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us &lt;math&gt;44*2+1=&lt;/math&gt; &lt;math&gt;\boxed{89}&lt;/math&gt;<br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_8&diff=135554 2006 AIME I Problems/Problem 8 2020-10-22T03:35:10Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> [[Hexagon]] &lt;math&gt; ABCDEF &lt;/math&gt; is divided into five [[rhombus]]es, &lt;math&gt; \mathcal{P, Q, R, S,} &lt;/math&gt; and &lt;math&gt; \mathcal{T,} &lt;/math&gt; as shown. Rhombuses &lt;math&gt; \mathcal{P, Q, R,} &lt;/math&gt; and &lt;math&gt; \mathcal{S} &lt;/math&gt; are [[congruent (geometry) | congruent]], and each has [[area]] &lt;math&gt; \sqrt{2006}. &lt;/math&gt; Let &lt;math&gt; K &lt;/math&gt; be the area of rhombus &lt;math&gt; \mathcal{T}&lt;/math&gt;. Given that &lt;math&gt; K &lt;/math&gt; is a [[positive integer]], find the number of possible values for &lt;math&gt; K&lt;/math&gt;.<br /> <br /> [[Image:2006AimeA8.PNG]]<br /> <br /> == Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; denote the common side length of the rhombi.<br /> Let &lt;math&gt;y&lt;/math&gt; denote one of the smaller interior [[angle]]s of rhombus &lt;math&gt; \mathcal{P} &lt;/math&gt;. Then &lt;math&gt;x^2\sin(y)=\sqrt{2006}&lt;/math&gt;. We also see that &lt;math&gt;K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y&lt;/math&gt;. Thus &lt;math&gt;K&lt;/math&gt; can be any positive integer in the [[interval]] &lt;math&gt;(0, 2\sqrt{2006})&lt;/math&gt;. <br /> &lt;math&gt;2\sqrt{2006} = \sqrt{8024}&lt;/math&gt; and &lt;math&gt;89^2 = 7921 &lt; 8024 &lt; 8100 = 90^2&lt;/math&gt;, so &lt;math&gt;K&lt;/math&gt; can be any [[integer]] between 1 and 89, inclusive. Thus the number of positive values for &lt;math&gt;K&lt;/math&gt; is &lt;math&gt;\boxed{089}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where &lt;math&gt;w*h=\sqrt{2006}&lt;/math&gt;. The height of rhombus T would be 2h, and the width would be &lt;math&gt;\sqrt{w^2-h^2}&lt;/math&gt;. Substitute the first equation to get &lt;math&gt;\sqrt{\frac{2006}{h^2}-h^2}&lt;/math&gt;. Then the area of the rhombus would be &lt;math&gt;2h * \sqrt{\frac{2006}{h^2}-h^2}&lt;/math&gt;. Combine like terms to get &lt;math&gt;2 * \sqrt{2006-h^4}&lt;/math&gt;. This expression equals an integer K. &lt;math&gt;2006-h^4&lt;/math&gt; specifically must be in the form &lt;math&gt;n^2/2&lt;/math&gt;. There is no restriction on h as long as it is a positive real number, so all we have to do is find all the positive possible values of &lt;math&gt;n^2&lt;/math&gt; within &lt;math&gt;2006-h^4&lt;/math&gt;. Now, quick testing shows that &lt;math&gt;44^2 &lt; 2006&lt;/math&gt; and &lt;math&gt;45^2&gt;2006&lt;/math&gt;, but we must also test &lt;math&gt;44.5^2&lt;/math&gt;, because the product of two will make it an integer. &lt;math&gt;44.5^2&lt;/math&gt; is also less than &lt;math&gt;2006&lt;/math&gt;, so we have numbers 1-44, times two because 0.5 can be added to each of time, plus 1, because 0.5 is also a valid value. (notice 0 is not valid because the height must be a positive number) That gives us &lt;math&gt;44*2+1=&lt;/math&gt;&lt;math&gt;\boxed{89}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2006_AIME_I_Problems/Problem_8&diff=135553 2006 AIME I Problems/Problem 8 2020-10-22T03:29:51Z <p>Jackshi2006: /* Solution */</p> <hr /> <div>== Problem ==<br /> [[Hexagon]] &lt;math&gt; ABCDEF &lt;/math&gt; is divided into five [[rhombus]]es, &lt;math&gt; \mathcal{P, Q, R, S,} &lt;/math&gt; and &lt;math&gt; \mathcal{T,} &lt;/math&gt; as shown. Rhombuses &lt;math&gt; \mathcal{P, Q, R,} &lt;/math&gt; and &lt;math&gt; \mathcal{S} &lt;/math&gt; are [[congruent (geometry) | congruent]], and each has [[area]] &lt;math&gt; \sqrt{2006}. &lt;/math&gt; Let &lt;math&gt; K &lt;/math&gt; be the area of rhombus &lt;math&gt; \mathcal{T}&lt;/math&gt;. Given that &lt;math&gt; K &lt;/math&gt; is a [[positive integer]], find the number of possible values for &lt;math&gt; K&lt;/math&gt;.<br /> <br /> [[Image:2006AimeA8.PNG]]<br /> <br /> == Solution 1==<br /> Let &lt;math&gt;x&lt;/math&gt; denote the common side length of the rhombi.<br /> Let &lt;math&gt;y&lt;/math&gt; denote one of the smaller interior [[angle]]s of rhombus &lt;math&gt; \mathcal{P} &lt;/math&gt;. Then &lt;math&gt;x^2\sin(y)=\sqrt{2006}&lt;/math&gt;. We also see that &lt;math&gt;K=x^2\sin(2y) \Longrightarrow K=2x^2\sin y \cdot \cos y \Longrightarrow K = 2\sqrt{2006}\cdot \cos y&lt;/math&gt;. Thus &lt;math&gt;K&lt;/math&gt; can be any positive integer in the [[interval]] &lt;math&gt;(0, 2\sqrt{2006})&lt;/math&gt;. <br /> &lt;math&gt;2\sqrt{2006} = \sqrt{8024}&lt;/math&gt; and &lt;math&gt;89^2 = 7921 &lt; 8024 &lt; 8100 = 90^2&lt;/math&gt;, so &lt;math&gt;K&lt;/math&gt; can be any [[integer]] between 1 and 89, inclusive. Thus the number of positive values for &lt;math&gt;K&lt;/math&gt; is &lt;math&gt;\boxed{089}&lt;/math&gt;.<br /> <br /> <br /> ==Solution 2==<br /> Call the side of each rhombus w. w is the width of the rhombus. Call the height h, where &lt;math&gt;w*h=\sqrt{2006}&lt;/math&gt;. The height of rhombus T would be 2h, and the width would be &lt;math&gt;\sqrt{w^2-h^2}&lt;/math&gt;. Substitute the first equation to get &lt;math&gt;\sqrt{\frac{2006}{h^2}-h^2}&lt;/math&gt;. Then the area of the rhombus would be &lt;math&gt;2h * \sqrt{\frac{2006}{h^2}-h^2}&lt;/math&gt;. Combine like terms to get &lt;math&gt;2 * \sqrt{2006-h^4}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=2006|n=I|num-b=7|num-a=9}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Trigonometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=2004_AIME_I_Problems/Problem_7&diff=134672 2004 AIME I Problems/Problem 7 2020-10-06T03:21:36Z <p>Jackshi2006: /* Solution 5 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt; C &lt;/math&gt; be the [[coefficient]] of &lt;math&gt; x^2 &lt;/math&gt; in the expansion of the product &lt;math&gt; (1 - x)(1 + 2x)(1 - 3x)\cdots(1 + 14x)(1 - 15x). &lt;/math&gt; Find &lt;math&gt; |C|. &lt;/math&gt;<br /> <br /> __TOC__<br /> == Solutions ==<br /> === Solution 1 ===<br /> Let our [[polynomial]] be &lt;math&gt;P(x)&lt;/math&gt;.<br /> <br /> It is clear that the coefficient of &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;P(x)&lt;/math&gt; is &lt;math&gt;-1 + 2 - 3 + \ldots + 14 - 15 = -8&lt;/math&gt;, so &lt;math&gt;P(x) = 1 -8x + Cx^2 + Q(x)&lt;/math&gt;, where &lt;math&gt;Q(x)&lt;/math&gt; is some polynomial [[divisibility | divisible]] by &lt;math&gt;x^3&lt;/math&gt;.<br /> <br /> Then &lt;math&gt;P(-x) = 1 + 8x + Cx^2 + Q(-x)&lt;/math&gt; and so &lt;math&gt;P(x)\cdot P(-x) = 1 + (2C - 64)x^2 + R(x)&lt;/math&gt;, where &lt;math&gt;R(x)&lt;/math&gt; is some polynomial divisible by &lt;math&gt;x^3&lt;/math&gt;.<br /> <br /> However, we also know &lt;math&gt;P(x)\cdot P(-x) = (1 - x)(1 + x)(1 +2x)(1 - 2x) \cdots (1 - 15x)(1 + 15x)&lt;/math&gt; &lt;math&gt;= (1 - x^2)(1 - 4x^2)\cdots(1 - 225x^2)&lt;/math&gt; &lt;math&gt;= 1 - (1 + 4 + \ldots + 225)x^2 + R(x)&lt;/math&gt;.<br /> <br /> Equating coefficients, we have &lt;math&gt;2C - 64 = -(1 + 4 + \ldots + 225) = -1240&lt;/math&gt;, so &lt;math&gt;-2C = 1176&lt;/math&gt; and &lt;math&gt;|C| = \boxed{588}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> Let &lt;math&gt;S&lt;/math&gt; be the [[set]] of integers &lt;math&gt;\{-1,2,-3,\ldots,14,-15\}&lt;/math&gt;. The coefficient of &lt;math&gt;x^2&lt;/math&gt; in the expansion is equal to the sum of the product of each pair of distinct terms, or &lt;math&gt;C = \sum_{1 \le i \neq j}^{15} S_iS_j&lt;/math&gt;. Also, we know that <br /> &lt;cmath&gt;\begin{align*}\left(\sum_{i=1}^{n} S_i\right)^2 &amp;= \left(\sum_{i=1}^{n} S_i^2\right) + 2\left(\sum_{1 \le i \neq j}^{15} S_iS_j\right)\\ (-8)^2 &amp;= \frac{15(15+1)(2\cdot 15+1)}{6} + 2C\end{align*}&lt;/cmath&gt;<br /> where the left-hand sum can be computed from:<br /> &lt;center&gt;&lt;math&gt;\sum_{i=1}^{15} S_i = S_{15} + \left(\sum_{i=1}^{7} S_{2i-1} + S_{2i}\right) = -15 + 7 = -8&lt;/math&gt;&lt;/center&gt;<br /> and the right-hand sum comes from the formula for the sum of the first &lt;math&gt;n&lt;/math&gt; perfect squares. Therefore, &lt;math&gt;|C| = \left|\frac{64-1240}{2}\right| = \boxed{588}&lt;/math&gt;.<br /> <br /> === Solution 3 (Bash)===<br /> <br /> Consider the set &lt;math&gt;[-1, 2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15]&lt;/math&gt;. Denote by &lt;math&gt;S&lt;/math&gt; all size 2 subsets of this set. Replace each element of &lt;math&gt;S&lt;/math&gt; by the product of the elements. Now, the quantity we seek is the sum of each element. Since consecutive elements add to &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;, we can simplify this to &lt;math&gt;|-1\cdot(-7)+2\cdot(-9)-3\cdot(-6)+4\cdot(-10)-5\cdot(-5)+\ldots+12\cdot(-14)-13\cdot(-1)+14\cdot(-15)|=|-588|=\boxed{588}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Let set &lt;math&gt;N&lt;/math&gt; be &lt;math&gt;\{-1, -3, \ldots -15\}&lt;/math&gt; and set &lt;math&gt;P&lt;/math&gt; be &lt;math&gt;\{2, 4, \ldots 14\}&lt;/math&gt;. The sum of the negative &lt;math&gt;x^2&lt;/math&gt; coefficients is the sum of the products of the elements in all two element sets such that one element is from &lt;math&gt;N&lt;/math&gt; and the other is from &lt;math&gt;P&lt;/math&gt;. Each summand is a term in the expansion of<br /> &lt;cmath&gt;(-1 - 3 - \ldots - 15)(2 + 4 + \ldots + 14)&lt;/cmath&gt;<br /> which equals &lt;math&gt;-56 * 64 = -(60^2 - 4^2) = -3584&lt;/math&gt;. The sum of the positive &lt;math&gt;x^2&lt;/math&gt; coefficients is the sum of the products of all two element sets such that the two elements are either both in &lt;math&gt;N&lt;/math&gt; or both in &lt;math&gt;P&lt;/math&gt;. By counting, the sum is &lt;math&gt;2992&lt;/math&gt;, so the sum of all &lt;math&gt;x^2&lt;/math&gt; coefficients is &lt;math&gt;-588&lt;/math&gt;. Thus, the answer is &lt;math&gt;\boxed{588}&lt;/math&gt;.<br /> <br /> <br /> == Solution 5==<br /> <br /> We can find out the coefficient of &lt;math&gt;x^2&lt;/math&gt; by multiplying every pair of two coefficients for &lt;math&gt;x&lt;/math&gt;. This means that we multiply &lt;math&gt;-1&lt;/math&gt; by &lt;math&gt;2,-3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15&lt;/math&gt; and &lt;math&gt;2&lt;/math&gt; by &lt;math&gt;3,4,-5,6,-7,8,-9,10,-11,12,-13,14,-15&lt;/math&gt;. and etc. This sum can be easily simplified and is equal to &lt;math&gt;(-1)(-7)+(-3)(-6)+(-5)(-5)+(-7)(-4)+(-9)(-3)+(-11)(-2)+(-13)(-1)+2(-9)+4(-10)+6(-11)+8(-12)+10(-13)+12(-14)+14(-15)&lt;/math&gt; or &lt;math&gt;588&lt;/math&gt;.