2011 AMC 12B Problems/Problem 5

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Problem

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 $N$?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$

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 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 clearly $60$, so the second smallest multiple of $60$ is $2\times60=120$. Therefore, the sum of the digits of $N$ is $1+2+0=\boxed{3\ \textbf{(A)}}$

See also

2011 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AMC 12 Problems and Solutions