Difference between revisions of "2010 AIME I Problems/Problem 3"

Revision as of 19:22, 4 July 2013

Problem

Suppose that $y = \frac34x$ and $x^y = y^x$ . The quantity $x + y$ can be expressed as a rational number $\frac {r}{s}$ , where $r$ and $s$ are relatively prime positive integers. Find $r + s$ .

Solution

We solve in general using $c$ instead of $3/4$ . Substituting $y = cx$ , we have:

$x^{cx} = (cx)^x \Longrightarrow (x^x)^c = c^x\cdot x^x$

Dividing by $x^x$ , we get $(x^x)^{c - 1} = c^x$ .

Taking the $x$ th root, $x^{c - 1} = c$ , or $x = c^{1/(c - 1)}$ .

In the case $c = \frac34$ , $x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}$ , $y = \frac {64}{27}$ , $x + y = \frac {256 + 192}{81} = \frac {448}{81}$ , yielding an answer of $448 + 81 = \boxed{529}$ .

Revision as of 15:52, 12 April 2012 (view source) 1=2 (talk \| contribs) m ← Older edit		Revision as of 19:22, 4 July 2013 (view source) Nathan wailes (talk \| contribs) Newer edit →
Line 17:		Line 17:

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

2010 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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 "2010 AIME I Problems/Problem 3"

Revision as of 19:22, 4 July 2013

Problem

Solution

See Also