Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Algebraic Equation.

## Algebraic Equation – AIME 2000

Suppose that x,y and z are three positive numbers that satisfy the equation xyz=1, \(x+\frac{1}{z}=5\) and \(y+\frac{1}{x}=29\) then \(z+\frac{1}{y}\)=\(\frac{m}{n}\) where m and n are relatively prime, find m+n

- is 107
- is 5
- is 840
- cannot be determined from the given information

**Key Concepts**

Algebra

Equations

Integers

## Check the Answer

Answer: is 5.

AIME, 2000, Question 7

Elementary Algebra by Hall and Knight

## Try with Hints

here \(x+\frac{1}{z}=5\) then1=z(5-x)=xyz putting xyz=1 gives 5-x=xy and \(y=(29-\frac{1}{x}\)) together gives 5-x=x\((29-\frac{1}{x}\)) then x=\(\frac{1}{5}\)

then y=29-5=24 and z=\(\frac{1}{5-x}\)=\(\frac{5}{24}\)

\(z+\frac{1}{y}\)=\(\frac{1}{4}\) then 1+4=5.

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## Other useful links

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