<br /> <br /> -David Camacho<br /> <br /> ===Solution 6===<br /> This is just another way of summing the subsets of 2 from &lt;math&gt;[-1, 2, -3, 4, -5, 6, -7, 8, -9, 10, -11, 12, -13, 14, -15]&lt;/math&gt;. Start from the right and multiply -15 to everything on its left. Use the distributive property and add all the 14 integers together to get 7. This gives us &lt;math&gt;-15 * 7&lt;/math&gt;. Doing this for 14 gives us &lt;math&gt;14 * -7&lt;/math&gt;, and for -13 we get &lt;math&gt;-13 * 6&lt;/math&gt;. This pattern repeats where every two integers will multiple 7, 6,... to 0. Combining and simplifying the pattern give us this: &lt;math&gt;-(29 * 7 + 25 * 6 + 21 * 5 + 17 * 4 + 13 * 3 + 9 * 2 + 5*1)&lt;/math&gt;. The expression gives us -588, or &lt;math&gt;C = \boxed{588}&lt;/math&gt;. This is a good solution because it guarantees we never add a product twice, and the pattern is simple to add by hand.<br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=2004|n=I|num-b=6|num-a=8}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_9&diff=133060 1989 AIME Problems/Problem 9 2020-09-03T23:41:51Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that &lt;math&gt;133^5+110^5+84^5+27^5=n^{5}&lt;/math&gt;. Find the value of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> Note that &lt;math&gt;n&lt;/math&gt; is even, since the &lt;math&gt;LHS&lt;/math&gt; consists of two odd and two even numbers. By [[Fermat's Little Theorem]], we know &lt;math&gt;{n^{5}}&lt;/math&gt; is congruent to &lt;math&gt;n&lt;/math&gt; [[modulo]] 5. Hence,<br /> &lt;center&gt;&lt;math&gt;3 + 0 + 4 + 2 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;4 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Continuing, we examine the equation modulo 3,<br /> &lt;center&gt;&lt;math&gt;1 - 1 + 0 + 0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Thus, &lt;math&gt;n&lt;/math&gt; is divisible by three and leaves a remainder of four when divided by 5. It's obvious that &lt;math&gt;n&gt;133&lt;/math&gt;, so the only possibilities are &lt;math&gt;n = 144&lt;/math&gt; or &lt;math&gt;n \geq 174&lt;/math&gt;. It quickly becomes apparent that 174 is much too large, so &lt;math&gt;n&lt;/math&gt; must be &lt;math&gt;\boxed{144}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> We can cheat a little bit and approximate, since we are dealing with such large numbers. As above, &lt;math&gt;n^5\equiv n\pmod{5}&lt;/math&gt;, and it is easy to see that &lt;math&gt;n^5\equiv n\pmod 2&lt;/math&gt;. Therefore, &lt;math&gt;133^5+110^5+84^5+27^5\equiv 3+0+4+7\equiv 4\pmod{10}&lt;/math&gt;, so the last digit of &lt;math&gt;n&lt;/math&gt; is 4.<br /> <br /> We notice that &lt;math&gt;133,110,84,&lt;/math&gt; and &lt;math&gt;27&lt;/math&gt; are all very close or equal to multiples of 27. We can rewrite &lt;math&gt;n^5&lt;/math&gt; as approximately equal to &lt;math&gt;27^5(5^5+4^5+3^5+1^5) = 27^5(4393)&lt;/math&gt;. This means &lt;math&gt;\frac{n}{27}&lt;/math&gt; must be close to &lt;math&gt;4393&lt;/math&gt;.<br /> <br /> 134 will obviously be too small, so we try 144. &lt;math&gt;\left(\frac{144}{27}\right)^5=\left(\frac{16}{3}\right)^5&lt;/math&gt;. Bashing through the division, we find that &lt;math&gt;\frac{1048576}{243}\approx 4315&lt;/math&gt;, which is very close to &lt;math&gt;4393&lt;/math&gt;. It is clear that 154 will not give any closer of an answer, given the rate that fifth powers grow, so we can safely assume that &lt;math&gt;\boxed{144}&lt;/math&gt; is the answer.<br /> <br /> <br /> ==Solution 3==<br /> In this solution we take advantage of the large numbers and utilize parity properties to give us a very good guess at the answer. The units digit of each of the four numbers is 3, 0, 4, and 7, respectively. This means the units digit of &lt;math&gt;n^5&lt;/math&gt; is 4. This tells us &lt;math&gt;n&lt;/math&gt; is even. Since we are dealing with enormous numbers, &lt;math&gt;n&lt;/math&gt; should not be that far from 133. &lt;math&gt;n&lt;/math&gt;'s unit digit is 0, 2, 4, 6, or 8. When to the power of 5 they each give 0, 2, 4, 6, and 8 as unit digits. This further clues us that &lt;math&gt;n&lt;/math&gt; ends in 4. <br /> <br /> &lt;math&gt;n&lt;/math&gt; is obviously larger than 133, so we start with 134. Now we need a way of distinguishing between numbers with unit digit 4. This can be done by simply solving up to the hundreds digit of &lt;math&gt;133^5&lt;/math&gt;, &lt;math&gt;110^5&lt;/math&gt;, &lt;math&gt;84^5&lt;/math&gt;, and &lt;math&gt;27^5&lt;/math&gt;, which isn't that difficult. For 133, all that has to be done is square it and take the last three digits, 689, and raise them to the power of 2 again, 721, then multiply this by 133. This gives us 893. Doing this for each tells us &lt;math&gt;n^5&lt;/math&gt; ends in 224. Testing 134 the same way we did with 133 gives us 424, 144 gives us 224, 154 gives us 024, 164 gives us 824, 624, 424, and finally 194 also gives 224.<br /> <br /> By simply using our judgement it's hard to imagine the sum of those powers of 5 to be &lt;math&gt;194^5&lt;/math&gt;. The answer is &lt;math&gt;\boxed{144}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_9&diff=133059 1989 AIME Problems/Problem 9 2020-09-03T23:41:32Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> One of Euler's conjectures was disproved in the 1960s by three American mathematicians when they showed there was a positive integer such that &lt;math&gt;133^5+110^5+84^5+27^5=n^{5}&lt;/math&gt;. Find the value of &lt;math&gt;n&lt;/math&gt;.<br /> <br /> == Solution 1 ==<br /> Note that &lt;math&gt;n&lt;/math&gt; is even, since the &lt;math&gt;LHS&lt;/math&gt; consists of two odd and two even numbers. By [[Fermat's Little Theorem]], we know &lt;math&gt;{n^{5}}&lt;/math&gt; is congruent to &lt;math&gt;n&lt;/math&gt; [[modulo]] 5. Hence,<br /> &lt;center&gt;&lt;math&gt;3 + 0 + 4 + 2 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;4 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Continuing, we examine the equation modulo 3,<br /> &lt;center&gt;&lt;math&gt;1 - 1 + 0 + 0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;<br /> &lt;center&gt;&lt;math&gt;0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;<br /> <br /> Thus, &lt;math&gt;n&lt;/math&gt; is divisible by three and leaves a remainder of four when divided by 5. It's obvious that &lt;math&gt;n&gt;133&lt;/math&gt;, so the only possibilities are &lt;math&gt;n = 144&lt;/math&gt; or &lt;math&gt;n \geq 174&lt;/math&gt;. It quickly becomes apparent that 174 is much too large, so &lt;math&gt;n&lt;/math&gt; must be &lt;math&gt;\boxed{144}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> We can cheat a little bit and approximate, since we are dealing with such large numbers. As above, &lt;math&gt;n^5\equiv n\pmod{5}&lt;/math&gt;, and it is easy to see that &lt;math&gt;n^5\equiv n\pmod 2&lt;/math&gt;. Therefore, &lt;math&gt;133^5+110^5+84^5+27^5\equiv 3+0+4+7\equiv 4\pmod{10}&lt;/math&gt;, so the last digit of &lt;math&gt;n&lt;/math&gt; is 4.<br /> <br /> We notice that &lt;math&gt;133,110,84,&lt;/math&gt; and &lt;math&gt;27&lt;/math&gt; are all very close or equal to multiples of 27. We can rewrite &lt;math&gt;n^5&lt;/math&gt; as approximately equal to &lt;math&gt;27^5(5^5+4^5+3^5+1^5) = 27^5(4393)&lt;/math&gt;. This means &lt;math&gt;\frac{n}{27}&lt;/math&gt; must be close to &lt;math&gt;4393&lt;/math&gt;.<br /> <br /> 134 will obviously be too small, so we try 144. &lt;math&gt;\left(\frac{144}{27}\right)^5=\left(\frac{16}{3}\right)^5&lt;/math&gt;. Bashing through the division, we find that &lt;math&gt;\frac{1048576}{243}\approx 4315&lt;/math&gt;, which is very close to &lt;math&gt;4393&lt;/math&gt;. It is clear that 154 will not give any closer of an answer, given the rate that fifth powers grow, so we can safely assume that &lt;math&gt;\boxed{144}&lt;/math&gt; is the answer.<br /> <br /> <br /> ==Solution 3==<br /> In this solution we take advantage of the large numbers and utilize parity properties to give us a very good guess at the answer. The units digit of each of the four numbers is 3, 0, 4, and 7, respectively. This means the units digit of &lt;math&gt;n^5&lt;/math&gt; is 4. This tells us &lt;math&gt;n&lt;/math&gt; is even. Since we are dealing with enormous numbers, &lt;math&gt;n&lt;/math&gt; should not be that far from 133. &lt;math&gt;n&lt;/math&gt;'s unit digit is 0, 2, 4, 6, or 8. When to the power of 5 they each give 0, 2, 4, 6, and 8 as unit digits. This further clues us that &lt;math&gt;n&lt;/math&gt; ends in 4. <br /> &lt;math&gt;n&lt;/math&gt; is obviously larger than 133, so we start with 134. Now we need a way of distinguishing between numbers with unit digit 4. This can be done by simply solving up to the hundreds digit of &lt;math&gt;133^5&lt;/math&gt;, &lt;math&gt;110^5&lt;/math&gt;, &lt;math&gt;84^5&lt;/math&gt;, and &lt;math&gt;27^5&lt;/math&gt;, which isn't that difficult. For 133, all that has to be done is square it and take the last three digits, 689, and raise them to the power of 2 again, 721, then multiply this by 133. This gives us 893. Doing this for each tells us &lt;math&gt;n^5&lt;/math&gt; ends in 224. Testing 134 the same way we did with 133 gives us 424, 144 gives us 224, 154 gives us 024, 164 gives us 824, 624, 424, and finally 194 also gives 224.<br /> By simply using our judgement it's hard to imagine the sum of those powers of 5 to be &lt;math&gt;194^5&lt;/math&gt;. The answer is &lt;math&gt;\boxed{144}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=8|num-a=10}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_4&diff=132741 1989 AIME Problems/Problem 4 2020-08-30T17:59:18Z <p>Jackshi2006: /* Solution */</p> <hr /> <div>== Problem ==<br /> If &lt;math&gt;a&lt;b&lt;c&lt;d&lt;e&lt;/math&gt; are [[consecutive]] [[positive]] [[integer]]s such that &lt;math&gt;b+c+d&lt;/math&gt; is a [[perfect square]] and &lt;math&gt;a+b+c+d+e&lt;/math&gt; is a [[perfect cube]], what is the smallest possible value of &lt;math&gt;c&lt;/math&gt;?