AMC 8 · 2013 · #10
Easy mode Grade 6Problem
Look at the two numbers 180 and 594.
The greatest common factor (GCF) is the biggest number that divides into both of them evenly.
The least common multiple (LCM) is the smallest number that both of them divide into evenly.
Now divide the LCM by the GCF.
What is the value of GCF of 180 and 594LCM of 180 and 594?
Pick an answer.
AMC 8 2013 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: Find the ratio $\dfrac{\operatorname{lcm}(180, 594)}{\gcd(180, 594)}$.
Givens: Two whole numbers: $180$ and $594$; The ratio we want is LCM divided by GCF (also called GCD); Answer choices: (A) $110$, (B) $165$, (C) $330$, (D) $625$, (E) $660$
Unknowns: The whole-number ratio $\operatorname{lcm}(180, 594) \div \gcd(180, 594)$
Understand
Restated: Find the ratio $\dfrac{\operatorname{lcm}(180, 594)}{\gcd(180, 594)}$.
Givens: Two whole numbers: $180$ and $594$; The ratio we want is LCM divided by GCF (also called GCD); Answer choices: (A) $110$, (B) $165$, (C) $330$, (D) $625$, (E) $660$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem
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 $\dfrac{\operatorname{lcm}}{\gcd} = \dfrac{a \cdot b}{\gcd^2}$, 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.
Execute — Answer: C
4.OA.B.4 Step 1 - Subproblem 1: prime factorize $180$.
- Break it down repeatedly into smaller factors until only primes remain.
💡 Finding the prime building blocks of a whole number is the Grade 4 "factors and multiples" idea.
4.OA.B.4 Step 2 - Subproblem 2: prime factorize $594$.
- Same procedure — peel off small primes until none are left.
💡 Repeatedly dividing by the smallest prime that fits is the standard factoring loop.
6.NS.B.4 Step 3 - Subproblem 3: read off the GCF.
- For each prime that appears in BOTH lists, keep the smaller exponent.
- $2$: $\min(2,1) = 1$.
- $3$: $\min(2,3) = 2$.
- The primes $5$ and $11$ appear in only one list, so they are skipped.
💡 GCF = "what the two numbers share" — only common primes, and only as many copies as both have.
6.NS.B.4 Step 4 - Subproblem 4: read off the LCM.
- For each prime that appears in EITHER list, keep the larger exponent.
- $2$: $\max(2,1) = 2$.
- $3$: $\max(2,3) = 3$.
- $5$: only in $180$, take $5^1$.
- $11$: only in $594$, take $11^1$.
💡 LCM = "the smallest number both divide into" — every prime needs enough copies for both.
6.EE.A.1 Step 5 - Subproblem 5: form the ratio.
- Don't multiply the LCM out first — divide the prime towers directly, subtracting exponents on matching primes.
💡 Dividing same-base powers means subtracting exponents — easier than multiplying out $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$ and then dividing by $18$.
4.OA.B.4 Step 6 Multiply the four small primes to finish.
💡 A single multiplication chain finishes the problem with no big numbers in sight.
4.OA.B.4 Subproblem 1: prime factorize $180$. Break it down repeatedly into smaller facto 4.OA.B.4 Subproblem 2: prime factorize $594$. Same procedure — peel off small primes unti 6.NS.B.4 Subproblem 3: read off the GCF. For each prime that appears in BOTH lists, keep 6.NS.B.4 Subproblem 4: read off the LCM. For each prime that appears in EITHER list, keep 6.EE.A.1 Subproblem 5: form the ratio. Don't multiply the LCM out first — divide the prim 4.OA.B.4 Multiply the four small primes to finish. Review
Reasonableness: Use the identity $\operatorname{lcm}(a, b) \cdot \gcd(a, b) = a \cdot b$ as an independent check: $\dfrac{\operatorname{lcm}}{\gcd} = \dfrac{a \cdot b}{\gcd^2} = \dfrac{180 \cdot 594}{18^2} = \dfrac{180}{18} \cdot \dfrac{594}{18} = 10 \cdot 33 = 330$. Two different computations agree on $330$, so answer (C) is solid.
Alternative: Tool #9 (Solve an Easier Related Problem) — try the same question on $12$ and $18$ first. $12 = 2^2 \cdot 3$, $18 = 2 \cdot 3^2$, so $\gcd = 2 \cdot 3 = 6$ and $\operatorname{lcm} = 2^2 \cdot 3^2 = 36$. The ratio is $36 / 6 = 6 = 2 \cdot 3$, which is exactly the product of the primes that appear with *different* exponents. The same rule on $180$ and $594$ keeps the primes $2, 3, 5, 11$ and gives $330$. Same answer (C).
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs for a whole number; recognize multiples and factors (Prime-factorizing $180 = 2^2 \cdot 3^2 \cdot 5$ and $594 = 2 \cdot 3^3 \cdot 11$, and finishing with the multiplication $2 \cdot 3 \cdot 5 \cdot 11 = 330$.)6.NS.B.4Find the greatest common factor and least common multiple of two whole numbers (Reading $\gcd(180, 594) = 18$ off the shared primes and $\operatorname{lcm}(180, 594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11$ off all primes used by either number.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Dividing the prime-power expressions by subtracting exponents on matching bases: $2^{2-1} \cdot 3^{3-2} \cdot 5 \cdot 11$.)
⭐ This 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.
⭐ This 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.
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