2024 AMC 12A Problem 4

Attempt Problem 4 of the 2024 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

4.

What is the least value of nn such that n!n! is a multiple of 2024?2024?
1111
2121
2222
2323
253253
Answer: D
Concepts:prime factorizationfactorial
Small hint:
Factor 2024=23⋅11⋅232024=2^3\cdot11\cdot23
Big hint:
n!n! must contain the prime 23,23, so n≥23n\ge23
Solution:
Factoring, 2024=23⋅11⋅23.2024=2^3\cdot11\cdot23. The factorial n!n! contains the prime 2323 only when n≥23.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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