<br /> <br /> == Solution ==<br /> Since the middle term of an [[arithmetic progression]] with an odd number of terms is the average of the series, we know &lt;math&gt;b + c + d = 3c&lt;/math&gt; and &lt;math&gt;a + b + c + d + e = 5c&lt;/math&gt;. Thus, &lt;math&gt;c&lt;/math&gt; must be in the form of &lt;math&gt;3 \cdot x^2&lt;/math&gt; based upon the first part and in the form of &lt;math&gt;5^2 \cdot y^3&lt;/math&gt; based upon the second part, with &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; denoting an [[integer]]s. &lt;math&gt;c&lt;/math&gt; is minimized if it’s [[prime factorization]] contains only &lt;math&gt;3,5&lt;/math&gt;, and since there is a cubed term in &lt;math&gt;5^2 \cdot y^3&lt;/math&gt;, &lt;math&gt;3^3&lt;/math&gt; must be a factor of &lt;math&gt;c&lt;/math&gt;. &lt;math&gt;3^35^2 = \boxed{675}&lt;/math&gt;, which works as the solution.<br /> <br /> <br /> ==Solution 2==<br /> Let &lt;math&gt;b&lt;/math&gt;, &lt;math&gt;c&lt;/math&gt;, &lt;math&gt;d&lt;/math&gt;, and &lt;math&gt;e&lt;/math&gt; equal &lt;math&gt;a+1&lt;/math&gt;, &lt;math&gt;a+2&lt;/math&gt;, &lt;math&gt;a+3&lt;/math&gt;, and &lt;math&gt;a+4&lt;/math&gt;, respectively. Call the square and cube &lt;math&gt;k^2&lt;/math&gt; and &lt;math&gt;m^3&lt;/math&gt;, where both k and m are integers. Then:<br /> <br /> &lt;math&gt;5a + 10 = m^3&lt;/math&gt;<br /> <br /> Now we know &lt;math&gt;m^3&lt;/math&gt; is a multiple of 125 and &lt;math&gt;m&lt;/math&gt; is a multiple of 5. The lower &lt;math&gt;m&lt;/math&gt; is, the lower the value of &lt;math&gt;c&lt;/math&gt; will be. Start from 5 and add 5 each time.<br /> <br /> &lt;math&gt;m = 5&lt;/math&gt; gives no solution for k<br /> <br /> &lt;math&gt;m = 10&lt;/math&gt; gives no solution for k<br /> <br /> &lt;math&gt;m = 15&lt;/math&gt; gives a solution for k.<br /> <br /> <br /> &lt;math&gt;10 + 5a = 15^3&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;2 + a = 675&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;c = \boxed{675}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_3&diff=132740 1989 AIME Problems/Problem 3 2020-08-30T17:52:43Z <p>Jackshi2006: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;n&lt;/math&gt; is a [[positive integer]] and &lt;math&gt;d&lt;/math&gt; is a single [[digit]] in [[base 10]]. Find &lt;math&gt;n&lt;/math&gt; if<br /> &lt;center&gt;&lt;math&gt;\frac{n}{810}=0.d25d25d25\ldots&lt;/math&gt;&lt;/center&gt;<br /> <br /> == Solution ==<br /> Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], &lt;math&gt;0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}&lt;/math&gt;. Thus &lt;math&gt;\frac{n}{810} = \frac{100d + 25}{999}&lt;/math&gt; so &lt;math&gt;n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}&lt;/math&gt;. Since 750 and 37 are [[relatively prime]], &lt;math&gt;4d + 1&lt;/math&gt; must be [[divisible]] by 37, and the only digit for which this is possible is &lt;math&gt;d = 9&lt;/math&gt;. Thus &lt;math&gt;4d + 1 = 37&lt;/math&gt; and &lt;math&gt;n = \boxed{750}&lt;/math&gt;.<br /> <br /> <br /> (Note: Any repeating sequence of &lt;math&gt;n&lt;/math&gt; digits that looks like &lt;math&gt;0.a_1a_2a_3...a_{n-1}a_na_1a_2...&lt;/math&gt; can be written as &lt;math&gt;\frac{a_1a_2...a_n}{10^n-1}&lt;/math&gt;, where &lt;math&gt;a_1a_2...a_n&lt;/math&gt; represents an &lt;math&gt;n&lt;/math&gt; digit number.)<br /> <br /> == Solution 2 ==<br /> To get rid of repeating decimals, we multiply the equation by 1000. We get &lt;math&gt;\frac{1000n}{810} = d25.d25d25...&lt;/math&gt; We subtract the original equation from the second to get &lt;math&gt;\frac{999n}{810}=d25&lt;/math&gt; We simplify to &lt;math&gt;\frac{37n}{30} = d25&lt;/math&gt; Since &lt;math&gt;\frac{37n}{30}&lt;/math&gt; is an integer, &lt;math&gt;n=(30)(5)(2k+1)&lt;/math&gt; because &lt;math&gt;37&lt;/math&gt; is relatively prime to &lt;math&gt;30&lt;/math&gt;, and d25 is divisible by &lt;math&gt;5&lt;/math&gt; but not &lt;math&gt;10&lt;/math&gt;. The only odd number that yields a single digit &lt;math&gt;d&lt;/math&gt; and 25 at the end of the three digit number is &lt;math&gt;k=2&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{750}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Similar to Solution 2, we start off by writing that &lt;math&gt; \frac{1000n}{810} = d25.d25d25 \dots&lt;/math&gt; .Then we subtract this from the original equation to get:<br /> <br /> &lt;cmath&gt; \frac{999n}{810} =d25 \Longrightarrow \frac{37n}{30} = d25 \Longrightarrow 37n = d25 \cdot 30&lt;/cmath&gt;<br /> <br /> Since n is an integer, we have that &lt;math&gt;37 \mid d25 \cdot 30&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;37&lt;/math&gt; is prime, we can apply Euclid's Lemma (which states that if &lt;math&gt;p&lt;/math&gt; is a prime and if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and if &lt;math&gt;p \mid ab&lt;/math&gt;, then &lt;math&gt;p \mid a&lt;/math&gt; or &lt;math&gt;p \mid b&lt;/math&gt;) to realize that &lt;math&gt;37 \mid d25 &lt;/math&gt;, since &lt;math&gt;37 \nmid 30&lt;/math&gt;. Then we can expand &lt;math&gt;d25&lt;/math&gt; as &lt;math&gt;25 \cdot (4d +1)&lt;/math&gt;. Since &lt;math&gt;37 \nmid 25 &lt;/math&gt;, by Euclid, we can arrive at &lt;math&gt;37 \mid 4d+1 \Longrightarrow d=9&lt;/math&gt;. From this we know that &lt;math&gt;n= 25 \cdot 30 = \boxed{750}&lt;/math&gt;. (This is true because &lt;math&gt;37n = 925 \cdot 30 \rightarrow n= 25 \cdot 30 = 750&lt;/math&gt;)<br /> <br /> ~qwertysri987<br /> <br /> <br /> ==Solution 4==<br /> Write out these equations:<br /> <br /> <br /> &lt;math&gt;\frac{n}{180} = \frac{d25}{999}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\frac{n}{30} = \frac{d25}{37}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;37n = 30(d25)&lt;/math&gt;<br /> <br /> <br /> Thus &lt;math&gt;n&lt;/math&gt; divides 25 and 30. The only solution for this under 1000 is &lt;math&gt;\boxed{750}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_3&diff=132739 1989 AIME Problems/Problem 3 2020-08-30T17:52:06Z <p>Jackshi2006: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;n&lt;/math&gt; is a [[positive integer]] and &lt;math&gt;d&lt;/math&gt; is a single [[digit]] in [[base 10]]. Find &lt;math&gt;n&lt;/math&gt; if<br /> &lt;center&gt;&lt;math&gt;\frac{n}{810}=0.d25d25d25\ldots&lt;/math&gt;&lt;/center&gt;<br /> <br /> == Solution ==<br /> Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], &lt;math&gt;0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}&lt;/math&gt;. Thus &lt;math&gt;\frac{n}{810} = \frac{100d + 25}{999}&lt;/math&gt; so &lt;math&gt;n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}&lt;/math&gt;. Since 750 and 37 are [[relatively prime]], &lt;math&gt;4d + 1&lt;/math&gt; must be [[divisible]] by 37, and the only digit for which this is possible is &lt;math&gt;d = 9&lt;/math&gt;. Thus &lt;math&gt;4d + 1 = 37&lt;/math&gt; and &lt;math&gt;n = \boxed{750}&lt;/math&gt;.<br /> <br /> <br /> (Note: Any repeating sequence of &lt;math&gt;n&lt;/math&gt; digits that looks like &lt;math&gt;0.a_1a_2a_3...a_{n-1}a_na_1a_2...&lt;/math&gt; can be written as &lt;math&gt;\frac{a_1a_2...a_n}{10^n-1}&lt;/math&gt;, where &lt;math&gt;a_1a_2...a_n&lt;/math&gt; represents an &lt;math&gt;n&lt;/math&gt; digit number.)<br /> <br /> == Solution 2 ==<br /> To get rid of repeating decimals, we multiply the equation by 1000. We get &lt;math&gt;\frac{1000n}{810} = d25.d25d25...&lt;/math&gt; We subtract the original equation from the second to get &lt;math&gt;\frac{999n}{810}=d25&lt;/math&gt; We simplify to &lt;math&gt;\frac{37n}{30} = d25&lt;/math&gt; Since &lt;math&gt;\frac{37n}{30}&lt;/math&gt; is an integer, &lt;math&gt;n=(30)(5)(2k+1)&lt;/math&gt; because &lt;math&gt;37&lt;/math&gt; is relatively prime to &lt;math&gt;30&lt;/math&gt;, and d25 is divisible by &lt;math&gt;5&lt;/math&gt; but not &lt;math&gt;10&lt;/math&gt;. The only odd number that yields a single digit &lt;math&gt;d&lt;/math&gt; and 25 at the end of the three digit number is &lt;math&gt;k=2&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{750}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Similar to Solution 2, we start off by writing that &lt;math&gt; \frac{1000n}{810} = d25.d25d25 \dots&lt;/math&gt; .Then we subtract this from the original equation to get:<br /> <br /> &lt;cmath&gt; \frac{999n}{810} =d25 \Longrightarrow \frac{37n}{30} = d25 \Longrightarrow 37n = d25 \cdot 30&lt;/cmath&gt;<br /> <br /> Since n is an integer, we have that &lt;math&gt;37 \mid d25 \cdot 30&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;37&lt;/math&gt; is prime, we can apply Euclid's Lemma (which states that if &lt;math&gt;p&lt;/math&gt; is a prime and if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and if &lt;math&gt;p \mid ab&lt;/math&gt;, then &lt;math&gt;p \mid a&lt;/math&gt; or &lt;math&gt;p \mid b&lt;/math&gt;) to realize that &lt;math&gt;37 \mid d25 &lt;/math&gt;, since &lt;math&gt;37 \nmid 30&lt;/math&gt;. Then we can expand &lt;math&gt;d25&lt;/math&gt; as &lt;math&gt;25 \cdot (4d +1)&lt;/math&gt;. Since &lt;math&gt;37 \nmid 25 &lt;/math&gt;, by Euclid, we can arrive at &lt;math&gt;37 \mid 4d+1 \Longrightarrow d=9&lt;/math&gt;. From this we know that &lt;math&gt;n= 25 \cdot 30 = \boxed{750}&lt;/math&gt;. (This is true because &lt;math&gt;37n = 925 \cdot 30 \rightarrow n= 25 \cdot 30 = 750&lt;/math&gt;)<br /> <br /> ~qwertysri987<br /> <br /> <br /> ==Solution 4==<br /> Write out the equations:<br /> <br /> <br /> &lt;math&gt;\frac{n}{180} = \frac{d25}{999}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\frac{n}{30} = \frac{d25}{37}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;37n = 30(d25)&lt;/math&gt;<br /> <br /> <br /> Thus &lt;math&gt;n&lt;/math&gt; divides 25 and 30. The only solution for this under 1000 is &lt;math&gt;\boxed{750}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_3&diff=132738 1989 AIME Problems/Problem 3 2020-08-30T17:51:34Z <p>Jackshi2006: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;n&lt;/math&gt; is a [[positive integer]] and &lt;math&gt;d&lt;/math&gt; is a single [[digit]] in [[base 10]]. Find &lt;math&gt;n&lt;/math&gt; if<br /> &lt;center&gt;&lt;math&gt;\frac{n}{810}=0.d25d25d25\ldots&lt;/math&gt;&lt;/center&gt;<br /> <br /> == Solution ==<br /> Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], &lt;math&gt;0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}&lt;/math&gt;. Thus &lt;math&gt;\frac{n}{810} = \frac{100d + 25}{999}&lt;/math&gt; so &lt;math&gt;n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}&lt;/math&gt;. Since 750 and 37 are [[relatively prime]], &lt;math&gt;4d + 1&lt;/math&gt; must be [[divisible]] by 37, and the only digit for which this is possible is &lt;math&gt;d = 9&lt;/math&gt;. Thus &lt;math&gt;4d + 1 = 37&lt;/math&gt; and &lt;math&gt;n = \boxed{750}&lt;/math&gt;.