2005 AMC 10A Problems/Problem 21

Revision as of 02:15, 31 July 2008 by Lulze (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 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 $n$ $\Rightarrow \boxed{B}$

See Also

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