Problem 3 · AMC 8 Stretch
Stretch
Number Theory
Fractions, Decimals & Percents
flip-and-divideuse-a-prime
A fraction is 'reducible' if its top and bottom share a common factor bigger than \(1\) (so it can be simplified). What is the smallest positive whole number \(n\) that makes \(\dfrac{n-12}{5n+23}\) a reducible fraction (and not equal to \(0\))?
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Answer: n=95
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Hint 1 of 4
A fraction reduces exactly when its flip reduces (same common factor on top and bottom). The flip \(\dfrac{5n+23}{n-12}\) is easier to study.
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Hint 2 of 4
Do the division: how many times does \(n-12\) go into \(5n+23\)? Five times \((n-12)\) is \(5n-60\), and \(5n+23-(5n-60)=83\). So \(\dfrac{5n+23}{n-12}=5+\dfrac{83}{n-12}\).
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Hint 3 of 4
So the only common factor that \(n-12\) can share with the top comes from the number \(83\). Since \(83\) is a prime number, \(n-12\) must be a multiple of \(83\).
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Approach: Flip the fraction, divide out the remainder, use that 83 is prime
- A fraction reduces exactly when its flip reduces, so study \(\dfrac{5n+23}{n-12}\).
- Divide: \(5\) copies of \((n-12)\) make \(5n-60\), and the leftover is \((5n+23)-(5n-60)=83\). So \(\dfrac{5n+23}{n-12}=5+\dfrac{83}{n-12}\).
- The fraction reduces exactly when \(n-12\) and \(83\) share a common factor bigger than \(1\). But \(83\) is prime, so its only factor bigger than \(1\) is \(83\) itself, meaning \(n-12\) must be a multiple of \(83\).
- The smallest positive \(n\) comes from \(n-12=83\), so \(n=95\).
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