Competition · AMC preparation · step 4 of 4
AMC 8 · 2013 · #10
Grade 6 number-theoryPick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question packs three jobs into one: find the GCF, find the LCM, then divide. Tool #7 (Identify Subproblems) makes those steps explicit — and we get all three from a single prime factorization of each number, so the work shares a foundation. Tool #9 (Easier Related Problem) supports the plan: trying a small case like 12 and 18 shows the clean shortcut lcm/gcd = (a · b)/gcd², which we can use as a check at the end. We prefer this concrete factor-tower approach over algebra (#13) because it keeps every step a small arithmetic move a Grade 6 student can verify.
Factor 180 into primes
Subproblem 1: prime-factorize 180 down to primes, giving 2² · 3² · 5.
Finding the prime building blocks of a whole number is the Grade 4 "factors and multiples" idea.
4.OA.B.4Identify SubproblemsFactor 594 into primes
Subproblem 2: same for 594 — peel off small primes to get 2 · 3³ · 11.
Repeatedly dividing by the smallest prime that fits is the standard factoring loop.
4.OA.B.4Identify SubproblemsRead off the GCF
Subproblem 3: for the GCF take the smaller exponent on each shared prime — 2¹ · 3² gives 18.
GCF = "what the two numbers share" — only common primes, and only as many copies as both have.
6.NS.B.4Identify SubproblemsRead off the LCM
Subproblem 4: for the LCM take the larger exponent and include primes from either list: 2² · 3³ · 5 · 11.
LCM = "the smallest number both divide into" — every prime needs enough copies for both.
The least common multiple of 180 and 594 is built by taking each prime factor to the larger of the two exponents it shows in the two numbers, giving 2² · 3³ · 5 · 11.
▸ Why?
The number 2² · 3³ · 5 · 11 is divisible by both 180 and 594, because for every prime it carries at least as many copies as each of those numbers uses.
▸ Why?
Divisibility can be checked prime by prime like this because every whole number splits into primes in exactly one way, so one number divides another only when all of the first number's prime copies are already present in the second.
▸ Why?
It is the smallest such common multiple, because each prime is handed only the larger of its two exponents — the fewest copies that still cover whichever of the two numbers demands more of that prime.
▸ Why?
Handing any prime fewer copies than that would leave whichever number demands the most of it unable to divide the result, so no smaller number can be a common multiple of both.
▸ Why?
A number divides another only when that other already contains every one of its prime copies, and each number's prime copies are fixed by the single way it factors into primes — so falling even one copy short of what a prime demands breaks the division.
Divide the LCM by the GCF
Subproblem 5: divide the towers by subtracting exponents on matching primes, leaving 2 · 3 · 5 · 11.
Dividing same-base powers means subtracting exponents — easier than multiplying out 2² · 3³ · 5 · 11 = 5940 and then dividing by 18.
6.EE.A.1Identify SubproblemsMultiply the leftover primes
Multiply the four primes: 2 · 3 · 5 · 11 = 330 → (C).
A single multiplication chain finishes the problem with no big numbers in sight.
4.OA.B.4Identify SubproblemsThis AMC 8 problem just needs Grade 6 prime-factor reasoning — factor each number once, then read off the GCF and LCM from the same tower.
- Factor 180 into primes
- Factor 594 into primes
- Read off the GCF
- Read off the LCM
- Divide the LCM by the GCF
- Multiply the leftover primes
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