2012 AMC 10A Problem 22

Attempt Problem 22 of the 2012 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2012 AMC 10A solutions, or check the answer key.

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22.

The sum of the first mm positive odd integers is 212212 more than the sum of the first nn positive even integers. What is the sum of all possible values of n?n?

255255

256256

257257

258258

259259

Answer: A
Concepts:summationdifference of squaresDiophantine Equation
Difficulty rating: 2260
Solution:

The first mm positive odd integers sum to m2m^2, and the first nn positive even integers sum to n(n+1)n(n+1). Thus m2=n(n+1)+212m^2=n(n+1)+212.

As a quadratic in nn, this has discriminant 14(212m2)=4m28471-4(212-m^2)=4m^2-847, which must be an odd square. Let p2=4m2847p^2=4m^2-847. Then (2m+p)(2mp)=847(2m+p)(2m-p)=847.

The positive factor pairs of 847847 are 8471847\cdot1, 1217121\cdot7, and 771177\cdot11. They give p=423,57,33p=423,57,33, respectively.

Because n=1+p2n=\dfrac{-1+p}{2}, the possible values of nn are 211,28,16211,28,16. Their sum is 255255.

Thus, A is the correct answer.

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