AMC 10 · 2019 · #7

Grade 6 arithmetic

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

lcmprime-factorizationdivisibility-rulesmultiples identify-subproblems ↑ Prerequisites: lcmprime-factorization

📏 Medium solution 💡 2 insights

Problem

Casper's wallet has exactly enough cents to buy 12 red pieces, 14 green pieces, 15 blue pieces, OR n purple pieces — every candy costs a whole number of cents and purple costs 20 cents. Find the smallest possible n.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

The hidden question is just "smallest whole-cent amount divisible by all four counts" — that's the LCM of 12, 14, 15, 20 (Tool #9 turns the candy story into a divisibility puzzle). Tool #6 / #3 lets us back-check by plugging each answer choice into 20n and asking whether it's divisible by 12, 14, 15. Tool #8 keeps us honest about cents vs. counts.

1STEP 1

Translate the story: the total in cents must be a multiple of 12, 14, 15, and 20, so the smallest is their lcm.

money = lcm(12, 14, 15, 20)

2STEP 2

Prime-factorize 12 = 2²·3, 14 = 2·7, 15 = 3·5, 20 = 2²·5, then keep each prime's highest power: 2² · 3 · 5 · 7.

2² · 3 · 5 · 7 = 4 · 3 · 5 · 7

3STEP 3

Multiply it out: 4·3·5·7 = 420 cents — Casper's smallest possible total.

lcm = 420 cents

4STEP 4

Purple is 20 cents each, so track units — cents ÷ (cents per piece) = pieces: n = 420 ÷ 20 = 21.

n =

\frac{420}{20}

= 21

5STEP 5

Back-check: the smaller candidate 18 needs 360 cents, but 360 isn't divisible by 14 — so 21 is the true smallest, choice (B).

n=21 → (B)

Answer

At 420 cents: red costs

\frac{420}{12}

= 35¢, green

\frac{420}{14}

= 30¢, blue

\frac{420}{15}

= 28¢, purple

\frac{420}{20}

= 21 pieces at 20¢ each. All whole numbers, all under a buck per candy — believable. So n = 21, answer (B).

💡Key takeaway

This AMC 10 problem only needs Grade 6 LCM you already know: Casper's cents must split exactly into 12, 14, 15, AND 20 pieces. The smallest such number is lcm(12,14,15,20) = 420 cents, so n = $\frac{420}{20}$ = 21.

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