2025 AMC 8

Complete problem set with solutions and individual problem pages

Problem 23 Hard

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.

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

  • E.

    4

Answer:B

The Condition (II) perfect square must end in "00" because \cdots 99+1=\cdots 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}}.

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