<br /> <br /> <br /> (Note: Any repeating sequence of &lt;math&gt;n&lt;/math&gt; digits that looks like &lt;math&gt;0.a_1a_2a_3...a_{n-1}a_na_1a_2...&lt;/math&gt; can be written as &lt;math&gt;\frac{a_1a_2...a_n}{10^n-1}&lt;/math&gt;, where &lt;math&gt;a_1a_2...a_n&lt;/math&gt; represents an &lt;math&gt;n&lt;/math&gt; digit number.)<br /> <br /> == Solution 2 ==<br /> To get rid of repeating decimals, we multiply the equation by 1000. We get &lt;math&gt;\frac{1000n}{810} = d25.d25d25...&lt;/math&gt; We subtract the original equation from the second to get &lt;math&gt;\frac{999n}{810}=d25&lt;/math&gt; We simplify to &lt;math&gt;\frac{37n}{30} = d25&lt;/math&gt; Since &lt;math&gt;\frac{37n}{30}&lt;/math&gt; is an integer, &lt;math&gt;n=(30)(5)(2k+1)&lt;/math&gt; because &lt;math&gt;37&lt;/math&gt; is relatively prime to &lt;math&gt;30&lt;/math&gt;, and d25 is divisible by &lt;math&gt;5&lt;/math&gt; but not &lt;math&gt;10&lt;/math&gt;. The only odd number that yields a single digit &lt;math&gt;d&lt;/math&gt; and 25 at the end of the three digit number is &lt;math&gt;k=2&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{750}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Similar to Solution 2, we start off by writing that &lt;math&gt; \frac{1000n}{810} = d25.d25d25 \dots&lt;/math&gt; .Then we subtract this from the original equation to get:<br /> <br /> &lt;cmath&gt; \frac{999n}{810} =d25 \Longrightarrow \frac{37n}{30} = d25 \Longrightarrow 37n = d25 \cdot 30&lt;/cmath&gt;<br /> <br /> Since n is an integer, we have that &lt;math&gt;37 \mid d25 \cdot 30&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;37&lt;/math&gt; is prime, we can apply Euclid's Lemma (which states that if &lt;math&gt;p&lt;/math&gt; is a prime and if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and if &lt;math&gt;p \mid ab&lt;/math&gt;, then &lt;math&gt;p \mid a&lt;/math&gt; or &lt;math&gt;p \mid b&lt;/math&gt;) to realize that &lt;math&gt;37 \mid d25 &lt;/math&gt;, since &lt;math&gt;37 \nmid 30&lt;/math&gt;. Then we can expand &lt;math&gt;d25&lt;/math&gt; as &lt;math&gt;25 \cdot (4d +1)&lt;/math&gt;. Since &lt;math&gt;37 \nmid 25 &lt;/math&gt;, by Euclid, we can arrive at &lt;math&gt;37 \mid 4d+1 \Longrightarrow d=9&lt;/math&gt;. From this we know that &lt;math&gt;n= 25 \cdot 30 = \boxed{750}&lt;/math&gt;. (This is true because &lt;math&gt;37n = 925 \cdot 30 \rightarrow n= 25 \cdot 30 = 750&lt;/math&gt;)<br /> <br /> ~qwertysri987<br /> <br /> <br /> ==Solution 4==<br /> Write out the equations:<br /> <br /> <br /> &lt;math&gt;\frac{n}{180} = \frac{d25}{999}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\frac{n}{30} = \frac{d25}{37}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;37n = 30(d25)&lt;/math&gt;<br /> <br /> <br /> Thus &lt;math&gt;n&lt;/math&gt; divides 25 and 30. The LCM is &lt;math&gt;\boxed{750}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_3&diff=132737 1989 AIME Problems/Problem 3 2020-08-30T17:51:22Z <p>Jackshi2006: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;n&lt;/math&gt; is a [[positive integer]] and &lt;math&gt;d&lt;/math&gt; is a single [[digit]] in [[base 10]]. Find &lt;math&gt;n&lt;/math&gt; if<br /> &lt;center&gt;&lt;math&gt;\frac{n}{810}=0.d25d25d25\ldots&lt;/math&gt;&lt;/center&gt;<br /> <br /> == Solution ==<br /> Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], &lt;math&gt;0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}&lt;/math&gt;. Thus &lt;math&gt;\frac{n}{810} = \frac{100d + 25}{999}&lt;/math&gt; so &lt;math&gt;n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}&lt;/math&gt;. Since 750 and 37 are [[relatively prime]], &lt;math&gt;4d + 1&lt;/math&gt; must be [[divisible]] by 37, and the only digit for which this is possible is &lt;math&gt;d = 9&lt;/math&gt;. Thus &lt;math&gt;4d + 1 = 37&lt;/math&gt; and &lt;math&gt;n = \boxed{750}&lt;/math&gt;.<br /> <br /> <br /> (Note: Any repeating sequence of &lt;math&gt;n&lt;/math&gt; digits that looks like &lt;math&gt;0.a_1a_2a_3...a_{n-1}a_na_1a_2...&lt;/math&gt; can be written as &lt;math&gt;\frac{a_1a_2...a_n}{10^n-1}&lt;/math&gt;, where &lt;math&gt;a_1a_2...a_n&lt;/math&gt; represents an &lt;math&gt;n&lt;/math&gt; digit number.)<br /> <br /> == Solution 2 ==<br /> To get rid of repeating decimals, we multiply the equation by 1000. We get &lt;math&gt;\frac{1000n}{810} = d25.d25d25...&lt;/math&gt; We subtract the original equation from the second to get &lt;math&gt;\frac{999n}{810}=d25&lt;/math&gt; We simplify to &lt;math&gt;\frac{37n}{30} = d25&lt;/math&gt; Since &lt;math&gt;\frac{37n}{30}&lt;/math&gt; is an integer, &lt;math&gt;n=(30)(5)(2k+1)&lt;/math&gt; because &lt;math&gt;37&lt;/math&gt; is relatively prime to &lt;math&gt;30&lt;/math&gt;, and d25 is divisible by &lt;math&gt;5&lt;/math&gt; but not &lt;math&gt;10&lt;/math&gt;. The only odd number that yields a single digit &lt;math&gt;d&lt;/math&gt; and 25 at the end of the three digit number is &lt;math&gt;k=2&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{750}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Similar to Solution 2, we start off by writing that &lt;math&gt; \frac{1000n}{810} = d25.d25d25 \dots&lt;/math&gt; .Then we subtract this from the original equation to get:<br /> <br /> &lt;cmath&gt; \frac{999n}{810} =d25 \Longrightarrow \frac{37n}{30} = d25 \Longrightarrow 37n = d25 \cdot 30&lt;/cmath&gt;<br /> <br /> Since n is an integer, we have that &lt;math&gt;37 \mid d25 \cdot 30&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;37&lt;/math&gt; is prime, we can apply Euclid's Lemma (which states that if &lt;math&gt;p&lt;/math&gt; is a prime and if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and if &lt;math&gt;p \mid ab&lt;/math&gt;, then &lt;math&gt;p \mid a&lt;/math&gt; or &lt;math&gt;p \mid b&lt;/math&gt;) to realize that &lt;math&gt;37 \mid d25 &lt;/math&gt;, since &lt;math&gt;37 \nmid 30&lt;/math&gt;. Then we can expand &lt;math&gt;d25&lt;/math&gt; as &lt;math&gt;25 \cdot (4d +1)&lt;/math&gt;. Since &lt;math&gt;37 \nmid 25 &lt;/math&gt;, by Euclid, we can arrive at &lt;math&gt;37 \mid 4d+1 \Longrightarrow d=9&lt;/math&gt;. From this we know that &lt;math&gt;n= 25 \cdot 30 = \boxed{750}&lt;/math&gt;. (This is true because &lt;math&gt;37n = 925 \cdot 30 \rightarrow n= 25 \cdot 30 = 750&lt;/math&gt;)<br /> <br /> ~qwertysri987<br /> <br /> <br /> ==Solution 4==<br /> Write out the equations:<br /> <br /> <br /> &lt;math&gt;\frac{n}{180} = \frac{d25}{999}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\frac{n}{30} = \frac{d25}{37}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;37n = 30(d25)&lt;/math&gt;<br /> <br /> <br /> Thus &lt;math&gt;n&lt;/math&gt; divides 25 and 30. The LCM is $\boxed{750}4.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_3&diff=132736 1989 AIME Problems/Problem 3 2020-08-30T17:51:08Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;n&lt;/math&gt; is a [[positive integer]] and &lt;math&gt;d&lt;/math&gt; is a single [[digit]] in [[base 10]]. Find &lt;math&gt;n&lt;/math&gt; if<br /> &lt;center&gt;&lt;math&gt;\frac{n}{810}=0.d25d25d25\ldots&lt;/math&gt;&lt;/center&gt;<br /> <br /> == Solution ==<br /> Repeating decimals represent [[rational number]]s. To figure out which rational number, we sum an [[infinite]] [[geometric series]], &lt;math&gt;0.d25d25d25\ldots = \sum_{i = 1}^\infty \frac{d25}{1000^n} = \frac{100d + 25}{999}&lt;/math&gt;. Thus &lt;math&gt;\frac{n}{810} = \frac{100d + 25}{999}&lt;/math&gt; so &lt;math&gt;n = 30\frac{100d + 25}{37} =750\frac{4d + 1}{37}&lt;/math&gt;. Since 750 and 37 are [[relatively prime]], &lt;math&gt;4d + 1&lt;/math&gt; must be [[divisible]] by 37, and the only digit for which this is possible is &lt;math&gt;d = 9&lt;/math&gt;. Thus &lt;math&gt;4d + 1 = 37&lt;/math&gt; and &lt;math&gt;n = \boxed{750}&lt;/math&gt;.<br /> <br /> <br /> (Note: Any repeating sequence of &lt;math&gt;n&lt;/math&gt; digits that looks like &lt;math&gt;0.a_1a_2a_3...a_{n-1}a_na_1a_2...&lt;/math&gt; can be written as &lt;math&gt;\frac{a_1a_2...a_n}{10^n-1}&lt;/math&gt;, where &lt;math&gt;a_1a_2...a_n&lt;/math&gt; represents an &lt;math&gt;n&lt;/math&gt; digit number.)<br /> <br /> == Solution 2 ==<br /> To get rid of repeating decimals, we multiply the equation by 1000. We get &lt;math&gt;\frac{1000n}{810} = d25.d25d25...&lt;/math&gt; We subtract the original equation from the second to get &lt;math&gt;\frac{999n}{810}=d25&lt;/math&gt; We simplify to &lt;math&gt;\frac{37n}{30} = d25&lt;/math&gt; Since &lt;math&gt;\frac{37n}{30}&lt;/math&gt; is an integer, &lt;math&gt;n=(30)(5)(2k+1)&lt;/math&gt; because &lt;math&gt;37&lt;/math&gt; is relatively prime to &lt;math&gt;30&lt;/math&gt;, and d25 is divisible by &lt;math&gt;5&lt;/math&gt; but not &lt;math&gt;10&lt;/math&gt;. The only odd number that yields a single digit &lt;math&gt;d&lt;/math&gt; and 25 at the end of the three digit number is &lt;math&gt;k=2&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{750}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> Similar to Solution 2, we start off by writing that &lt;math&gt; \frac{1000n}{810} = d25.d25d25 \dots&lt;/math&gt; .Then we subtract this from the original equation to get:<br /> <br /> &lt;cmath&gt; \frac{999n}{810} =d25 \Longrightarrow \frac{37n}{30} = d25 \Longrightarrow 37n = d25 \cdot 30&lt;/cmath&gt;<br /> <br /> Since n is an integer, we have that &lt;math&gt;37 \mid d25 \cdot 30&lt;/math&gt;.