Prime factorizing $N$, we see $N=2^{3} \cdot 3^{5} \cdot 5 \cdot 7 \cdot 17^{2}$. The sum of $N$ \textquotesingle 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 $$ a=\left(1+3+3^{2}+3^{3}+3^{4}+3^{5}\right)(1+5)(1+7)\left(1+17+17^{2}\right) $$ and the total sum of divisors is $$ (1+2+4+8)\left(1+3+3^{2}+3^{3}+3^{4}+3^{5}\right)(1+5)(1+7)\left(1+17+17^{2}\right)=15 a . $$ Thus, our ratio is $$ \frac{a}{15 a-a}=\frac{a}{14 a}=\text { (C) } 1: 14 \text {. } $$ 
Let $N=34 \cdot 34 \cdot 63 \cdot 270$. What is the ratio of the sum of the odd divisors of $N$ to the sum of the even divisors of $N$ ? 
[A] 1:16 
[B] 1:15 
[C] 1:14 
[D] 1:8 
[E] 1:3 