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

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== Problem ==
 
== Problem ==
The director of a marching band asks the band members to line up in rows of four, but one is left over. Then she tries to line them up in rows of six, but three are left over. Finally, she tries to line them up in rows of seven, but four are left over.
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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.
What is the maximum number of members that the band can have?
 
 
 
 
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Revision as of 17:53, 16 December 2019

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 $n > 14$ 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 $\boxed{294}$.

Solution 2

Define the number of rows/columns of the square formation as $s$, and the number of rows of the rectangular formation $r$ (so there are $r - 7$ columns). Thus, $s^2 + 5 = r(r-7) \Longrightarrow r^2 - 7r - s^2 - 5 = 0$. The quadratic formula yields $r = \frac{7 \pm \sqrt{49 - 4(1)(-s^2 - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}$. $\sqrt{4s^2 + 69}$ must be an integer, say $x$. Then $4s^2 + 69 = x^2$ and $(x + 2s)(x - 2s) = 69$. The factors of $69$ are $(1,69), (3,23)$; $x$ is maximized for the first case. Thus, $x = \frac{69 + 1}{2} = 35$, and $r = \frac{7 \pm 35}{2} = 21, -14$. The latter obviously can be discarded, so there are $21$ rows and $21 - 7 = 14$ columns, making the answer $294$.

Solution 3

The number of members is $m^2+5=n(n+7)$ for some $n$ and $m$. Multiply both sides by $4$ and complete the square to get $4m^2+69=(2n+7)^2$. Thus, we have $69=((2n+7)+2m)((2n+7)-2m)$. Since we want to maximize $n$, set the first factor equal to $69$ and the second equal to $1$. Solving gives $n=14$, so the answer is $14\cdot21=294$.


See also

2005 AIME I (ProblemsAnswer KeyResources)
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

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