Difference between revisions of "2000 AIME I Problems/Problem 6"

Revision as of 18:48, 4 July 2013

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

$\begin{eqnarray*} \frac{x+y}{2} &=& \sqrt{xy} + 2\\ x+y-4 &=& 2\sqrt{xy}\\ y - 2\sqrt{xy} + x &=& 4\\ \sqrt{y} - \sqrt{x} &=& \pm 2\end{eqnarray*}$

Because $y > x$ , we only consider $+2$ .

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 $\boxed{997}$ .

Revision as of 09:32, 20 June 2011 (view source) Talkinaway (talk \| contribs) (→‎Solution) ← Older edit		Revision as of 18:48, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
Line 22:		Line 22:

	[[Category:Intermediate Algebra Problems]]		[[Category:Intermediate Algebra Problems]]
		+	{{MAA Notice}}

2000 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

Page

Toolbox

Search

Difference between revisions of "2000 AIME I Problems/Problem 6"

Revision as of 18:48, 4 July 2013

Problem

Solution

See also