AMC 10 · 2021 · #7
Grade 6 number-theoryPick an answer.
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.
Factor into primes
Break all four into primes.
Grade 6 prime factorization: break each factor into primes, then merge.
6.NS.B.4Identify SubproblemsSeparate the powers of two
Even divisors are odd ones times a power of two.
Grade 4 divisors: a small case verifies the structure before generalizing.
4.OA.B.4Solve An Easier Related ProblemWrite the even sum via the odd sum
The even sum is a multiple of the odd sum.
Grade 6 distributive thinking: factor (power of 2) out of every divisor.
Every even divisor is a power of two times an odd divisor, so the power of two factors right out.
▸ Why?
Every number has exactly one prime recipe, so the twos separate cleanly from the rest.
▸ Why?
A factor shared by every term can be lifted out of the sum, leaving the odd part behind.
Cancel the odd sum
The odd sum cancels entirely.
Grade 6 ratios: the unknown S_odd cancels, leaving a pure ratio.
6.RP.A.1Identify SubproblemsRead the ratio
The ratio is 1 to 14.
Grade 6 ratio matching: pick the listed ratio that equals 1 : 14.
6.RP.A.3Eliminate PossibilitiesThis AMC 12 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).