Problem 1 · AMC 8 Stretch
Core
Number Theory
Fractions, Decimals & Percents
bound-a-variablefind-factor-pairs
Find every pair of different positive whole numbers \(a\) and \(b\) (with \(a>b\)) so that \(\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{9}\).
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Answer: (a,b)=(90,10) and (36,12)
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Hint 1 of 4
Think about sizes first. If \(b\) were \(9\) or smaller, then \(\tfrac1b\) would already be \(\tfrac19\) or bigger, leaving nothing for \(\tfrac1a\). So both \(a\) and \(b\) must be bigger than \(9\).
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Hint 2 of 4
Clear the fractions. Multiply everything by \(9ab\) to get \(9b+9a=ab\). Rearrange to \(ab-9a-9b=0\).
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Hint 3 of 4
Here is the classic trick: add \(81\) to both sides so the left side factors. You get \(ab-9a-9b+81=81\), which is \((a-9)(b-9)=81\).
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Approach: Add 81 and factor (Simon's Favorite Factoring), then search factor pairs
- Both numbers must be bigger than 9: if \(b\le 9\) then \(\tfrac1b\ge\tfrac19\), which already uses up all of \(\tfrac19\), leaving nothing for \(\tfrac1a\). So \(a>b>9\).
- Clear fractions by multiplying \(\tfrac1a+\tfrac1b=\tfrac19\) by \(9ab\): \(9b+9a=ab\), so \(ab-9a-9b=0\).
- Add \(81\) to both sides so the left factors: \(ab-9a-9b+81=81\), i.e. \((a-9)(b-9)=81\).
- List factor pairs of \(81\) with the bigger factor going to \(a\): \(a-9=81, b-9=1\Rightarrow a=90, b=10\); and \(a-9=27, b-9=3\Rightarrow a=36, b=12\). (The split \(9\times9\) gives \(a=b=18\), but we need \(a>b\), so skip it.)
- Check: \(\tfrac{1}{90}+\tfrac{1}{10}=\tfrac{1}{90}+\tfrac{9}{90}=\tfrac{10}{90}=\tfrac19\), and \(\tfrac{1}{36}+\tfrac{1}{12}=\tfrac{1}{36}+\tfrac{3}{36}=\tfrac{4}{36}=\tfrac19\). So the pairs are \((a,b)=(90,10)\) and \((36,12)\).
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