<br /> <br /> Since &lt;math&gt;37&lt;/math&gt; is prime, we can apply Euclid's Lemma (which states that if &lt;math&gt;p&lt;/math&gt; is a prime and if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers and if &lt;math&gt;p \mid ab&lt;/math&gt;, then &lt;math&gt;p \mid a&lt;/math&gt; or &lt;math&gt;p \mid b&lt;/math&gt;) to realize that &lt;math&gt;37 \mid d25 &lt;/math&gt;, since &lt;math&gt;37 \nmid 30&lt;/math&gt;. Then we can expand &lt;math&gt;d25&lt;/math&gt; as &lt;math&gt;25 \cdot (4d +1)&lt;/math&gt;. Since &lt;math&gt;37 \nmid 25 &lt;/math&gt;, by Euclid, we can arrive at &lt;math&gt;37 \mid 4d+1 \Longrightarrow d=9&lt;/math&gt;. From this we know that &lt;math&gt;n= 25 \cdot 30 = \boxed{750}&lt;/math&gt;. (This is true because &lt;math&gt;37n = 925 \cdot 30 \rightarrow n= 25 \cdot 30 = 750&lt;/math&gt;)<br /> <br /> ~qwertysri987<br /> <br /> <br /> ==Solution 4==<br /> &lt;math&gt;\frac{n}{180} = \frac{d25}{999}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;\frac{n}{30} = \frac{d25}{37}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;37n = 30(d25)&lt;/math&gt;<br /> <br /> <br /> Thus &lt;math&gt;n&lt;/math&gt; divides 25 and 30. The LCM is$\boxed{750}4.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1989|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_9&diff=132690 1990 AIME Problems/Problem 9 2020-08-29T22:35:25Z <p>Jackshi2006: /* Solution 2 Recursion */</p> <hr /> <div>== Problem ==<br /> A [[fair]] coin is to be tossed &lt;math&gt;10_{}^{}&lt;/math&gt; times. Let &lt;math&gt;i/j^{}_{}&lt;/math&gt;, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find &lt;math&gt;i+j_{}^{}&lt;/math&gt;. <br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Clearly, at least &lt;math&gt;5&lt;/math&gt; tails must be flipped; any less, then by the [[Pigeonhole Principle]] there will be heads that appear on consecutive tosses. <br /> <br /> Consider the case when &lt;math&gt;5&lt;/math&gt; tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled &lt;math&gt;(H)&lt;/math&gt;:<br /> <br /> :&lt;math&gt;(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)&lt;/math&gt;<br /> <br /> There are six slots for the heads to be placed, but only &lt;math&gt;5&lt;/math&gt; heads remaining. Thus, using [[stars-and-bars]] there are &lt;math&gt;{6\choose5}&lt;/math&gt; possible combinations of 6 heads. Continuing this pattern, we find that there are &lt;math&gt;\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144&lt;/math&gt;. There are a total of &lt;math&gt;2^{10}&lt;/math&gt; possible flips of &lt;math&gt;10&lt;/math&gt; coins, making the probability &lt;math&gt;\frac{144}{1024} = \frac{9}{64}&lt;/math&gt;. Thus, our solution is &lt;math&gt;9 + 64 = \boxed{073}&lt;/math&gt;.<br /> <br /> === Solution 2 (Recursion)===<br /> Call the number of ways of flipping &lt;math&gt;n&lt;/math&gt; coins and not receiving any consecutive heads &lt;math&gt;S_n&lt;/math&gt;. Notice that tails must be received in at least one of the first two flips.<br /> <br /> If the first coin flipped is a T, then the remaining &lt;math&gt;n-1&lt;/math&gt; flips must fall under one of the configurations of &lt;math&gt;S_{n-1}&lt;/math&gt;.<br /> <br /> If the first coin flipped is a H, then the second coin must be a T. There are then &lt;math&gt;S_{n-2}&lt;/math&gt; configurations.<br /> <br /> Thus, &lt;math&gt;S_n = S_{n-1} + S_{n-2}&lt;/math&gt;. By counting, we can establish that &lt;math&gt;S_1 = 2&lt;/math&gt; and &lt;math&gt;S_2 = 3&lt;/math&gt;. Therefore, &lt;math&gt;S_3 = 5,\ S_4 = 8&lt;/math&gt;, forming the [[Fibonacci sequence]]. Listing them out, we get &lt;math&gt;2,3,5,8,13,21,34,55,89,144&lt;/math&gt;, and the 10th number is &lt;math&gt;144&lt;/math&gt;. Putting this over &lt;math&gt;2^{10}&lt;/math&gt; to find the probability, we get &lt;math&gt;\frac{9}{64}&lt;/math&gt;. Our solution is &lt;math&gt;9+64=\boxed{073}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> We can also split the problem into casework.<br /> <br /> Case 1: 0 Heads<br /> <br /> There is only one possibility.<br /> <br /> Case 2: 1 Head<br /> <br /> There are 10 possibilities.<br /> <br /> Case 3: 2 Heads<br /> <br /> There are 36 possibilities.<br /> <br /> Case 4: 3 Heads<br /> <br /> There are 56 possibilities.<br /> <br /> Case 5: 4 Heads<br /> <br /> There are 35 possibilities.<br /> <br /> Case 6: 5 Heads<br /> <br /> There are 6 possibilities.<br /> <br /> We have &lt;math&gt;1+10+36+56+35+6=144&lt;/math&gt;, and there are &lt;math&gt;1024&lt;/math&gt; possible outcomes, so the probability is &lt;math&gt;\frac{144}{1024}=\frac{9}{64}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{073}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{-RootThreeOverTwo}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_9&diff=132689 1990 AIME Problems/Problem 9 2020-08-29T22:35:06Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> A [[fair]] coin is to be tossed &lt;math&gt;10_{}^{}&lt;/math&gt; times. Let &lt;math&gt;i/j^{}_{}&lt;/math&gt;, in lowest terms, be the [[probability]] that heads never occur on consecutive tosses. Find &lt;math&gt;i+j_{}^{}&lt;/math&gt;. <br /> <br /> __TOC__<br /> == Solution ==<br /> === Solution 1 ===<br /> Clearly, at least &lt;math&gt;5&lt;/math&gt; tails must be flipped; any less, then by the [[Pigeonhole Principle]] there will be heads that appear on consecutive tosses. <br /> <br /> Consider the case when &lt;math&gt;5&lt;/math&gt; tails occur. The heads must fall between the tails such that no two heads fall between the same tails, and must fall in the positions labeled &lt;math&gt;(H)&lt;/math&gt;:<br /> <br /> :&lt;math&gt;(H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)\ T\ (H)&lt;/math&gt;<br /> <br /> There are six slots for the heads to be placed, but only &lt;math&gt;5&lt;/math&gt; heads remaining. Thus, using [[stars-and-bars]] there are &lt;math&gt;{6\choose5}&lt;/math&gt; possible combinations of 6 heads. Continuing this pattern, we find that there are &lt;math&gt;\sum_{i=6}^{11} {i\choose{11-i}} = {6\choose5} + {7\choose4} + {8\choose3} + {9\choose2} + {{10}\choose1} + {{11}\choose0} = 144&lt;/math&gt;. There are a total of &lt;math&gt;2^{10}&lt;/math&gt; possible flips of &lt;math&gt;10&lt;/math&gt; coins, making the probability &lt;math&gt;\frac{144}{1024} = \frac{9}{64}&lt;/math&gt;. Thus, our solution is &lt;math&gt;9 + 64 = \boxed{073}&lt;/math&gt;.<br /> <br /> === Solution 2 Recursion===<br /> Call the number of ways of flipping &lt;math&gt;n&lt;/math&gt; coins and not receiving any consecutive heads &lt;math&gt;S_n&lt;/math&gt;. Notice that tails must be received in at least one of the first two flips.<br /> <br /> If the first coin flipped is a T, then the remaining &lt;math&gt;n-1&lt;/math&gt; flips must fall under one of the configurations of &lt;math&gt;S_{n-1}&lt;/math&gt;.<br /> <br /> If the first coin flipped is a H, then the second coin must be a T. There are then &lt;math&gt;S_{n-2}&lt;/math&gt; configurations.<br /> <br /> Thus, &lt;math&gt;S_n = S_{n-1} + S_{n-2}&lt;/math&gt;. By counting, we can establish that &lt;math&gt;S_1 = 2&lt;/math&gt; and &lt;math&gt;S_2 = 3&lt;/math&gt;. Therefore, &lt;math&gt;S_3 = 5,\ S_4 = 8&lt;/math&gt;, forming the [[Fibonacci sequence]]. Listing them out, we get &lt;math&gt;2,3,5,8,13,21,34,55,89,144&lt;/math&gt;, and the 10th number is &lt;math&gt;144&lt;/math&gt;. Putting this over &lt;math&gt;2^{10}&lt;/math&gt; to find the probability, we get &lt;math&gt;\frac{9}{64}&lt;/math&gt;. Our solution is &lt;math&gt;9+64=\boxed{073}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> We can also split the problem into casework.<br /> <br /> Case 1: 0 Heads<br /> <br /> There is only one possibility.<br /> <br /> Case 2: 1 Head<br /> <br /> There are 10 possibilities.<br /> <br /> Case 3: 2 Heads<br /> <br /> There are 36 possibilities.<br /> <br /> Case 4: 3 Heads<br /> <br /> There are 56 possibilities.<br /> <br /> Case 5: 4 Heads<br /> <br /> There are 35 possibilities.<br /> <br /> Case 6: 5 Heads<br /> <br /> There are 6 possibilities.<br /> <br /> We have &lt;math&gt;1+10+36+56+35+6=144&lt;/math&gt;, and there are &lt;math&gt;1024&lt;/math&gt; possible outcomes, so the probability is &lt;math&gt;\frac{144}{1024}=\frac{9}{64}&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{073}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\textbf{-RootThreeOverTwo}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_3&diff=132685 1990 AIME Problems/Problem 3 2020-08-29T19:21:00Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P_1^{}&lt;/math&gt; be a [[Regular polygon|regular]] &lt;math&gt;r~\mbox{gon}&lt;/math&gt; and &lt;math&gt;P_2^{}&lt;/math&gt; be a regular &lt;math&gt;s~\mbox{gon}&lt;/math&gt; &lt;math&gt;(r\geq s\geq 3)&lt;/math&gt; such that each [[interior angle]] of &lt;math&gt;P_1^{}&lt;/math&gt; is &lt;math&gt;\frac{59}{58}&lt;/math&gt; as large as each interior angle of &lt;math&gt;P_2^{}&lt;/math&gt;. What's the largest possible value of &lt;math&gt;s_{}^{}&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> The formula for the interior angle of a regular sided [[polygon]] is &lt;math&gt;\frac{(n-2)180}{n}&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}&lt;/math&gt;. Cross multiplying and simplifying, we get &lt;math&gt;\frac{58(r-2)}{r} = \frac{59(s-2)}{s}&lt;/math&gt;. Cross multiply and combine like terms again to yield &lt;math&gt;58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs&lt;/math&gt;. Solving for &lt;math&gt;r&lt;/math&gt;, we get &lt;math&gt;r = \frac{116s}{118 - s}&lt;/math&gt;.<br /> <br /> &lt;math&gt;r \ge 0&lt;/math&gt; and &lt;math&gt;s \ge 0&lt;/math&gt;, making the [[numerator]] of the [[fraction]] positive. To make the [[denominator]] [[positive]], &lt;math&gt;s &lt; 118&lt;/math&gt;; the largest possible value of &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;117&lt;/math&gt;.<br /> <br /> This is achievable because the denominator is &lt;math&gt;1&lt;/math&gt;, making &lt;math&gt;r&lt;/math&gt; a positive number &lt;math&gt;116 \cdot 117&lt;/math&gt; and &lt;math&gt;s = \boxed{117}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Like above, use the formula for the interior angles of a regular sided [[polygon]].