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

Revision as of 08:45, 11 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 11: / Line 11: @@
 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>.
+<math>\frac{12}{n+1}</math> is an integer when <math>n+1</math> is a [[divisor |factor]] of <math>12</math>.
 The factors of <math>12</math> are <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>6</math>, and <math>12</math>.