Question

一位行进乐队指挥想把队员排成一个包含所有队员、且没有空位的队形。如果他们排成正方形队形，会剩下 $5$ 名队员。指挥发现，如果他们排成一个行数比列数多 $7$ 的矩形队形，就可以达到要求。求这个乐队最多可能有多少名队员。

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 finds that if they are arranged in a rectangular formation with $7$ more rows than columns, the desired result can be obtained. Find the maximum number of members this band can have.

答案

答案：294

知识点：丢番图方程配方法平方差

难度评级：2230

解答：

设乐队有 $n$ 名队员。正方形队形给出 $n = s^2 + 5$ ，而列数为 $x$ 的矩形队形给出 $n = x(x + 7)$ 。将 $x^2 + 7x = s^2 + 5$ 乘以 $4$ 并配方，得 $(2x + 7)^2 - (2s)^2 = 69,$ 所以 $(2x + 7 - 2s)$ $\cdot (2x + 7 + 2s) = 69$ 。

将 $69 = 1 \cdot 69 = 3 \cdot 23$ ，且把较大的因数写在后面：由 $1 \cdot 69$ 得 $2x + 7 = 35$ 且 $2s = 34$ ，所以 $x = 14$ ， $s = 17$ ，于是 $n = 17^2 + 5 = 294$ 。由 $3 \cdot 23$ 得 $2x + 7 = 13$ 且 $2s = 10$ ，所以 $x = 3$ ， $s = 5$ ，于是 $n = 30$ 。

最大值是 $294$ 可由 $21 \times 14$ 的矩形达到。

Let the band have $n$ members, with $n = s^2 + 5$ for the square formation and $n = x(x + 7)$ for the rectangular formation with $x$ columns. Multiplying $x^2 + 7x = s^2 + 5$ by $4$ and completing the square gives $(2x + 7)^2 - (2s)^2 = 69,$ so $(2x + 7 - 2s)$ $\cdot (2x + 7 + 2s) = 69.$

Writing $69 = 1 \cdot 69 = 3 \cdot 23$ with the larger factor second: from $1 \cdot 69$ we get $2x + 7 = 35$ and $2s = 34,$ so $x = 14,$ $s = 17,$ and $n = 17^2 + 5 = 294.$ From $3 \cdot 23$ we get $2x + 7 = 13$ and $2s = 10,$ so $x = 3,$ $s = 5,$ and $n = 30.$

The maximum is $294,$ achieved by a $21 \times 14$ rectangle.

2005 AIME I 第 4 题

答案：294

难度评级：2230

解答：

其他年份的第 4 题