Problem:
Let N=34⋅34⋅63⋅270. What is the ratio of the sum of the odd divisors of N to the sum of the even divisors of N ?
Answer Choices:
A. 1:16
B. 1:15
C. 1:14
D. 1:8
E. 1:3
Solution:
Prime factorize N to get N=23⋅35⋅5⋅7⋅172. For each odd divisor n of N, there exist even divisors 2n,4n,8n of N, therefore the ratio is 1:(2+4+8)= (C)1:14
OR
Prime factorizing N, we see N=23⋅35⋅5⋅7⋅172. The sum of N 's odd divisors are the sum of the factors of N without $2 $, and the sum of the even divisors is the sum of the odds subtracted by the total sum of divisors. The sum of odd divisors is given by
a=(1+3+32+33+34+35)(1+5)(1+7)(1+17+172)
and the total sum of divisors is
(1+2+4+8)(1+3+32+33+34+35)(1+5)(1+7)(1+17+172)=15a
Thus, our ratio is
15a−aa=14aa= (C)1:14
The problems on this page are the property of the MAA's American Mathematics Competitions