AMC10 2005 A

AMC10 2005 A · Q21

AMC10 2005 A · Q21. It mainly tests Divisibility & factors, GCD & LCM.

For how many positive integers $n$ does $1 + 2 + \cdots + n$ evenly divide $6n$?

有且仅有几个正整数$n$使得$1 + 2 + \cdots + n$能整除$6n$？

(A) 3 3

(B) 5 5

(D) 9 9

(E) 11 11

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) Because $1+2+\cdots+n=\dfrac{n(n+1)}{2}$, $1+2+\cdots+n$ divides the positive integer $6n$ if and only if $\dfrac{6n}{n(n+1)/2}=\dfrac{12}{n+1}$ is an integer. There are five such positive values of $n$, namely, $1,2,3,5,$ and $11$.

（B）因为 $1+2+\cdots+n=\dfrac{n(n+1)}{2}$，当且仅当 $\dfrac{6n}{n(n+1)/2}=\dfrac{12}{n+1}$ 是整数时，$1+2+\cdots+n$ 整除正整数 $6n$。满足条件的正整数 $n$ 有五个，分别是 $1,2,3,5,$ 和 $11$。

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