2011 AMC 12B Problem 5

Attempt Problem 5 of the 2011 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2011 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

Find the least common multiple of

1

through

6

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

A number divisible by every integer from

1

6

must be a multiple of

\operatorname{lcm}(1,2,3,4,5,6)=60.

The second smallest positive multiple of

60

120,

whose digit sum is

1+2+0=3.

Thus, the correct answer is A.

Problem 5 in other years