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2021 AMC 12B Problem 7

Problem 7 of 25EasierNumber Theory

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?

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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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Tagged: prime factorization · sum of factors

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