2010 AIME I Problems/Problem 5

Problem

Positive integers $a$ , $b$ , $c$ , and $d$ satisfy $a > b > c > d$ , $a + b + c + d = 2010$ , and $a^2 - b^2 + c^2 - d^2 = 2010$ . Find the number of possible values of $a$ .

Solution

Using the difference of squares, $2010 = (a^2 - b^2) + (c^2 - d^2) = (a + b)(a - b) + (c + d)(c - d)\geq a + b + c + d = 2010$ , where equality must hold so $b = a - 1$ and $d = c - 1$ . Then we see $a = 1004$ is maximal and $a = 504$ is minimal, so the answer is $\boxed{501}$ .

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2010 AIME I Problems/Problem 5

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