AMC12 2011 B

AMC12 2011 B · Q5

AMC12 2011 B · Q5. It mainly tests Primes & prime factorization, GCD & LCM.

Let $N$ be the second smallest positive integer that is divisible by every positive integer less than $7$. What is the sum of the digits of $N$?

设 $N$ 是能被所有小于 $7$ 的正整数整除的第二小的正整数。$N$ 的各位数字之和是多少？

(A) 3 3

(B) 4 4

(D) 6 6

(E) 9 9

Answer

Correct choice: (A)

正确答案：（A）

Solution

$N$ must be divisible by every positive integer less than $7$, or $1, 2, 3, 4, 5,$ and $6$. Each number that is divisible by each of these is a multiple of their least common multiple. $LCM(1,2,3,4,5,6)=60$, so each number divisible by these is a multiple of $60$. The smallest multiple of $60$ is $60$, so the second smallest multiple of $60$ is $2\times60=120$. Therefore, the sum of the digits of $N$ is $1+2+0=\boxed{3\ \textbf{(A)}}$

$N$ 必须能被所有小于 $7$ 的正整数整除，即 $1, 2, 3, 4, 5,$ 和 $6$。所有能被这些数整除的数都是它们最小公倍数的倍数。$LCM(1,2,3,4,5,6)=60$，所以每个满足条件的数都是 $60$ 的倍数。$60$ 的最小倍数是 $60$，因此第二小的倍数是 $2\times60=120$。所以 $N$ 的数位和为 $1+2+0=\boxed{3\ \textbf{(A)}}$。

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