Problem 7 · AMC 8 Stretch
Core
Fractions, Decimals & Percents
Arithmetic & Operations
solve-a-simpler-problem
Compute \(\dfrac{3}{17} + \dfrac{6}{13}\). First warm up with the easier sum \(\dfrac{1}{2} + \dfrac{1}{3}\), thinking of adding fractions as combining amounts measured in the same unit. Give your answer as a fraction in lowest terms.
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Answer: 141/221
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Hint 1 of 4
If the numbers feel scary, do a smaller version first: how do you add \(\frac{1}{2} + \frac{1}{3}\)?
Still stuck? Show hint 2 →
Hint 2 of 4
You can only add when both pieces are measured in the SAME size. For halves and thirds, what size works? (Sixths!) Rewrite both over \(6\).
Still stuck? Show hint 3 →
Hint 3 of 4
For seventeenths and thirteenths, a size that works for both is \(17 \times 13 = 221\). Rewrite each fraction with denominator \(221\).
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Approach: Solve a simpler analogous problem, then use a common unit (denominator)
- Adding fractions just means combining 'so many of one size piece.' Warm-up: for \(\frac{1}{2} + \frac{1}{3}\) use sixths — \(\frac{1}{2} = \frac{3}{6}\) and \(\frac{1}{3} = \frac{2}{6}\), so the sum is \(\frac{5}{6}\).
- Same idea, bigger numbers. A common size for seventeenths and thirteenths is \(17 \times 13 = 221\). Then \(\frac{3}{17} = \frac{39}{221}\) and \(\frac{6}{13} = \frac{102}{221}\).
- Add the tops: \(\frac{39}{221} + \frac{102}{221} = \frac{141}{221}\).
- Since \(221 = 13 \times 17\) and \(141 = 3 \times 47\) share no common factor, \(\frac{141}{221}\) is already in lowest terms.
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