Question

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

What is $a?$

$249$

$250$

$251$

$252$

$253$

Answer: E

Concepts:algebraic manipulation system of equations Diophantine Equation

Difficulty rating: 2090

Solution:

Adding the two equations gives $2a^2 + 2b^2 + 2c^2$ $- 2ab - 2bc - 2ca = 14,$ that is, $\begin{aligned} &(a-b)^2 + (b-c)^2 \\ &\quad {}+ (c-a)^2 = 14. \end{aligned}$

The only way to write $14$ as a sum of three squares is $9 + 4 + 1.$ Since $a \ge b \ge c,$ we get $a - c = 3,$ with either $(a-b, b-c) = (2,1)$ or $(1,2).$

Substituting $(a, b, c) = (c+3, c+1, c)$ into the first equation gives $3(2c+3) + 2(c+1) = 2011,$ so $c = 250$ and $(a, b, c) = (253, 251, 250).$ The other case has no integer solution.

Thus, the correct answer is E.

2012 AMC 12A Problem 21

Answer: E

Difficulty rating: 2090

Solution:

Problem 21 in Other Years