Question

设 $M$ 为最大的整数，使得 $M + 1213$ 和 $M + 3773$ 都是完全平方数。 $M$ 的个位数字是多少？

Let $M$ be the greatest integer such that both $M + 1213$ and $M + 3773$ are perfect squares. What is the units digit of $M?$

$1$

$2$

$3$

$6$

$8$

答案：E

知识点：完全平方数平方差奇偶性最优化

难度评级：1600

解答：

设 $M + 1213 = y^2$ 、 $M + 3773 = x^2$ 。相减得 $x^2 - y^2 = 2560$ ，即 $(x - y)(x + y) = 2560$ 。所以两个因子同奇偶，而乘积为偶数，二者都为偶数。写 $x - y = 2s$ 、 $x + y = 2t$ ，则 $st = 640$ 。要使 $M$ 最大，即使 $y = t - s$ 最大，需要 $s$ 尽量小。取 $s = 1, t = 640$ ，得 $y = 639$ 。因此 $M = 639^2 - 1213 = 407108$ ，个位数字为 $8$ ，正确答案是 E。

Set $M + 1213 = y^2$ and $M + 3773 = x^2.$ Subtracting, $x^2 - y^2 = 2560,$ so $(x - y)(x + y) = 2560.$ The two factors share a parity, and their product is even, so both are even: write $x - y = 2s,$ $x + y = 2t,$ with $st = 640.$ To make $M$ as large as possible we want $y = t - s$ as large as possible, so $s$ as small as possible. Take $s = 1, t = 640,$ giving $y = 639.$ Then $M = 639^2 - 1213 = 407108,$ whose units digit is $8.$ Thus, E is the correct answer.

2024 AMC 10A 第 15 题

答案：E

难度评级：1600

解答：

其他年份的第 15 题