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 $\boxed{294}$ .

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 $14\cdot21=294$ .

Solution 4

Partially completing the square

Geometrically: Split up the formation of $n + 7$ rows and $n$ columns into a square of $n$ rows and $n$ columns and a separate rectangle of the dimensions $7$ rows by $n$ columns. We want to take the rows from the rectangle and add them to the square to get another square and $5$ left over. If we attach exactly $2$ rows on the top and exactly $2$ rows on the side of the $n$ x $n$ square, then we have an $(n + 2)$ x $(n + 2)$ square that's missing a $2$ x $2$ corner. For the remaining $3n$ to fill this square plus the $5$ extra members, $n$ must be $3$ . If we instead plaster exactly $3$ rows from the $7$ x $n$ formation to two adjacent sides of the $n$ x $n$ square, we have an $(n + 3)$ x $(n + 3)$ formation that's missing a $3$ x $3$ corner. For the remaining row of length $n$ to fill this plus five, $n = 14$ . Plugging these in, we find $n = 14$ has a much higher count of members: $(n + 7)n; n = 14 --> 21(14) = 294$

Algebraically: We have $n^2 + 7n = m$ , where $m$ is the number of members in the band and $n$ is a positive integer. We partially complete the square for $n$ to get $n^2 + 7n = (n + 1)^2 + 5n - 1 = (n + 2)^2 + 3n - 4 = (n + 3)^2 + n - 9$ Our goal is to get $n^2 + 7n = y^2 + 5$ because we want $m$ to be $5$ more than a perfect square. From the above, $5n - 1 = 5$ means $n$ isn't an integer, $3n - 4 = 5$ means that $n = 3$ , and $n - 9 = 5$ means that $n = 14$ . Out of these, $n = 14$ is associated with the highest number of members in the band, so $m = (14^2) + 7(14) = 294$

Solution 5

Let there be $m$ members and $n$ members for the square and $c$ for the number of columns of the other formation. We have $n^2 +5 = c(c+7) \implies n^2+5 = \left(c+\frac{7}{2}\right)^2 -\frac{49}{4} \implies n^2 - \left(c+\frac{7}{2}\right)^2 = -\frac{69}{4} \implies \left(n-c-\frac{7}{2}\right)\left(n + c +\frac{7}{2}\right) \implies (2n-2c-7)(2n+2n+7) = -69.$

To maximize this we let $2n+2c+7 = 68$ and $2n-2c-7 = 1.$ Solving we find $n = 17$ so the desired number of members is $17^2 + 5 = \boxed{294}.$

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
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