2025 AMC 8 第 23 题

先试着解答 2025 AMC 8 第 23 题,然后核对你的答案与视频与精心整理的解答,解答来自 LIVE by Po-Shen Loh。你也可以参加完整限时模拟考试、查看全部 2025 AMC 8 解答,或核对答案。

所有题目均经美国数学协会(MAA)官方合法授权使用。

23.

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 99。
(II) 这个数比一个完全平方数小 11。
(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 9.9.
(II) The number is 11 less than a perfect square.
(III) The number is the product of exactly two prime numbers.
00
11
22
33
44
答案:B
知识点:平方差质数
小提示:
比它大一的平方数必须以 0000 结尾
The square one larger must end in 0000
大提示:
检查哪些 1010 的倍数两边相邻数都是质数
Check multiples of 1010 whose two neighbors are prime
视频讲解:
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文字解答:
这个数形如 XX99XX99,所以大 11 的完全平方数以 0000 结尾,因此它是一个以 00 结尾的数的平方。设这个平方为 a2a^2。
为了让 a2−1a^2 - 1 是一个 44 位数且末两位为 9999,aa 只能是 {40,50,60,70,80,90,100}\{40, 50, 60, 70, 80, 90, 100\}。
由于 a2−1=(a−1)(a+1)a^2 - 1 = (a-1)(a+1),还需要 a−1a-1 和 a+1a+1 都是质数。也就是说,要在 {40,50,…,100}\{40, 50, \dots, 100\} 附近寻找相邻的质数对。
逐一检查,只有 5959 和 6161 可行,因此这样的数恰有 11 个。答案是 B。
The number has the form XX99,XX99, and so the perfect square that is 11 more ends in 00,00, and so it is the square of a number ending in 0.0. Suppose that square is a2.a^2.
In order for a2−1a^2 - 1 to be a 44-digit number ending in 9999, the only possibilities for aa are {40,50,60,70,80,90,100}.\{40, 50, 60, 70, 80, 90, 100\}.
Since a2−1=(a−1)(a+1),a^2 - 1 = (a-1)(a+1), we also need both a−1a-1 and a+1a+1 to be prime. We are then looking for pairs of prime numbers that are right around {40,50,…,100}.\{40, 50, \dots, 100\}.
Going through all the possibilities, the only ones that work are 5959 and 6161, and so there is exactly 11 way to do this. The answer is B.
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