2025 AMC 8 Problems/Problem 23

Problem

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.

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$

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 Difference of Squares[1], $n^2-1 = (n+1)(n-1)$ . So:

$40^2-1 = (39)(41)$
$50^2-1 = (49)(51)$
$60^2-1 = (59)(61)$
$70^2-1 = (69)(71)$
$80^2-1 = (79)(81)$
$90^2-1 = (89)(91)$

On this list, the only number that is the product of $2$ prime numbers is $60^2-1 = (59)(61)$ , so the answer is $\boxed{\text{(B)\ 1}}$ .

~Soupboy0

Video Solution 1 by SpreadTheMathLove

https://www.youtube.com/watch?v=jTTcscvcQmI

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/VP7g-s8akMY?si=wexxSYnEz2IcjeIb&t=3539 ~hsnacademy

Video Solution by Thinking Feet

https://youtu.be/PKMpTS6b988

Video Solution by Dr. David

https://youtu.be/jn-qIwv57nQ

2025 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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