Difference between revisions of "2005 AIME I Problems/Problem 4"
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=== Solution 3 === | === Solution 3 === | ||
− | The number of members=<math>m^2+5=n(n+7)</math> for some <math>n</math> and <math>m</math>. Multiply both sides by 4 and complete the square to get <math>4m^2+69=(2n+7)^2</math>. Thus, we have <math>69=((2n+7)+2m)((2n+7)-2m)</math>. Sine we want to maximize <math>n</math>, set the first factor equal to 69 and the second equal to 1. Solving gives <math>n=14</math>, so the answer is <math>14 | + | The number of members=<math>m^2+5=n(n+7)</math> for some <math>n</math> and <math>m</math>. Multiply both sides by 4 and complete the square to get <math>4m^2+69=(2n+7)^2</math>. Thus, we have <math>69=((2n+7)+2m)((2n+7)-2m)</math>. Sine we want to maximize <math>n</math>, set the first factor equal to 69 and the second equal to 1. Solving gives <math>n=14</math>, so the answer is <math>14\cdot21=294</math>. |
== 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}} |
Revision as of 20:22, 29 December 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 .
Solution 3
The number of members= for some and . Multiply both sides by 4 and complete the square to get . Thus, we have . Sine we want to maximize , set the first factor equal to 69 and the second equal to 1. 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 |