2021 AMC 12B Spring Problem 7

Attempt Problem 7 of the 2021 AMC 12B Spring below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2021 AMC 12B Spring solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

7.

Let N=34⋅34⋅63⋅270.N=34\cdot 34\cdot 63\cdot 270. What is the ratio of the sum of the odd divisors of NN to the sum of the even divisors of N?N?
1:161:16
1:151:15
1:141:14
1:81:8
1:31:3
Answer: C
Concepts:prime factorizationsum of factors
Small hint:
Find the exact power of 22 dividing NN
Big hint:
If MM is the odd part of NN and 2k2^k the even part, every even divisor is 2j2^j times an odd divisor
Solution:
Factoring, 34=2⋅17,34=2\cdot 17, 63=32⋅7,63=3^2\cdot 7, and 270=2⋅33⋅5,270=2\cdot 3^3\cdot 5, so N=23⋅35⋅5⋅7⋅172.N=2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2.
Let MM be the odd part 35⋅5⋅7⋅172.3^5\cdot 5\cdot 7\cdot 17^2. The sum of all divisors is (1+2+4+8) σ(M)(1+2+4+8)\,\sigma(M) =15 σ(M).=15\,\sigma(M).
The odd divisors sum to σ(M),\sigma(M), so the even divisors sum to 15 σ(M)−σ(M)=14 σ(M).15\,\sigma(M)-\sigma(M)=14\,\sigma(M).
The ratio is σ(M):14 σ(M)=1:14.\sigma(M):14\,\sigma(M)=1:14.
Thus, the correct answer is C.
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