AMC12 2024 A

AMC12 2024 A · Q4

AMC12 2024 A · Q4. It mainly tests Primes & prime factorization, Number theory misc.

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

求最小的 $n$，使得 $n!$ 是 $2024$ 的倍数。

(A) 11 11

(B) 21 21

(D) 23 23

(E) 253 253

Answer

Correct choice: (D)

正确答案：（D）

Solution

Note that $2024=2^3\cdot11\cdot23$ in the prime factorization. Since $23!$ is a multiple of $2^3, 11,$ and $23,$ we conclude that $23!$ is a multiple of $2024.$ Therefore, we have $n=\boxed{\textbf{(D) } 23}.$ Remark Memorizing the prime factorization of the current year is useful for the AMC 8/10/12 Exams. Remark

注意 $2024=2^3\cdot11\cdot23$ 是其质因数分解。因为 $23!$ 是 $2^3$、$11$ 和 $23$ 的倍数，所以 $23!$ 是 $2024$ 的倍数。因此，$n=\boxed{\textbf{(D) } 23}$。 Remark Memorizing the prime factorization of the current year is useful for the AMC 8/10/12 Exams。 Remark

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