Question

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

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 $N?$

$3$

$4$

$5$

$6$

$9$

求 $1$ 到 $6$ 的最小公倍数。

Find the least common multiple of $1$ through $6$

满足条件的数是这个最小公倍数的倍数；取第二小的正倍数。

The valid numbers are the multiples of that value; take the second one

能被 $1$ 到 $6$ 每个整数整除的数必须是 $\operatorname{lcm}(1,2,3,4,5,6)=60$ 的倍数。

$60$ 的第二小正倍数是 $120$ ，其数字和为 $1+2+0=3$ 。

所以正确答案是 A。

A number divisible by every integer from $1$ to $6$ must be a multiple of $\operatorname{lcm}(1,2,3,4,5,6)=60.$

The second smallest positive multiple of $60$ is $120,$ whose digit sum is $1+2+0=3.$

Thus, the correct answer is A.