Question

有多少个四位数同时满足以下三个性质？

(I) 十位数字和个位数字都是 $9$ 。

(II) 这个数比一个完全平方数小 $1$ 。

(III) 这个数是恰好两个质数的乘积。

How many four-digit numbers have all three of the following properties?

(I) The tens digit and ones digit are both $9.$

(II) The number is $1$ less than a perfect square.

(III) The number is the product of exactly two prime numbers.

$0$

$1$

$2$

$3$

$4$

答案：B

知识点：平方差质数

难度评级：1770

视频讲解：

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文字解答：

这个数形如 $XX99$ ，所以大 $1$ 的完全平方数以 $00$ 结尾，因此它是一个以 $0$ 结尾的数的平方。设这个平方为 $a^2$ 。

为了让 $a^2 - 1$ 是一个 $4$ 位数且末两位为 $99$ ， $a$ 只能是 $\{40, 50, 60, 70, 80, 90, 100\}$ 。

由于 $a^2 - 1 = (a-1)(a+1)$ ，还需要 $a-1$ 和 $a+1$ 都是质数。也就是说，要在 $\{40, 50, \dots, 100\}$ 附近寻找相邻的质数对。

逐一检查，只有 $59$ 和 $61$ 可行，因此这样的数恰有 $1$ 个。答案是 B。

The number has the form $XX99,$ and so the perfect square that is $1$ more ends in $00,$ and so it is the square of a number ending in $0.$ Suppose that square is $a^2.$

In order for $a^2 - 1$ to be a $4$ -digit number ending in $99$ , the only possibilities for $a$ are $\{40, 50, 60, 70, 80, 90, 100\}.$

Since $a^2 - 1 = (a-1)(a+1),$ we also need both $a-1$ and $a+1$ to be prime. We are then looking for pairs of prime numbers that are right around $\{40, 50, \dots, 100\}.$

Going through all the possibilities, the only ones that work are $59$ and $61$ , and so there is exactly $1$ way to do this. The answer is B.

2025 AMC 8 第 23 题

答案：B

难度评级：1770

视频讲解：

文字解答：

其他年份的第 23 题