AMC10 2024 A
AMC10 2024 A · Q5
AMC10 2024 A · Q5. It mainly tests Primes & prime factorization, GCD & LCM.
What is the least value of $n$ such that $n!$ is a multiple of $2024$?
最小的 $n$ 使得 $n!$ 是 $2024$ 的倍数的值是多少？
(A)
11
11
(B)
21
21
(C)
22
22
(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 $2024=2^3\cdot11\cdot23$ 的质因数分解。由于 $23!$ 包含 $2^3$、$11$ 和 $23$ 的倍数，故 $23!$ 是 $2024$ 的倍数。因此，$n=\boxed{\textbf{(D) } 23}$。
备注
记住当年份的质因数分解对 AMC 8/10/12 考试很有用。
备注
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