2005 AMC 10A Problems/Problem 21

Problem

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

$\textbf{(A) } 3\qquad \textbf{(B) } 5\qquad \textbf{(C) } 7\qquad \textbf{(D) } 9\qquad \textbf{(E) } 11$

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 $\boxed{\textbf{(B) }5}$.

Video Solution

CHECK OUT Video Solution: https://youtu.be/WRv86DHa3zY

See also

2005 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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