AMC 10 · 2020 · #9

Grade 6 arithmetic

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Chained Ratio Linking

lcmratio-proportionmultiples guess-and-checkpattern-recognition ↑ Prerequisites: lcmratio-proportion

📏 Short solution 💡 2 insights

Problem

One bench seats either 7 adults or 11 children (alone). Lining up N benches end to end, we want an equal number of adults and children to fill all the bench space exactly — no empty room. What is the smallest positive whole number N that lets this work?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Guess and Check

Five candidates in the answer list. Tool #6 (Guess and Check) plugs each N into the bench-balance check directly. Tool #3 (Eliminate) discards any N that forces a fractional number of people. Tool #5 (Pattern) gives the underlying rule: k must be a common multiple of both 7 (adults' bench ratio) and 11 (children's bench ratio), and the smallest such k is LCM(7, 11) = 77, leading to N = $\frac{77}{7} + \frac{77}{11}$ = 11 + 7 = 18.

1STEP 1

Turn seating into bench-fractions: k adults fill $\frac{k}{7}$ benches, k children fill $\frac{k}{11}$ benches, so $\frac{k}{7} + \frac{k}{11} = N$ .

\frac{k}{7} + \frac{k}{11} = N

2STEP 2

N is whole only if k clears both 7 and 11, so k must be their common multiple — the smallest is LCM(7, 11) = 77.

k = LCM(7, 11) = 77

3STEP 3

Put k = 77 into the bench equation: N = $\frac{77}{7} + \frac{77}{11}$ = 11 + 7 = 18.

N =

\frac{77}{7} + \frac{77}{11}

= 11 + 7 = 18

4STEP 4

Check a smaller N: N = 9 forces k = $\frac{9·77}{18}$ = 38.5 — fractional people, so any non-multiple of 18 fails.

N = 9 → k = 38.5 — fractional people, impossible

5STEP 5

N = 18 matches choice (B).

N = 18 → (B)

Answer

Verify the arrangement physically. 18 benches contain 77 adults (filling 11 benches) and 77 children (filling 7 benches). Adult count = child count = 77 ✓. Total bench space used = 11 + 7 = 18 benches ✓. No bench is partially used. All conditions met. Also note 7 + 11 = 18 — a neat shortcut: with a adults per bench and c children per bench, the answer is a + c whenever gcd(a, c) = 1, because k = ac gives N = c + a.

💡Key takeaway

This AMC 10 problem only needs Grade 6 LCM you already know — the same head count k must be a whole number of 7-adult benches AND 11-child benches, so k = LCM(7,11) = 77, and the total benches are $\frac{77}{7} + \frac{77}{11}$ = 11 + 7 = 18.

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