2025 AMC 10A Problem 12

Below is the professionally curated solution for Problem 12 of the 2025 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10A solutions, or check the answer key.

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Concepts:caseworkmultiplication principle

Difficulty rating: 1560

12.

Carlos uses a 44-digit passcode to unlock his computer. In his passcode, exactly one digit is even, exactly one (possibly different) digit is prime, and no digit is 0.0. How many 44-digit passcodes satisfy these conditions?

176176

192192

432432

464464

608608

Solution:

No digit is 0,0, so digits run from 11 to 9.9. Put the single even digit in the first slot for now and multiply by 44 at the end to place it. Split on that even digit. If it's the prime 2,2, then the three odd digits all have to be non-prime, so each is 11 or 9,9, giving 23=82^3 = 8 ways. Otherwise the even digit is 4,6,4, 6, or 88 (33 choices), and exactly one of the odd digits is prime, worth 3,5,3, 5, or 77 (33 choices) in one of the 33 odd positions, while the other two odds come from {1,9}\{1, 9\} (222^2 ways). That's 3334=108.3 \cdot 3 \cdot 3 \cdot 4 = 108. Altogether, 4(8+108)=464.4(8 + 108) = 464. Therefore, the answer is D.

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