2011 AMC 12B 第 5 题

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.

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