Problem 23 · 2025 AMC 8
Hard
Number Theory
primesdifference-of-squaresprime-test
How many four-digit numbers have all three of the following properties?
- The tens digit and ones digit are both 9.
- The number is 1 less than a perfect square.
- The number is the product of exactly two prime numbers.
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Answer: B — Exactly 1.
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Hint 1 of 2
Translate each clue into a structural fact. Ending in 99 means "add 1" gives a number ending in 00 — and a perfect square ending in 00 forces its square root to be a multiple of 10. That alone cuts the candidates to a handful.
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Hint 2 of 2
Now the number is (10k)2 − 1 = (10k − 1)(10k + 1), already factored. "Product of exactly two primes" means both of those factors must themselves be prime — you're hunting twin primes straddling a multiple of 10.
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Approach: turn the clues into structure: square ends in 00, then seek twin primes
- Decode the conditions instead of testing thousands of numbers. "Ends in 99" + 1 = ends in 00, and a square ending in 00 must be (10k)2. So the number is (10k)2 − 1 = (10k − 1)(10k + 1) — a difference of squares, factored for free.
- Four-digit size forces 10k ∈ {40, 50, 60, 70, 80, 90, 100}, just 7 cases.
- "Product of exactly two primes" now means both 10k − 1 and 10k + 1 are prime. Check: 39×41 (39 = 3×13, no), 49×51 (49 = 72, no), 59×61 (both prime ✓), 69×71 (69 = 3×23, no), 79×81 (81 = 34, no), 89×91 (91 = 7×13, no), 99×101 (99 = 9×11, no).
- Only 59 × 61 = 3599 survives, so exactly 1.
- Why this transfers: "one less than a perfect square" should instantly trigger the difference-of-squares factoring a2 − 1 = (a−1)(a+1) — turning a vague property into a concrete factorization you can prime-test.
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