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