AMC 10 · 2019 · #5
Grade 6 number-theoryPick an answer.
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.
What the money must satisfy
The money is a multiple of every count.
"Exactly buys k candies" just means the total is a multiple of k.
6.NS.B.4Solve An Easier Related ProblemBreak into primes
Break each number into primes.
LCM grabs the strongest dose of each prime from the bunch.
A least common multiple grabs the strongest dose of each prime from the bunch.
▸ Why?
A common multiple must contain every prime at least as often as any of the numbers does.
▸ Why?
Each number has exactly one prime recipe, so the primes can be compared one at a time.
Take the least common multiple
Include the purple price in the least common multiple.
420 cents = $4.20 — a believable pocket-change amount.
5.NBT.B.5Solve An Easier Related ProblemDivide by the purple price
Divide by that price.
Total cents divided by cost per candy gives candy count.
5.NBT.B.6Analyze The UnitsMatch the choice
The result is 21.
Try the smaller choice first; if it fails the divisibility test, the next answer survives.
4.OA.B.4Eliminate PossibilitiesThis AMC 12 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 = 420/20 = 21.