2005 AMC 10A Problems/Problem 21

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 1

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.

$\frac{12}{n+1}$ 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 $5\Longrightarrow \mathrm{(B)}$

Solution 2

The sum of the first $n$ positive integers is $\frac{n(n+1)}{2}$ . If this is to divide $6n$ , then there exists a positive integer $k$ such that:

$k \cdot \frac{n(n+1)}{2} = 6n$

$k(n+1) = 12$

Therefore, $k$ and $n+1$ are divisors of $12$ . There are $6$ divisors of $12$ , which are $1, 2, 3, 4, 6, 12$ . The divisors which multiply to $12$ can be assigned to $k$ and $n+1$ in either order. However, when $1$ is assigned to $n+1$ , then $n=0$ , which is not possible, because $n$ must be positive. Therefore, we have $6-1=5$ values of $n$ $\Rightarrow \boxed{B}$

Page

Toolbox

Search

2005 AMC 10A Problems/Problem 21

Contents

Problem

Solution 1

Solution 2

See Also