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.

A. 0

B. 1

C. 2

D. 3

E. 4

1. 2025 AMC 8 官方解析

Answer (B): Suppose $x$ is a number that has all three properties. By property (I), $x + 1$ is a multiple of 100. By property (II), $x + 1$ is a perfect square, so $x + 1 = n^2$ for some integer $n$. These facts together imply $n$ must be a multiple of 10. Because $x$ has four digits, the possible values of $x = n^2 - 1$ are
$x = 40^2 - 1,$
$x = 50^2 - 1,$
$x = 60^2 - 1,$
$x = 70^2 - 1,$
$x = 80^2 - 1,$
$x = 90^2 - 1, \text{ or}$
$x = 100^2 - 1$
These are the only possible values because $30^2 - 1$ is a 3-digit number and $110^2 - 1$ is a 5-digit number.

For any value of $n$, we can factor $n^2 - 1 = (n - 1)(n + 1)$. Moreover, $n - 1$, $n$, and $n + 1$ are consecutive integers, so one of them is divisible by 3. If any of the above values of $x$ is factored as $x = (n - 1)(n + 1)$, then $n - 1$ and $n + 1$ must be prime by property (III). In particular, neither $n - 1$ nor $n + 1$ is divisible by 3. Thus $n$ must be divisible by 3. This reduces the above list to
$x = 60^2 - 1 = (60 - 1)(60 + 1) = 59 \cdot 61 \text{ and } x = 90^2 - 1 = (90 - 1)(90 + 1) = 89 \cdot 91$
Only $60^2 - 1 = 59 \cdot 61$ is the product of two prime numbers because $91 = 7 \cdot 13$. Thus, there is only one number that has all three properties.