AMC8 2025

AMC8 2025 · Q23

AMC8 2025 · Q23. It mainly tests Primes & prime factorization, Perfect squares & cubes.

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.

有多少个四位数具有以下三个性质？ (I) 十位和个位都是9。 (II) 该数比一个完全平方少1。 (III) 该数恰好是两个素数的乘积。

(A) \ 0 \ 0

(B) \ 1 \ 1

(D) \ 3 \ 3

(E) \ 4 \ 4

Answer

Correct choice: (B)

正确答案：（B）

Solution

The "Condition (II)" perfect square must end in "$00$" because $...99+1=...00$ Condition (I). Four-digit perfect squares ending in "$00$" are ${40, 50, 60, 70, 80, 90}$. Condition (II) also says the number is in the form $n^2-1$. By the Difference of Squares, $n^2-1 = (n+1)(n-1)$. Hence: - $40^2-1 = (39)(41)$ - $50^2-1 = (49)(51)$ - $60^2-1 = (59)(61)$ - $70^2-1 = (69)(71)$ - $80^2-1 = (79)(81)$ - $90^2-1 = (89)(91)$ On this list, the only number that is the product of $2$ prime numbers (condition $3$) is $60^2-1 = (59)(61)$, so the answer is $\boxed{\text{(B) 1}}$.

“条件(II)”的完全平方必须以“$00$”结尾，因为$...99+1=...00$ 条件(I)。以“$00$”结尾的四位完全平方是${40, 50, 60, 70, 80, 90}$。条件(II)还说明该数的形式为$n^2-1$。由平方差公式，$n^2-1 = (n+1)(n-1)$。因此： - $40^2-1 = (39)(41)$ - $50^2-1 = (49)(51)$ - $60^2-1 = (59)(61)$ - $70^2-1 = (69)(71)$ - $80^2-1 = (79)(81)$ - $90^2-1 = (89)(91)$ 在这个列表中，唯一是2个素数乘积（条件3）的数是$60^2-1 = (59)(61)$，因此答案是$\boxed{\text{(B) 1}}$。

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