AMC10 2005 10A, Problem 21

For how many positive integers $n$ does $1+2+...+n$ evenly divide from $6n$?

Solution: If $1+2+...+n$ evenly divides $6n$, then $\frac{6n}{1+2+...+n}$ is an integer. Since $1+2+...+n = \frac{n(n+1)}{2}$ we may substitute the RHS in the above fraction. So the problem asks us for how many positive integers $n$ is $\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}$ an integer, or equivalently when $k(n+1) = 12$ for a positive integer $k$. $\frac{12}{n+1}$ is an integer when $n+1$ is a factor of $12$. The factors of $12$ are $1, 2, 3, 4, 6,$ and $12$, so the possible values of $n$ are $0, 1, 2, 3, 5,$ and $11$. But since $0$ isn't a positive integer, only $1, 2, 3, 5,$ and $11$ are the possible values of $n$. Therefore the number of possible values of $n$ is $\fbox{5}$.

Answer: 5