Difference between revisions of "2005 AIME I Problems/Problem 4"

 
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== Solution ==
 
== Solution ==
 +
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.
  
 
== See also ==
 
== See also ==
 +
* [[2005 AIME I Problems/Problem 3 | Previous problem]]
 +
* [[2005 AIME I Problems/Problem 5 | Next problem]]
 
* [[2005 AIME I Problems]]
 
* [[2005 AIME I Problems]]

Revision as of 18:05, 17 January 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

If $n > 14$ then $n^2 + 6n + 14 < n^2 + 7n < n^2 + 8n + 21$ and so $(n + 3)^2 + 5 < n(n + 7) < (n + 4)^2 + 5$. If $n$ is an integer there are no numbers which are 5 more than a perfect square strictly between $(n + 3)^2 + 5$ and $(n + 4)^2 + 5$. Thus, if the number of columns is $n$, the number of students is $n(n + 7)$ which must be 5 more than a perfect square, so $n \leq 14$. In fact, when $n = 14$ we have $n(n + 7) = 14\cdot 21 = 294 = 17^2 + 5$, so this number works and no larger number can. Thus, the answer is 294.

See also

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