AMC 10 · 2019 · #7
Grade 6 arithmeticProblem
Each piece of candy in a store costs a whole number of cents. Casper has exactly enough money to buy either 12 pieces of red candy, 14 pieces of green candy, 15 pieces of blue candy, or n pieces of purple candy. A piece of purple candy costs 20 cents. What is the smallest possible value of n?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $n$.
Givens: Each candy price is a whole number of cents; Money = $12 \times (\text{red price}) = 14 \times (\text{green price}) = 15 \times (\text{blue price}) = n \times 20$; Choices: (A) $18$, (B) $21$, (C) $24$, (D) $25$, (E) $28$
Unknowns: The smallest positive integer $n$
Understand
Restated: 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 $n$.
Givens: Each candy price is a whole number of cents; Money = $12 \times (\text{red price}) = 14 \times (\text{green price}) = 15 \times (\text{blue price}) = n \times 20$; Choices: (A) $18$, (B) $21$, (C) $24$, (D) $25$, (E) $28$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #6 Guess and Check, #3 Eliminate Possibilities, #8 Analyze the Units
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.
Execute — Answer: B
6.NS.B.4 Step 1 - Translate: Casper's total money in cents is a multiple of $12$, $14$, $15$, and (since purple costs $20$) $20$.
- The smallest such money is $\operatorname{lcm}(12, 14, 15, 20)$.
💡 "Exactly buys $k$ candies" just means the total is a multiple of $k$.
6.NS.B.4 Step 2 - Prime factor each count: $12 = 2^2 \cdot 3$, $14 = 2 \cdot 7$, $15 = 3 \cdot 5$, $20 = 2^2 \cdot 5$.
- Take the highest power of each prime that appears.
💡 LCM grabs the strongest dose of each prime from the bunch.
5.NBT.B.5 Step 3 - Multiply: $4 \cdot 3 = 12$, $12 \cdot 5 = 60$, $60 \cdot 7 = 420$.
- So Casper has $420$ cents.
💡 $420$ cents = $\$4.20$ — a believable pocket-change amount.
5.NBT.B.6 Step 4 - Purple costs $20$ cents each, so $n = 420 \div 20$.
- Track the units: $\frac{\text{cents}}{\text{cents per piece}} = \text{pieces}$.
💡 Total cents divided by cost per candy gives candy count.
4.OA.B.4 Step 5 - Spot-check against the choices.
- $n = 21$ matches (B).
- Sanity test the smaller candidate $n = 18$: would need money $= 360$ cents, but $360$ is NOT divisible by $14$ ($360/14$ is not whole).
- So $18$ fails — $21$ really is the smallest.
💡 Try the smaller choice first; if it fails the divisibility test, the next answer survives.
6.NS.B.4 Translate: Casper's total money in cents is a multiple of $12$, $14$, $15$, and 6.NS.B.4 Prime factor each count: $12 = 2^2 \cdot 3$, $14 = 2 \cdot 7$, $15 = 3 \cdot 5$, 5.NBT.B.5 Multiply: $4 \cdot 3 = 12$, $12 \cdot 5 = 60$, $60 \cdot 7 = 420$. So Casper has 5.NBT.B.6 Purple costs $20$ cents each, so $n = 420 \div 20$. Track the units: $\frac{\tex 4.OA.B.4 Spot-check against the choices. $n = 21$ matches (B). Sanity test the smaller ca Review
Reasonableness: At $420$ cents: red costs $420/12 = 35\text{¢}$, green $420/14 = 30\text{¢}$, blue $420/15 = 28\text{¢}$, purple $420/20 = 21$ pieces at $20\text{¢}$ each. All whole numbers, all under a buck per candy — believable. So $n = 21$, answer (B).
Alternative: Tool #6 (Guess & Check across choices): for each $n \in \{18, 21, 24, 25, 28\}$ compute $20n$ and ask whether it is divisible by $12$, $14$, and $15$. $20\cdot 18 = 360$ fails ($360/14$ not whole). $20\cdot 21 = 420$ works ($420/12=35$, $420/14=30$, $420/15=28$). Done.
CCSS standards used (min grade 6)
6.NS.B.4Find greatest common factor and least common multiple of two numbers (Recognising the answer is $\operatorname{lcm}(12, 14, 15, 20)$ and computing it from prime factorisations.)5.NBT.B.5Fluently multiply multi-digit whole numbers (Multiplying $4 \cdot 3 \cdot 5 \cdot 7$ to get the LCM $= 420$ cents.)5.NBT.B.6Find whole-number quotients with up to four-digit dividends and two-digit divisors (Computing $420 \div 20 = 21$ to get the candy count.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Checking divisibility of each candidate money amount by $12, 14, 15$.)
⭐ 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 $\operatorname{lcm}(12,14,15,20) = 420$ cents, so $n = 420/20 = 21$.
⭐ 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 $\operatorname{lcm}(12,14,15,20) = 420$ cents, so $n = 420/20 = 21$.
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