Problem 24 · 2012 Math Kangaroo
Stretch
Number Theory
divisibilityfactorization
In a game with fractions I am allowed to carry out two operations, namely either increase the numerator by 8 or increase the denominator by 7 without simplifying during the game. Starting with the fraction 78 after n such operations I again obtain a fraction with equal value. What is the smallest value of n?
Show answer
Answer: D — 113
Show hints
Hint 1 of 2
Say you add 8 to the numerator \(p\) times and 7 to the denominator \(q\) times.
Still stuck? Show hint 2 →
Hint 2 of 2
Set the new fraction equal to \(\tfrac{7}{8}\) and clear fractions; you'll get a clean ratio between \(p\) and \(q\).
Show solution
Approach: set the new fraction equal and solve for the move counts
- After \(p\) numerator-moves and \(q\) denominator-moves the fraction is \(\frac{7+8p}{8+7q}\).
- Setting it equal to \(\tfrac{7}{8}\) gives \(8(7+8p) = 7(8+7q)\), which simplifies to \(64p = 49q\).
- Since \(\gcd(64,49)=1\), the smallest whole solution is \(p = 49\), \(q = 64\), so \(n = p+q = 113\), choice D.
Mark:
· log in to save