Difference between revisions of "2005 AIME I Problems/Problem 4"
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== Problem == | == Problem == | ||
− | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are 5 members left over. The director realizes that if he arranges the group in a formation with 7 more rows than columns, there are no members left over. Find the maximum number of members this band can have. | + | The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are <math>5</math> members left over. The director realizes that if he arranges the group in a formation with <math>7</math> more rows than columns, there are no members left over. Find the maximum number of members this band can have. |
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | If <math>n > 14</math> then <math>n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21</math> and so <math>(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5</math>. If <math>n</math> is an [[integer]] there are no numbers which are 5 more than a [[perfect square]] strictly between <math>(n + 3)^2 + 5</math> and <math>(n + 4)^2 + 5</math>. Thus, if the number of columns is <math>n</math>, the number of students is <math>n(n + 7)</math> which must be 5 more than a perfect square, so <math>n \leq 14</math>. In fact, when <math>n = 14</math> we have <math>n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5</math>, so this number works and no larger number can. Thus, the answer is 294. | + | If <math>n > 14</math> then <math>n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21</math> and so <math>(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5</math>. If <math>n</math> is an [[integer]] there are no numbers which are 5 more than a [[perfect square]] strictly between <math>(n + 3)^2 + 5</math> and <math>(n + 4)^2 + 5</math>. Thus, if the number of columns is <math>n</math>, the number of students is <math>n(n + 7)</math> which must be 5 more than a perfect square, so <math>n \leq 14</math>. In fact, when <math>n = 14</math> we have <math>n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5</math>, so this number works and no larger number can. Thus, the answer is <math>\boxed{294}</math>. |
=== Solution 2 === | === Solution 2 === | ||
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== See also == | == See also == | ||
{{AIME box|year=2005|n=I|num-b=3|num-a=5}} | {{AIME box|year=2005|n=I|num-b=3|num-a=5}} | ||
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+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 17:36, 26 April 2008
Problem
The director of a marching band wishes to place the members into a formation that includes all of them and has no unfilled positions. If they are arranged in a square formation, there are members left over. The director realizes that if he arranges the group in a formation with more rows than columns, there are no members left over. Find the maximum number of members this band can have.
Solution
Solution 1
If then and so . If is an integer there are no numbers which are 5 more than a perfect square strictly between and . Thus, if the number of columns is , the number of students is which must be 5 more than a perfect square, so . In fact, when we have , so this number works and no larger number can. Thus, the answer is .
Solution 2
Define the number of rows/columns of the square formation as , and the number of rows of the rectangular formation (so there are columns). Thus, . The quadratic formula yields . must be an integer, say . Then and . The factors of are ; is maximized for the first case. Thus, , and . The latter obviously can be discarded, so there are rows and columns, making the answer .
Solution 3
The number of members is for some and . Multiply both sides by and complete the square to get . Thus, we have . Since we want to maximize , set the first factor equal to and the second equal to . Solving gives , so the answer is .
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |