AMC 8 · 2013 · #10

Grade 6 number-theory

Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation

lcmgcdprime-factorization identify-subproblems ↑ Prerequisites: prime-factorizationmultiplesfactors

📏 Medium solution 💡 3 insights

Problem

Find the ratio

\frac{<span class="mk-c">lcm(180, 594)</span>}{<span class="mk-c">gcd(180, 594)</span>}

Pick an answer.

(A)

110

(B)

165

(C)

330

(D)

625

(E)

660

AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

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.

1STEP 1

Subproblem 1: prime-factorize 180 down to primes, giving 2² · 3² · 5.

180 = 2 · 90 = 2 · 2 · 45 = 2² · 9 · 5 = 2² · 3² · 5

2STEP 2

Subproblem 2: same for 594 — peel off small primes to get 2 · 3³ · 11.

594 = 2 · 297 = 2 · 3 · 99 = 2 · 3 · 3 · 33 = 2 · 3³ · 11

3STEP 3

Subproblem 3: for the GCF take the smaller exponent on each shared prime — 2¹ · 3² gives 18.

gcd(180, 594) = 2¹ · 3² = 2 · 9 = 18

4STEP 4

Subproblem 4: for the LCM take the larger exponent and include primes from either list: 2² · 3³ · 5 · 11.

lcm(180, 594) = 2² · 3³ · 5 · 11

5STEP 5

Subproblem 5: divide the towers by subtracting exponents on matching primes, leaving 2 · 3 · 5 · 11.

\frac{lcm}{gcd}

\frac{2² · 3³ · 5 · 11}{2¹ · 3²}

= 2²⁻¹ · 3³⁻² · 5 · 11 = 2 · 3 · 5 · 11

6STEP 6

Multiply the four primes: 2 · 3 · 5 · 11 = 330 → (C).

2 · 3 · 5 · 11 = 6 · 55 = 330 → (C)

Answer

330

Use the identity lcm(a, b) · gcd(a, b) = a · b as an independent check:

\frac{lcm}{gcd}

\frac{a · b}{gcd²}

\frac{180 · 594}{18²}

\frac{180}{18}

\frac{594}{18}

= 10 · 33 = 330. Two different computations agree on 330, so answer (C) is solid.

💡Key takeaway

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.

More like this