Difference between revisions of "2005 AIME I Problems/Problem 4"
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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 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. | ||
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== Solution == | == Solution == | ||
+ | === 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 294. | ||
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+ | === Solution 2 === | ||
+ | Define the number of rows/columns of the square formation as <math>s</math>, and the number of rows of the rectangular formation <math>r</math> (so there are <math>r - 7</math> columns). Thus, <math>s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0</math>. The [[quadratic formula]] yields <math>r = \frac{7 \pm \sqrt{49 - 4(1)(-s - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}</math>. <math>\displaystyle \sqrt{4s^2 + 69}</math> must be an [[integer]], say <math>x</math>. Then <math>4s^2 + 69 = x^2</math> and <math>(x + 2s)(x - 2s) = 69</math>. The factors of <math>69</math> are <math>(1,69), (3,23)</math>; <math>x</math> is maximized for the first case. Thus, <math>x = \frac{69 + 1}{2} = 35</math>, and <math>r = \frac{7 \pm 35}{2} = 21, -14</math>. The latter obviously can be discarded, so there are <math>21</math> rows and <math>21 - 7 = 14</math> columns, making the answer <math>294</math>. | ||
== See also == | == See also == | ||
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Revision as of 15:30, 4 March 2007
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.
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 294.
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 .
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 |