AMC 8 2025 Problem 23
AMC8 2025 年第 23 题
整除与因数★★★★☆
How many four-digit numbers have all three of the following properties?
(I) The tens 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.
请问有多少个四位数同时满足以下三个条件？
（一）十位上的数字和个位上的数字都是9。
（二）该数比一个完全平方数小1。
（三）该数正好是两个质数的乘积。
- A.0
- B.1
- C.2
- D.3
- E.4
正确答案 · Correct Answer
B
解析 · Solution
解法1
常规解法·三个条件一条一条筛
设这个四位数为 。
条件（一）： 末两位是 99，所以 N+1 末两位是 00。
条件（二）：N+1 是完全平方数。末尾带 00 的平方数一定是「整十数的平方」（平方数末尾的 0 成对出现）。 是四位数，N+1 从 1100 到 10000，其中整十数的平方有 ，即 ，——注意 N=9999 时 也过前两关，别漏掉它。
条件（三）：（平方差公式）。 要恰好是两个质数之积，而这个乘积已经摆在眼前——所以必须两个因子都是质数（只要有一个是合数，还能继续拆，质因数就不止两个）。逐对检查：
39×41：39=3×13 ✗；49×51： ✗；59×61：都是质数 ✓；69×71：69=3×23 ✗；79×81： ✗；89×91：91=7×13 ✗；99×101：99=9×11 ✗。
只有 N=59×61=3599 全过。共 1 个，选 B。
易错提醒：91 是本题头号陷阱——它长得像质数，其实 91=7×13；把 89×91 也数进去就会错答 2。值得背下几个「长得像质数的合数」：91=7×13、51=3×17、57=3×19、87=3×29。
巧解·找整十数两侧的孪生质数
一步推出 N=(10m-1)(10m+1)：一对相差 2 的数之积。「恰两个质数」就是要求这一对是孪生质数（相差 2 的质数对）。问题浓缩成： 这些整十数里，谁的左右两边同时是质数？只有 60 的两侧 59、61。所以只有 N=59×61=3599 这 1 个。
前置（平方差公式）：；本题用它把 拆成 (10m+1)(10m-1)。
Solution 1
*(Elementary — filter by the three conditions, one at a time.)*
Let the four-digit number be .
Condition (I): ends in 99, so N+1 ends in 00.
Condition (II): N+1 is a perfect square. A perfect square ending in 00 must be the square of a multiple of 10 (trailing zeros of a square come in pairs). Since is four-digit, N+1 runs from 1100 to 10000, so for — note that N=9999 gives , which also passes the first two conditions; don't skip it.
Condition (III): by difference of squares. For to be a product of exactly two primes, both factors must be prime (if either were composite it would split further, giving more than two prime factors). Check each pair:
39×41: 39=3×13 ✗; 49×51: ✗; 59×61: both prime ✓; 69×71: 69=3×23 ✗; 79×81: ✗; 89×91: 91=7×13 ✗; 99×101: 99=9×11 ✗.
Only N=59×61=3599 survives all three. Exactly 1, choice B.
Watch out: 91 is the top trap here — it looks prime, but 91=7×13; counting 89×91 too leads to the wrong answer 2. Worth memorizing the usual "fake primes": 91=7×13, 51=3×17, 57=3×19, 87=3×29.
*(Slick — twin primes flanking a multiple of ten.)*
From we get N=(10m-1)(10m+1) in one step: a product of two numbers differing by 2. "Exactly two primes" forces this pair to be twin primes. The question shrinks to: which of the multiples of ten has primes on both sides? Only 60, flanked by 59 and 61. Hence just 1 such number, N=59×61=3599.
Prerequisite (difference of squares): ; here it splits into (10m+1)(10m-1).
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