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2000 AIME I Problems/Problem 6

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Problem

For how many ordered pairs (x,y) of integers is it true that 0 < x < y < 10^{6} and that the arithmetic mean of x and y is exactly 2 more than the geometric mean of x and y?

Solution

\begin{eqnarray*}\frac{x+y}{2} &=& \sqrt{xy} + 2\\x+y-4 &=& 2\sqrt{xy}\\y - 2\sqrt{xy} + x &=& 4\\\sq...

For simplicity, we can count how many valid pairs of (\sqrt{x},\sqrt{y}) that satisfy our equation.

The maximum that \sqrt{y} can be is 10^3 - 1 = 999 because \sqrt{y} must be an integer (this is because \sqrt{y} - \sqrt{x} = 2, an integer). Then \sqrt{x} = 997, and we continue this downward until \sqrt{y} = 3, in which case \sqrt{x} = 1. The number of pairs of (\sqrt{x},\sqrt{y}), and so (x,y) is then \boxed{997}.

See also

2000 AIME I (ProblemsResources)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
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