Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Exponents and Equations.

## Exponents and Equations – AIME 2010

Suppose that y=\(\frac{3x}{4}\) 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.

.

- is 107
- is 529
- is 840
- cannot be determined from the given information

**Key Concepts**

Algebra

Equations

Number Theory

## Check the Answer

But try the problem first…

Answer: is 529.

AIME, 2010, Question 3.

Elementary Number Theory by Sierpinsky

## Try with Hints

First hint

y=\(\frac{3x}{4}\) into \(x^{y}=y^{x}\) and \(x^{\frac{3x}{4}}\)=\((\frac{3x}{4})^{x}\) implies \(x^{\frac{3x}{4}}\)=\((\frac{3}{4})^{x}x^{x}\) implies \(x^{-x}{4}\)=\((\frac{3}{4})^{x}\) implies \(x^{\frac{-1}{4}}=\frac{3}{4}\) implies \(x=\frac{256}{81}\).

Second Hint

y=\(\frac{3x}{4}=\frac{192}{81}\).

Final Step

x+y=\(\frac{448}{81}\) then 448+81=529.

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

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