2005 AMC 10A Problems/Problem 21

Problem

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

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

Solution

By a standard result, $1+2+\dotsb+n = \frac{n(n+1)}{2}$ , so this will evenly divide $6n$ precisely if $\frac{6n}{\left(\frac{n(n+1)}{2}\right)} = \frac{12}{n+1}$ is an integer, or equivalently, $(n+1)$ is a factor of $12$ .

The positive factors of $12$ are $1$ , $2$ , $3$ , $4$ , $6$ , and $12$ (i.e. $6$ factors in total), but as $n$ must be a positive integer, we must have $n+1 \geq 1+1 = 2$ . Therefore, the factor $1$ cannot be used, but all the others can, giving $6-1 = \boxed{\textbf{(B) } 5}$ possible values of $n$ (namely, $n$ can be $2-1 = 1$ , $3-1 = 2$ , $4-1 = 3$ , $6-1 = 5$ , or $12-1 = 11$ ).

Video Solution

https://youtu.be/WRv86DHa3zY

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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2005 AMC 10A Problems/Problem 21

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Problem

Solution

Video Solution

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