<br /> <br /> <br /> &lt;math&gt;\frac{(r-2)180}{r} = \frac{59}{58} * \frac{(s-2)180}{s}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * 180 * (s-2) * r = 58 * 180 * (r-2) * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * (rs - 2r) = 58 * (rs - 2s)&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs - 118r = -116s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs = 118r-116s&lt;/math&gt;<br /> <br /> <br /> This equation tells us &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;118r&lt;/math&gt;. If &lt;math&gt;s&lt;/math&gt; specifically divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is &lt;math&gt;s=59&lt;/math&gt;, which does give a solution: &lt;math&gt;s=59, r=116&lt;/math&gt;. Although, the problem asks for &lt;math&gt;s&lt;/math&gt;, not &lt;math&gt;r&lt;/math&gt;. The only conceivable reasoning behind this is that &lt;math&gt;r&lt;/math&gt; is greater than 1000. This prompts us to look into the second case, where &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;r&lt;/math&gt;. Make &lt;math&gt;r = s * k&lt;/math&gt;. Rewrite the equation using this new information.<br /> <br /> <br /> &lt;math&gt;s * s * k = 118 * s * k - 116 * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s * k = 118 * k - 116&lt;/math&gt;<br /> <br /> <br /> Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum: 116.<br /> <br /> <br /> &lt;math&gt;s * 116 = 118 * 116 - 116&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = 118 - 1&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = \boxed{117}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_3&diff=132684 1990 AIME Problems/Problem 3 2020-08-29T19:20:39Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P_1^{}&lt;/math&gt; be a [[Regular polygon|regular]] &lt;math&gt;r~\mbox{gon}&lt;/math&gt; and &lt;math&gt;P_2^{}&lt;/math&gt; be a regular &lt;math&gt;s~\mbox{gon}&lt;/math&gt; &lt;math&gt;(r\geq s\geq 3)&lt;/math&gt; such that each [[interior angle]] of &lt;math&gt;P_1^{}&lt;/math&gt; is &lt;math&gt;\frac{59}{58}&lt;/math&gt; as large as each interior angle of &lt;math&gt;P_2^{}&lt;/math&gt;. What's the largest possible value of &lt;math&gt;s_{}^{}&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> The formula for the interior angle of a regular sided [[polygon]] is &lt;math&gt;\frac{(n-2)180}{n}&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}&lt;/math&gt;. Cross multiplying and simplifying, we get &lt;math&gt;\frac{58(r-2)}{r} = \frac{59(s-2)}{s}&lt;/math&gt;. Cross multiply and combine like terms again to yield &lt;math&gt;58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs&lt;/math&gt;. Solving for &lt;math&gt;r&lt;/math&gt;, we get &lt;math&gt;r = \frac{116s}{118 - s}&lt;/math&gt;.<br /> <br /> &lt;math&gt;r \ge 0&lt;/math&gt; and &lt;math&gt;s \ge 0&lt;/math&gt;, making the [[numerator]] of the [[fraction]] positive. To make the [[denominator]] [[positive]], &lt;math&gt;s &lt; 118&lt;/math&gt;; the largest possible value of &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;117&lt;/math&gt;.<br /> <br /> This is achievable because the denominator is &lt;math&gt;1&lt;/math&gt;, making &lt;math&gt;r&lt;/math&gt; a positive number &lt;math&gt;116 \cdot 117&lt;/math&gt; and &lt;math&gt;s = \boxed{117}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Like above, use the formula for the interior angles of a regular sided [[polygon]].<br /> <br /> <br /> &lt;math&gt;\frac{(r-2)180}{r} = \frac{59}{58} * \frac{(s-2)180}{s}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * 180 * (s-2) * r = 58 * 180 * (r-2) * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * (rs - 2r) = 58 * (rs - 2s)&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs - 118r = -116s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs = 118r-116s&lt;/math&gt;<br /> <br /> <br /> This equation tells us &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;118r&lt;/math&gt;. If &lt;math&gt;s&lt;/math&gt; specifically divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is &lt;math&gt;s=59&lt;/math&gt;, which does give a solution: &lt;math&gt;s=59, r=116&lt;/math&gt;. Although, the problem asks for &lt;math&gt;s&lt;/math&gt;, not &lt;math&gt;r&lt;/math&gt;. The only conceivable reasoning behind this is that &lt;math&gt;r&lt;/math&gt; is greater than 1000. This prompts us to look into the second case, where &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;r&lt;/math&gt;. Make &lt;math&gt;r = s * k&lt;/math&gt;. Rewrite the equation using this new information.<br /> <br /> <br /> &lt;math&gt;s * s * k = 118 * s * k - 116 * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s * k = 118 * k - 116&lt;/math&gt;<br /> <br /> <br /> Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum, 116.<br /> <br /> <br /> &lt;math&gt;s * 116 = 118 * 116 - 116&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = 118 - 1&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = \boxed{117}&lt;/math&gt;<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1990_AIME_Problems/Problem_3&diff=132683 1990 AIME Problems/Problem 3 2020-08-29T19:20:20Z <p>Jackshi2006: /* Solution */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;P_1^{}&lt;/math&gt; be a [[Regular polygon|regular]] &lt;math&gt;r~\mbox{gon}&lt;/math&gt; and &lt;math&gt;P_2^{}&lt;/math&gt; be a regular &lt;math&gt;s~\mbox{gon}&lt;/math&gt; &lt;math&gt;(r\geq s\geq 3)&lt;/math&gt; such that each [[interior angle]] of &lt;math&gt;P_1^{}&lt;/math&gt; is &lt;math&gt;\frac{59}{58}&lt;/math&gt; as large as each interior angle of &lt;math&gt;P_2^{}&lt;/math&gt;. What's the largest possible value of &lt;math&gt;s_{}^{}&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> The formula for the interior angle of a regular sided [[polygon]] is &lt;math&gt;\frac{(n-2)180}{n}&lt;/math&gt;. <br /> <br /> Thus, &lt;math&gt;\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}&lt;/math&gt;. Cross multiplying and simplifying, we get &lt;math&gt;\frac{58(r-2)}{r} = \frac{59(s-2)}{s}&lt;/math&gt;. Cross multiply and combine like terms again to yield &lt;math&gt;58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs&lt;/math&gt;. Solving for &lt;math&gt;r&lt;/math&gt;, we get &lt;math&gt;r = \frac{116s}{118 - s}&lt;/math&gt;.<br /> <br /> &lt;math&gt;r \ge 0&lt;/math&gt; and &lt;math&gt;s \ge 0&lt;/math&gt;, making the [[numerator]] of the [[fraction]] positive. To make the [[denominator]] [[positive]], &lt;math&gt;s &lt; 118&lt;/math&gt;; the largest possible value of &lt;math&gt;s&lt;/math&gt; is &lt;math&gt;117&lt;/math&gt;.<br /> <br /> This is achievable because the denominator is &lt;math&gt;1&lt;/math&gt;, making &lt;math&gt;r&lt;/math&gt; a positive number &lt;math&gt;116 \cdot 117&lt;/math&gt; and &lt;math&gt;s = \boxed{117}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Like above, use the formula for the interior angles of a regular sided [[polygon]].<br /> <br /> <br /> &lt;math&gt;\frac{(r-2)180}{r} = \frac{59}{58} * \frac{(s-2)180}{s}&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * 180 * (s-2) * r = 58 * 180 * (r-2) * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;59 * (rs - 2r) = 58 * (rs - 2s)&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs - 118r = -116s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;rs = 118r-116s&lt;/math&gt;<br /> <br /> <br /> This equation tells us &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;118r&lt;/math&gt;. If &lt;math&gt;s&lt;/math&gt; specifically divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is &lt;math&gt;s=59&lt;/math&gt;, which does give a solution: &lt;math&gt;s=59, r=116&lt;/math&gt;. Although, the problem asks for &lt;math&gt;s&lt;/math&gt;, not &lt;math&gt;r&lt;/math&gt;. The only conceivable reasoning behind this is that &lt;math&gt;r&lt;/math&gt; is greater than 1000. This prompts us to look into the second case, where &lt;math&gt;s&lt;/math&gt; divides &lt;math&gt;r&lt;/math&gt;. Make &lt;math&gt;r = s * k&lt;/math&gt;. Rewrite the equation using this new information.<br /> <br /> <br /> &lt;math&gt;s * s * k = 118 * s * k - 116 * s&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s * k = 118 * k - 116&lt;/math&gt;<br /> <br /> <br /> Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum, 116.<br /> <br /> <br /> &lt;math&gt;s * 116 = 118 * 116 - 116&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = 118 - 1&lt;/math&gt;<br /> <br /> <br /> &lt;math&gt;s = \boxed{117}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AIME box|year=1990|num-b=2|num-a=4}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1991_AIME_Problems/Problem_6&diff=132625 1991 AIME Problems/Problem 6 2020-08-27T21:51:01Z <p>Jackshi2006: /* Solution */</p> <hr /> <div>== Problem ==<br /> Suppose &lt;math&gt;r^{}_{}&lt;/math&gt; is a [[real number]] for which<br /> &lt;div style=&quot;text-align:center&quot;&gt;&lt;math&gt;<br /> \left\lfloor r + \frac{19}{100} \right\rfloor + \left\lfloor r + \frac{20}{100} \right\rfloor + \left\lfloor r + \frac{21}{100} \right\rfloor + \cdots + \left\lfloor r + \frac{91}{100} \right\rfloor = 546.<br /> &lt;/math&gt;&lt;/div&gt;<br /> Find &lt;math&gt;\lfloor 100r \rfloor&lt;/math&gt;. (For real &lt;math&gt;x^{}_{}&lt;/math&gt;, &lt;math&gt;\lfloor x \rfloor&lt;/math&gt; is the [[floor function|greatest integer]] less than or equal to &lt;math&gt;x^{}_{}&lt;/math&gt;.)<br /> <br /> == Solution ==<br /> There are &lt;math&gt;91 - 19 + 1 = 73&lt;/math&gt; numbers in the [[sequence]]. Since the terms of the sequence can be at most &lt;math&gt;1&lt;/math&gt; apart, all of the numbers in the sequence can take one of two possible values. Since &lt;math&gt;\frac{546}{73} = 7 R 35&lt;/math&gt;, the values of each of the terms of the sequence must be either &lt;math&gt;7&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt;. As the remainder is &lt;math&gt;35&lt;/math&gt;, &lt;math&gt;8&lt;/math&gt; must take on &lt;math&gt;35&lt;/math&gt; of the values, with &lt;math&gt;7&lt;/math&gt; being the value of the remaining &lt;math&gt;73 - 35 = 38&lt;/math&gt; numbers. The 39th number is &lt;math&gt;\lfloor r+\frac{19 + 39 - 1}{100}\rfloor= \lfloor r+\frac{57}{100}\rfloor&lt;/math&gt;, which is also the first term of this sequence with a value of &lt;math&gt;8&lt;/math&gt;, so &lt;math&gt;8 \le r + \frac{57}{100} &lt; 8.01&lt;/math&gt;. Solving shows that &lt;math&gt;\frac{743}{100} \le r &lt; \frac{744}{100}&lt;/math&gt;, so &lt;math&gt;743\le 100r &lt; 744&lt;/math&gt;, and &lt;math&gt;\lfloor 100r \rfloor = \boxed{743}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AIME box|year=1991|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132466 1992 AIME Problems/Problem 4 2020-08-24T17:25:32Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ==Solution 2==<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio &lt;math&gt;N&lt;/math&gt;. &lt;math&gt;N = \dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt;. Because we have the ratio, &lt;math&gt;N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt; = &lt;math&gt;3:4:5&lt;/math&gt;,<br /> <br /> &lt;math&gt;\frac{x-t}{t+1} = \frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}&lt;/math&gt;.