Difference between revisions of "2005 AMC 10A Problems/Problem 21"

Revision as of 15:36, 2 August 2006

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

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

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 $5\Longrightarrow \mathrm{(B)}$

@@ Line 9: / Line 9: @@
 Since <math> 1+2+...+n = \frac{n(n+1)}{2} </math> we may substitute the [[RHS]] in the above [[fraction]].
-So the problem asks us for how many [[postive integer]]s <math>n</math> is <math>\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}</math> an integer.
+So the problem asks us for how many [[positive integer]]s <math>n</math> is <math>\frac{6n}{\frac{n(n+1)}{2}}=\frac{12}{n+1}</math> an integer.
 <math>\frac{12}{n+1}</math> is an integer when <math>n+1</math> is a [[factor]] of <math>12</math>.