Question

The sum of the first $m$ positive odd integers is $212$ more than the sum of the first $n$ positive even integers. What is the sum of all possible values of $n?$

$255$

$256$

$257$

$258$

$259$

Solution:

The first $m$ positive odd integers sum to $m^2$ , and the first $n$ positive even integers sum to $n(n+1)$ . Thus $m^2=n(n+1)+212$ .

As a quadratic in $n$ , this has discriminant $1-4(212-m^2)=4m^2-847$ , which must be an odd square. Let $p^2=4m^2-847$ . Then $(2m+p)(2m-p)=847$ .

The positive factor pairs of $847$ are $847\cdot1$ , $121\cdot7$ , and $77\cdot11$ . They give $p=423,57,33$ , respectively.

Because $n=\dfrac{-1+p}{2}$ , the possible values of $n$ are $211,28,16$ . Their sum is $255$ .

Thus, A is the correct answer.

2012 AMC 10A Problem 22

Solution:

Problem 22 in Other Years