AMC 10 · 2021 · #12

Grade 6 rate-ratio

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

prime-factorizationdivisor-sumdivisor-countratio-proportion identify-subproblemseasier-related-problem ↑ Prerequisites: prime-factorizationdivisor-count

📏 Medium solution 💡 3 insights

Problem

Let N = 34 · 34 · 63 · 270. Find the ratio (sum of odd divisors of N) : (sum of even divisors of N).

Pick an answer.

(A)

~1 : 16

(B)

~1 : 15

(C)

~1 : 14

(D)

~1 : 8

(E)

~1 : 3

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems) splits the work into three pieces: (a) prime-factorize N to find the power of 2 in it; (b) sum the odd divisors using only the odd prime part; (c) get the sum of even divisors by subtracting from the total. Tool #9 (Easier Problem) checks the key insight on a tiny case like N = 2³ · 3 first, where you can list every divisor by hand. Tool #5 (Pattern) recognizes that every divisor is (power of 2) × (odd divisor), so the even-divisor sum is (2 + 4 + 8) = 14 times the odd-divisor sum. Tool #3 (Eliminate) confirms (C) against the five ratio choices.

1STEP 1

Prime-factorize each factor and collect like primes; the power of two in N is 2³.

N = (2² · 17²)(3² · 7)(2 · 3³ · 5) = 2³ · 3⁵ · 5 · 7 · 17²

2STEP 2

Test M = 24 = 2³ · 3: odd sum 4, even sum 56, ratio 1 : 14 — every even divisor is an odd one times 2, 4, or 8.

sum even(24) = (2+4+8)(1+3) = 14 · 4 = 56

3STEP 3

Every divisor is a power of 2 (from 2⁰ to 2³) times an odd divisor, so the even-divisor sum is (2 + 4 + 8) S_odd = 14 S_odd.

S_even = (2 + 4 + 8) S_odd = 14 S_odd

4STEP 4

Form the ratio: S_odd : S_even = S_odd : 14 S_odd = 1 : 14; the value of S_odd cancels.

\frac{S_odd}{S_even}

\frac{1}{14}

5STEP 5

Among the choices, 1 : 14 is (C); the rest come from miscounting the power of 2 or adding 2⁰ to the even sum.

1 : 14 → (C)

Answer

~1 : 14

Sanity check the power of 2: 34 · 34 contributes 2², and 270 contributes one more 2, so N has exactly 2³. Total divisor sum formula σ(2³) = 1 + 2 + 4 + 8 = 15 factors as 1 + 14, matching odd-vs-even. The ratio 1 : 14 is the same as for any number of the form 2³ · (odd), which makes sense: only the power of 2 in N controls this ratio.

💡Key takeaway

This AMC 10 problem only needs Grade 6 ratio sense you already know! Prime-factorize N = 2³ · 3⁵ · 5 · 7 · 17² — only the 2³ part matters. Every even divisor is (2 or 4 or 8) × (odd divisor), so the even-divisor sum is (2+4+8) = 14 times the odd-divisor sum. Ratio 1 : 14, answer (C).

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