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Algebraic Equation | AIME I, 2000 Question 7

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

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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