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2024 AMC 12A Problem 4

Problem 4 of 25EasierNumber TheoryCounting & Probability

What is the least value of nn such that n!n! is a multiple of 2024?2024?

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Solution

Factoring, 2024=231123.2024=2^3\cdot11\cdot23. The factorial n!n! contains the prime 2323 only when n23.n\ge23. At n=23,n=23, the product 23!23! already includes 23, 11,23,\ 11, and plenty of factors of 2,2, so 23!23! is a multiple of 2024.2024. Thus, the correct answer is D.

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Concepts: prime factorization · factorial

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.