<br /> Solve the equation to get get &lt;math&gt; t= 26 &lt;/math&gt; and &lt;math&gt;x = \boxed{062}&lt;/math&gt;.<br /> <br /> https://www.wolframalpha.com/input/?i=%28x-t%29%2F%28t%2B1%29++%3D+4%2F3%2C++%28x-t%29%28x-t-1%29%2F%28%28t%2B1%29%28t%2B2%29%29+%3D+5%2F3<br /> <br /> <br /> -Solution and LaTeX by jackshi2006<br /> <br /> <br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132465 1992 AIME Problems/Problem 4 2020-08-24T17:25:17Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ==Solution 2==<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio &lt;math&gt;N&lt;/math&gt;. &lt;math&gt;N = \dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt;. Because we have the ratio, &lt;math&gt;N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt; = &lt;math&gt;3:4:5&lt;/math&gt;.<br /> <br /> &lt;math&gt;\frac{x-t}{t+1} = \frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}&lt;/math&gt;.<br /> Solve the equation to get get &lt;math&gt; t= 26 &lt;/math&gt; and &lt;math&gt;x = \boxed{062}&lt;/math&gt;.<br /> <br /> https://www.wolframalpha.com/input/?i=%28x-t%29%2F%28t%2B1%29++%3D+4%2F3%2C++%28x-t%29%28x-t-1%29%2F%28%28t%2B1%29%28t%2B2%29%29+%3D+5%2F3<br /> <br /> <br /> -Solution and LaTeX by jackshi2006<br /> <br /> <br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132464 1992 AIME Problems/Problem 4 2020-08-24T17:25:09Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ==Solution 2==<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio &lt;math&gt;N&lt;/math&gt;. &lt;math&gt;N = \dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt;. Because we have the ratio, &lt;math&gt;N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt; = &lt;math&gt;3:4:5&lt;/math&gt;.<br /> <br /> &lt;math&gt;\frac{x-t}{t+1} = \frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}&lt;/math&gt;.<br /> Solve the equation to get get &lt;math&gt; t= 26 &lt;/math&gt; and &lt;math&gt;x = \boxed{062}&lt;/math&gt;.<br /> <br /> https://www.wolframalpha.com/input/?i=%28x-t%29%2F%28t%2B1%29++%3D+4%2F3%2C++%28x-t%29%28x-t-1%29%2F%28%28t%2B1%29%28t%2B2%29%29+%3D+5%2F3<br /> <br /> <br /> -Solution and LaTeX by jacksshi2006<br /> <br /> <br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132463 1992 AIME Problems/Problem 4 2020-08-24T17:15:21Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ==Solution 2==<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio &lt;math&gt;N&lt;/math&gt;. &lt;math&gt;N = \dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt;. Because we have the ratio, &lt;math&gt;N : N * \frac{x-t}{t+1} : N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt; = &lt;math&gt;3:4:5&lt;/math&gt;.<br /> <br /> &lt;math&gt;\frac{x-t}{t+1} = \frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}&lt;/math&gt;.<br /> Solve the equation to get get &lt;math&gt; t= 26 &lt;/math&gt; and &lt;math&gt;x = \boxed{062}&lt;/math&gt;.<br /> <br /> https://www.wolframalpha.com/input/?i=%28x-t%29%2F%28t%2B1%29++%3D+4%2F3%2C++%28x-t%29%28x-t-1%29%2F%28%28t%2B1%29%28t%2B2%29%29+%3D+5%2F3<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132462 1992 AIME Problems/Problem 4 2020-08-24T17:14:08Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ==Solution 2==<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio &lt;math&gt;N&lt;/math&gt;. &lt;math&gt;N = \dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt;. Because we have the 3:4:5 ratio, &lt;math&gt;\frac{x-t}{t+1} = \frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}&lt;/math&gt;.<br /> Solve the equation to get get &lt;math&gt; t= 26 &lt;/math&gt; and &lt;math&gt;x = \boxed{062}&lt;/math&gt;.<br /> <br /> https://www.wolframalpha.com/input/?i=%28x-t%29%2F%28t%2B1%29++%3D+4%2F3%2C++%28x-t%29%28x-t-1%29%2F%28%28t%2B1%29%28t%2B2%29%29+%3D+5%2F3<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132461 1992 AIME Problems/Problem 4 2020-08-24T17:13:28Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ==Solution 2==<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio N. This is &lt;math&gt;\dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt;. Because we have the 3:4:5 ratio, &lt;math&gt;\frac{x-t}{t+1} = \frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}&lt;/math&gt;.<br /> Solve the equation to get get &lt;math&gt; t= 26 &lt;/math&gt; and &lt;math&gt;x = \boxed{062}&lt;/math&gt;.<br /> <br /> https://www.wolframalpha.com/input/?i=%28x-t%29%2F%28t%2B1%29++%3D+4%2F3%2C++%28x-t%29%28x-t-1%29%2F%28%28t%2B1%29%28t%2B2%29%29+%3D+5%2F3<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132460 1992 AIME Problems/Problem 4 2020-08-24T17:13:20Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ==Solution 2==<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio N. This is &lt;math&gt;\dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt;. Because we have the 3:4:5 ratio, &lt;math&gt;\frac{x-t}{t+1} = \frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}&lt;/math&gt;.<br /> Solve the equation to get get &lt;math&gt; t= 26 &lt;/math&gt; and &lt;math&gt;x = \boxed{062}&lt;/math&gt;.<br /> https://www.wolframalpha.com/input/?i=%28x-t%29%2F%28t%2B1%29++%3D+4%2F3%2C++%28x-t%29%28x-t-1%29%2F%28%28t%2B1%29%28t%2B2%29%29+%3D+5%2F3<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132459 1992 AIME Problems/Problem 4 2020-08-24T17:13:03Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ==Solution 2==<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio N. This is &lt;math&gt;\dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt;. Because we have the 3:4:5 ratio, &lt;math&gt;\frac{x-t}{t+1} = \frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}&lt;/math&gt;.<br /> Solve the equation to get get &lt;math&gt; t= 26 &lt;/math&gt; and &lt;math&gt;x = \boxed{062}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> <br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132458 1992 AIME Problems/Problem 4 2020-08-24T17:12:53Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ==Solution 2==<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio N. This is &lt;math&gt;\dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt;. Because we have the 3:4:5 ratio, &lt;math&gt;\frac{x-t}{t+1} = \frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}&lt;/math&gt;.<br /> Solve the equation to get get &lt;math&gt; t= 26 &lt;/math&gt; and &lt;math&gt;x = \boxed{062}&lt;/math&gt;.<br /> <br /> -jackshi2006<br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132457 1992 AIME Problems/Problem 4 2020-08-24T17:11:57Z <p>Jackshi2006: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ==Solution 2==<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio N. This is &lt;math&gt;\dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt;. Because we have the 3:4:5 ratio, &lt;math&gt;\frac{x-t}{t+1} = \frac{4}{3}&lt;/math&gt; and &lt;math&gt;\frac{(x-t)*(x-t-1)}{(t+1)*(t+2)} = \frac{5}{3}&lt;/math&gt;.<br /> Solve the equation to get get &lt;math&gt; t= 26 &lt;/math&gt; and &lt;math&gt;x = 62&lt;/math&gt;.<br /> <br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1992_AIME_Problems/Problem_4&diff=132456 1992 AIME Problems/Problem 4 2020-08-24T17:04:58Z <p>Jackshi2006: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> In Pascal's Triangle, each entry is the sum of the two entries above it. In which row of [[Pascal's Triangle]] do three consecutive entries occur that are in the ratio &lt;math&gt;3: 4: 5&lt;/math&gt;?<br /> <br /> == Solution 1==<br /> <br /> Consider what the ratio means. Since we know that they are consecutive terms, we can say<br /> &lt;cmath&gt;\frac{\dbinom{n}{k-1}}{3} = \frac{\dbinom{n}{k}}{4} = \frac{\dbinom{n}{k+1}}{5}.&lt;/cmath&gt;<br /> <br /> Taking the first part, and using our expression for &lt;math&gt;n&lt;/math&gt; choose &lt;math&gt;k&lt;/math&gt;,<br /> &lt;cmath&gt; \frac{n!}{3(k-1)!(n-k+1)!} = \frac{n!}{4k!(n-k)!}&lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(k-1)!(n-k+1)!} = \frac{1}{4k!(n-k)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{3(n-k+1)} = \frac{1}{4k} &lt;/cmath&gt;<br /> &lt;cmath&gt; n-k+1 = \frac{4k}{3} &lt;/cmath&gt;<br /> &lt;cmath&gt; n = \frac{7k}{3} - 1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{3(n+1)}{7} = k &lt;/cmath&gt;<br /> Then, we can use the second part of the equation.<br /> &lt;cmath&gt; \frac{n!}{4k!(n-k)!} = \frac{n!}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4k!(n-k)!} = \frac{1}{5(k+1)!(n-k-1)!} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{1}{4(n-k)} = \frac{1}{5(k+1)} &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4(n-k)}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5}-\frac{4k}{5} = k+1 &lt;/cmath&gt;<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9k}{5} +1. &lt;/cmath&gt;<br /> Since we know &lt;math&gt;k = \frac{3(n+1)}{7}&lt;/math&gt; we can plug this in, giving us<br /> &lt;cmath&gt; \frac{4n}{5} = \frac{9\left(\frac{3(n+1)}{7}\right)}{5} +1 &lt;/cmath&gt;<br /> &lt;cmath&gt; 4n = 9\left(\frac{3(n+1)}{7}\right)+5 &lt;/cmath&gt;<br /> &lt;cmath&gt; 7(4n - 5) = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; 28n - 35 = 27n+27 &lt;/cmath&gt;<br /> &lt;cmath&gt; n = 62 &lt;/cmath&gt;<br /> We can also evaluate for &lt;math&gt;k&lt;/math&gt;, and find that &lt;math&gt;k = \frac{3(62+1)}{7} = 27.&lt;/math&gt; Since we want &lt;math&gt;n&lt;/math&gt;, however, our final answer is &lt;math&gt;\boxed{062.