AMC 10 · 2020 · #9

Grade 6 arithmetic

Archetype Multiplicative Ratio with Integrality Constraints

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

📏 Short solution 💡 2 insights

Problem

A single bench section at a school event can hold either $7$ adults or $11$ children. When $N$ bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of $N?$

$\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: $1$ bench $=$ $7$ adult-seats $=$ $11$ child-seats; $N$ benches in a row, completely full; Total adults seated $=$ total children seated $=$ some positive whole number, call it $k$; Choices: (A) $9$, (B) $18$, (C) $27$, (D) $36$, (E) $77$

Unknowns: The smallest positive integer $N$

Understand

Plan

Primary tool: #6 Guess and Check

Secondary: #3 Eliminate Possibilities, #5 Look for a Pattern

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 $\text{LCM}(7, 11) = 77$, leading to $N = 77/7 + 77/11 = 11 + 7 = 18$.

Execute — Answer: B

#5 Look for a Pattern 6.RP.A.3 Step 1

Translate the seating into bench-fractions.
If $k$ adults sit, they fill $\dfrac{k}{7}$ benches (one bench per $7$ adults).
If $k$ children sit, they fill $\dfrac{k}{11}$ benches.
Together they fill exactly $N$ benches.

$$\dfrac{k}{7} + \dfrac{k}{11} = N$$

💡 $7$ adults $=$ $1$ bench, so $k$ adults $=$ $k/7$ benches. Same logic for children.

#5 Look for a Pattern 6.NS.B.4 Step 2

For $N$ to be a positive whole number, $k$ must clear both denominators — i.e.
$k$ must be a common multiple of $7$ and $11$.
The smallest such $k$ is $\text{LCM}(7, 11) = 77$ (since $7$ and $11$ are both prime, with no common factor).

$$k = \text{LCM}(7, 11) = 77$$

💡 Two prime numbers $\Rightarrow$ LCM is just their product.

#6 Guess and Check 3.OA.C.7 Step 3

Plug $k = 77$ back into the bench equation.

$$N = \dfrac{77}{7} + \dfrac{77}{11} = 11 + 7 = 18$$

💡 $77$ adults need $11$ benches, $77$ children need $7$ benches — together $18$ benches, exactly full.

#3 Eliminate Possibilities 6.EE.B.7 Step 4

Verify smaller candidates can't work.
$N = 9$ would need $k$ adults filling some benches and $k$ children filling the rest, with $\dfrac{k}{7} + \dfrac{k}{11} = 9 \Rightarrow k = \dfrac{9 \cdot 77}{18} = 38.5$ — not a whole person.
Same fractional failure happens for any $N$ that isn't a multiple of $18$.

$N = 9 \Rightarrow k = 38.5$ — fractional people, impossible

💡 Any $N$ less than $18$ forces non-whole-number people — physically impossible.

#3 Eliminate Possibilities 4.NBT.A.2 Step 5

$N = 18$ matches choice (B).

$$N = 18\;\Rightarrow\;\textbf{(B)}$$

💡 Read the matching answer choice.

[1] #5 6.RP.A.3 Translate the seating into bench-fractions. If $k$ adults sit, they fill $\dfrac

[2] #5 6.NS.B.4 For $N$ to be a positive whole number, $k$ must clear both denominators — i.e. $

[3] #6 3.OA.C.7 Plug $k = 77$ back into the bench equation.

[4] #3 6.EE.B.7 Verify smaller candidates can't work. $N = 9$ would need $k$ adults filling some

[5] #3 4.NBT.A.2 $N = 18$ matches choice (B).

Review

Reasonableness: 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 $N = c + a$.

Alternative: Tool #6 (pure Guess and Check) on each choice. Test $N = 18$: $\dfrac{k}{7} + \dfrac{k}{11} = 18 \Rightarrow k \cdot \dfrac{18}{77} = 18 \Rightarrow k = 77$ ✓. Test $N = 9$: $k = 77/2 = 38.5$ ✗. Test $N = 27$: $k = 77 \cdot 27/18 = 115.5$ ✗. Test $N = 36$: $k = 154$ ✓ but $N = 18$ is smaller. So minimum is (B).

CCSS standards used (min grade 6)

6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Converting '$k$ adults at $7$/bench' into a bench-count $\dfrac{k}{7}$, and similarly for children.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Computing $\text{LCM}(7, 11) = 77$ as the smallest $k$ that makes both bench fractions whole.)
3.OA.C.7 Fluently multiply and divide within 100 (Evaluating $77/7 = 11$ and $77/11 = 7$ to get $N = 18$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving $\dfrac{k}{7} + \dfrac{k}{11} = N$ for $k$ to check smaller candidates.)
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Matching the computed value $18$ to choice (B).)

⭐ 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 $k = \text{LCM}(7,11) = 77$, and the total benches are $77/7 + 77/11 = 11 + 7 = 18$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Review

CCSS standards used (min grade 6)

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