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2012 AMC 10A Problem 24

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2012 AMC 10A Problem 24AMC 10 Step by Step

Name: Date: October 28, 2026

Algebraic Manipulation · difficulty 4 of 5 · about 8 min · position P21-25

Let $a$ , $b$ , and $c$ be positive integers with $a\ge$ $b\ge$ $c$ such that $\begin{align*}a^2-b^2-c^2+ab&=2011\text{ and}\\ a^2+3b^2+3c^2-3ab-2ac-2bc&=-1997.\end{align*}$

What is $a$ ?

A) $249$
B) $250$
C) $251$
D) $252$
E) $253$

2012 AMC 10A Problem 24

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