2005 AIME I Problems/Problem 4

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 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 $r = \frac{7 \pm \sqrt{49 - 4(1)(-s - 5)}}{2} = \frac{7 \pm \sqrt{4s^2 + 69}}{2}$ . $\displaystyle \sqrt{4s^2 + 69}$ 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 $294$ .

2005 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2005 AIME I Problems/Problem 4

Problem

Contents

Solution

Solution 1

Solution 2

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