AMC10 2005 B

AMC10 2005 B · Q22

AMC10 2005 B · Q22. It mainly tests Primes & prime factorization, Number theory misc.

For how many positive integers $n$ less than or equal to 24 is $n!$ evenly divisible by $1 + 2 + \cdots + n$?

对于多少个正整数$n\le 24$，$n!$能被$1 + 2 + \cdots + n$整除？

(A) 8 8

(B) 12 12

(D) 17 17

(E) 21 21

Answer

Correct choice: (C)

正确答案：（C）

Solution

(C) Since $1+2+\cdots+n=\dfrac{n(n+1)}{2}$, the condition is equivalent to having an integer value for $\dfrac{n!}{n(n+1)/2}$. This reduces, when $n\ge 1$, to having an integer value for $\dfrac{2(n-1)!}{n+1}$. This fraction is an integer unless $n+1$ is an odd prime. There are 8 odd primes less than or equal to 25, so there are $24-8=16$ numbers less than or equal to 24 that satisfy the condition.

（C）因为 $1+2+\cdots+n=\dfrac{n(n+1)}{2}$，所以该条件等价于下面这个式子的值为整数： $\dfrac{n!}{n(n+1)/2}$。当 $n\ge 1$ 时，这可化简为要求下式为整数： $\dfrac{2(n-1)!}{n+1}$。除非 $n+1$ 是奇素数，否则该分式为整数。不超过 25 的奇素数有 8 个，因此不超过 24 且满足条件的数有 $24-8=16$ 个。

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