2005 AMC 10A Problems/Problem 21

Revision as of 21:08, 30 October 2020 by Thestudyofeverything (talk | contribs) (Solution)

Problem

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

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

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 $0$ isn't a positive integer, so only $1$, $2$, $3$, $5$, and $11$ are possible values of $n$. Therefore the number of possible values of $n$ is $5\Longrightarrow \boxed{\mathrm{(B)}}$.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png