}&lt;/math&gt; ~LaTeX by ciceronii<br /> <br /> <br /> ===Solution 2===<br /> Call the row x, and the number from the leftmost side t. Call the first term in the ratio N. This is &lt;math&gt;\dbinom{x}{t}&lt;/math&gt;. The next term is &lt;math&gt;N * \frac{x-t}{t+1}&lt;/math&gt;, and the final term is &lt;math&gt;N * \frac{(x-t)*(x-t-1)}{(t+1)*(t+2)}&lt;/math&gt; <br /> <br /> {{AIME box|year=1992|num-b=3|num-a=5}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Number Theory Problems]]<br /> {{MAA Notice}}<br /> MU WDWDOIOIJDWOIJDWIWOJDIWOJDIJOWDIJOWDIO 10x*****</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_6&diff=132002 1994 AIME Problems/Problem 6 2020-08-17T23:57:19Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> The graphs of the equations<br /> &lt;center&gt;&lt;math&gt;y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,&lt;/math&gt;&lt;/center&gt;<br /> are drawn in the coordinate plane for &lt;math&gt;k=-10,-9,-8,\ldots,9,10.\,&lt;/math&gt; These 63 lines cut part of the plane into equilateral triangles of side &lt;math&gt;2/\sqrt{3}.\,&lt;/math&gt; How many such triangles are formed?<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon. <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> picture pica, picb, picc;<br /> int i;<br /> for(i=-10;i&lt;=10;++i){<br /> if((i%10) == 0){draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i),black+0.7);}<br /> else{draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i));}<br /> }<br /> picb = rotate(120,origin)*pica;<br /> picc = rotate(240,origin)*pica;<br /> add(pica);add(picb);add(picc);<br /> &lt;/asy&gt;<br /> <br /> Solving the above equations for &lt;math&gt;k=\pm 10&lt;/math&gt;, we see that the hexagon in question is regular, with side length &lt;math&gt;\frac{20}{\sqrt{3}}&lt;/math&gt;. Then, the number of triangles within the hexagon is simply the ratio of the area of the hexagon to the area of a regular triangle. Since the ratio of the area of two similar figures is the square of the ratio of their side lengths, we see that the ratio of the area of one of the six equilateral triangles composing the regular hexagon to the area of a unit regular triangle is just &lt;math&gt;\left(\frac{20/\sqrt{3}}{2/\sqrt{3}}\right)^2 = 100&lt;/math&gt;. Thus, the total number of unit triangles is &lt;math&gt;6 \times 100 = 600&lt;/math&gt;.<br /> <br /> There are &lt;math&gt;6 \cdot 10&lt;/math&gt; equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is &lt;math&gt;600+60 = \boxed{660}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> There are three types of lines: horizontal, upward-slanting diagonal, and downward-slanting diagonal. There are &lt;math&gt;21&lt;/math&gt; of each type of line, and a unit equilateral triangle is determined by exactly one of each type of line. Given an upward-slanting diagonal and a downward-slanting diagonal, they determine exactly two unit equilateral triangles as shown below.<br /> &lt;asy&gt;<br /> size(60);<br /> pair u=rotate(60)*(1,0),d=rotate(-60)*(1,0),h=(1,0);<br /> draw((0,0)--4*u^^-2*h+4*u--(-2*h+4*u+4*d));<br /> draw(u--2*u+d,dotted);<br /> draw(3*u--3*u-h,dotted);<br /> &lt;/asy&gt;<br /> Therefore, if all horizontal lines are drawn, there will be a total of &lt;math&gt;2\cdot 21^2=882&lt;/math&gt; unit equilateral triangles. Of course, we only draw &lt;math&gt;21&lt;/math&gt; horizontal lines, so we are overcounting the triangles that are caused by the undrawn horizontal lines. In the below diagram, we draw the diagonal lines and the highest and lowest horizontal lines.<br /> &lt;asy&gt;<br /> size(200);<br /> pair u=rotate(60)*(2/sqrt(3),0),d=rotate(-60)*(2/sqrt(3),0),h=(2/sqrt(3),0);<br /> for (int i=0;i&lt;21;++i) {path v=(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d;draw(shift(0,-2*i)*v);}<br /> for (int i=0;i&lt;21;++i) {path v=rotate(180)*((-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d);draw(shift(0,2*i)*v);}<br /> draw((-15,-10)--(15,-10));<br /> draw((-15,10)--(15,10));<br /> &lt;/asy&gt;<br /> <br /> We see that the lines &lt;math&gt;y=-21,-20,\dots, -11&lt;/math&gt; and &lt;math&gt;y=11,12,\dots,21&lt;/math&gt; would complete several of the &lt;math&gt;882&lt;/math&gt; unit equilateral triangles. In fact, we can see that the lines &lt;math&gt;y=-21,-20,\dots,-11&lt;/math&gt; complete &lt;math&gt;1,2,(1+3),(2+4),(3+5),(4+6),\dots,(9+11)&lt;/math&gt; triangles, or &lt;math&gt;111&lt;/math&gt; triangles. The positive horizontal lines complete the same number of triangles, hence the answer is &lt;math&gt;882-2\cdot 111=\boxed{660}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3 Elementary Counting===<br /> Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is 20 units, and by extending horizontal lines to the sides of the hexagon. This shows that for every side of the hexagon there are 10 spaces. Therefore the side length of the biggest triangle (imagine one of the overlapping triangles in the Star of David) is 30. The total number of triangles in the hexagon can be found by finding the number of triangles in the extended triangle and subtracting the 3 corner triangles. This gives us &lt;math&gt;30^2 - 10^2 - 10^2- 10^2 = 600&lt;/math&gt;. That is the number of triangles in the hexagon. The remaining triangles form in groups of 10 on the exterior of each side. &lt;math&gt;600 + 6 * 10 = \boxed{660}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1994|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006 https://artofproblemsolving.com/wiki/index.php?title=1994_AIME_Problems/Problem_6&diff=132001 1994 AIME Problems/Problem 6 2020-08-17T23:56:21Z <p>Jackshi2006: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> The graphs of the equations<br /> &lt;center&gt;&lt;math&gt;y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,&lt;/math&gt;&lt;/center&gt;<br /> are drawn in the coordinate plane for &lt;math&gt;k=-10,-9,-8,\ldots,9,10.\,&lt;/math&gt; These 63 lines cut part of the plane into equilateral triangles of side &lt;math&gt;2/\sqrt{3}.\,&lt;/math&gt; How many such triangles are formed?<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon. <br /> <br /> &lt;asy&gt;<br /> size(200);<br /> picture pica, picb, picc;<br /> int i;<br /> for(i=-10;i&lt;=10;++i){<br /> if((i%10) == 0){draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i),black+0.7);}<br /> else{draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i));}<br /> }<br /> picb = rotate(120,origin)*pica;<br /> picc = rotate(240,origin)*pica;<br /> add(pica);add(picb);add(picc);<br /> &lt;/asy&gt;<br /> <br /> Solving the above equations for &lt;math&gt;k=\pm 10&lt;/math&gt;, we see that the hexagon in question is regular, with side length &lt;math&gt;\frac{20}{\sqrt{3}}&lt;/math&gt;. Then, the number of triangles within the hexagon is simply the ratio of the area of the hexagon to the area of a regular triangle. Since the ratio of the area of two similar figures is the square of the ratio of their side lengths, we see that the ratio of the area of one of the six equilateral triangles composing the regular hexagon to the area of a unit regular triangle is just &lt;math&gt;\left(\frac{20/\sqrt{3}}{2/\sqrt{3}}\right)^2 = 100&lt;/math&gt;. Thus, the total number of unit triangles is &lt;math&gt;6 \times 100 = 600&lt;/math&gt;.<br /> <br /> There are &lt;math&gt;6 \cdot 10&lt;/math&gt; equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is &lt;math&gt;600+60 = \boxed{660}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> There are three types of lines: horizontal, upward-slanting diagonal, and downward-slanting diagonal. There are &lt;math&gt;21&lt;/math&gt; of each type of line, and a unit equilateral triangle is determined by exactly one of each type of line. Given an upward-slanting diagonal and a downward-slanting diagonal, they determine exactly two unit equilateral triangles as shown below.<br /> &lt;asy&gt;<br /> size(60);<br /> pair u=rotate(60)*(1,0),d=rotate(-60)*(1,0),h=(1,0);<br /> draw((0,0)--4*u^^-2*h+4*u--(-2*h+4*u+4*d));<br /> draw(u--2*u+d,dotted);<br /> draw(3*u--3*u-h,dotted);<br /> &lt;/asy&gt;<br /> Therefore, if all horizontal lines are drawn, there will be a total of &lt;math&gt;2\cdot 21^2=882&lt;/math&gt; unit equilateral triangles. Of course, we only draw &lt;math&gt;21&lt;/math&gt; horizontal lines, so we are overcounting the triangles that are caused by the undrawn horizontal lines. In the below diagram, we draw the diagonal lines and the highest and lowest horizontal lines.<br /> &lt;asy&gt;<br /> size(200);<br /> pair u=rotate(60)*(2/sqrt(3),0),d=rotate(-60)*(2/sqrt(3),0),h=(2/sqrt(3),0);<br /> for (int i=0;i&lt;21;++i) {path v=(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d;draw(shift(0,-2*i)*v);}<br /> for (int i=0;i&lt;21;++i) {path v=rotate(180)*((-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d);draw(shift(0,2*i)*v);}<br /> draw((-15,-10)--(15,-10));<br /> draw((-15,10)--(15,10));<br /> &lt;/asy&gt;<br /> <br /> We see that the lines &lt;math&gt;y=-21,-20,\dots, -11&lt;/math&gt; and &lt;math&gt;y=11,12,\dots,21&lt;/math&gt; would complete several of the &lt;math&gt;882&lt;/math&gt; unit equilateral triangles. In fact, we can see that the lines &lt;math&gt;y=-21,-20,\dots,-11&lt;/math&gt; complete &lt;math&gt;1,2,(1+3),(2+4),(3+5),(4+6),\dots,(9+11)&lt;/math&gt; triangles, or &lt;math&gt;111&lt;/math&gt; triangles. The positive horizontal lines complete the same number of triangles, hence the answer is &lt;math&gt;882-2\cdot 111=\boxed{660}&lt;/math&gt;.<br /> <br /> <br /> ===Solution 3===<br /> Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is 20 units, and by extending horizontal lines to the sides of the hexagon. This shows that for every side of the hexagon there are 10 spaces. Therefore the side length of the biggest triangle (imagine one of the overlapping triangles in the Star of David) is 30. The total number of triangles in the hexagon can be found by finding the number of triangles in the extended triangle and subtracting the 3 corner triangles. This gives us &lt;math&gt;30^2 - 10^2 - 10^2- 10^2 = 600&lt;/math&gt;. That is the number of triangles in the hexagon. The remaining triangles form in groups of 10 on the exterior of each side. &lt;math&gt;600 + 6 * 10 = \boxed{660}&lt;/math&gt;.<br /> <br /> <br /> -jackshi2006<br /> <br /> == See also ==<br /> {{AIME box|year=1994|num-b=5|num-a=7}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]<br /> [[Category:Intermediate Geometry Problems]]<br /> {{MAA Notice}}</div> Jackshi2006