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April 17, 2020

Exponents and Equations | AIME I, 2010 Question 3

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



Number Theory

Check the Answer

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


Final Step

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

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