Difference between revisions of "2010 AIME I Problems/Problem 3"
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In the case <math>c = \frac34</math>, <math>x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}</math>, <math>y = \frac {64}{27}</math>, <math>x + y = \frac {256 + 192}{81} = \frac {448}{81}</math>, yielding an answer of <math>448 + 81 = \boxed{529}</math>. | In the case <math>c = \frac34</math>, <math>x = \Bigg(\frac34\Bigg)^{ - 4} = \Bigg(\frac43\Bigg)^4 = \frac {256}{81}</math>, <math>y = \frac {64}{27}</math>, <math>x + y = \frac {256 + 192}{81} = \frac {448}{81}</math>, yielding an answer of <math>448 + 81 = \boxed{529}</math>. | ||
| − | == See | + | == See Also == |
{{AIME box|year=2010|num-b=2|num-a=4|n=I}} | {{AIME box|year=2010|num-b=2|num-a=4|n=I}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 15:52, 12 April 2012
Problem
Suppose that 
and 
. The quantity 
can be expressed as a rational number 
, where 
and 
are relatively prime positive integers. Find 
.
Solution
We solve in general using 
instead of 
. Substituting 
, we have:
![\[x^{cx} = (cx)^x \Longrightarrow (x^x)^c = c^x\cdot x^x\]](http://latex.artofproblemsolving.com/6/d/8/6d806bc47a9f272be605a24af932316a92ed65fd.png)
Dividing by 
, we get 
.
Taking the 
th root, 
, or 
.
In the case 
, 
, 
, 
, yielding an answer of 
.
See Also
| 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 | ||
