2021 AMC 10B Problem 12

Step 1: try a small example.

1. Take $24 = 8 \times 3$. Its odd divisors are $1, 3$, with sum $4$.
2. Its even divisors are $2, 6$ (the odd ones times $2$), $4, 12$ (times $4$) and $8, 24$ (times $8$). Their sum is $2 \cdot 4 + 4 \cdot 4 + 8 \cdot 4 = 14 \cdot 4 = 56$.
3. So for $24$ the ratio is $4 : 56 = 1 : 14$. The pattern: every even divisor is an odd divisor times $2$, $4$ or $8$.

Step 2: count the 2s in $N$.

4. Break each factor into primes: $34 = 2 \cdot 17$, $63 = 3^2 \cdot 7$, $270 = 2 \cdot 3^3 \cdot 5$.
5. So $N = 34 \cdot 34 \cdot 63 \cdot 270$ has $2 \cdot 2 \cdot 2 = 2^3$ in it, and the rest is odd:
$$N = 2^3 \times M, \qquad M = 3^5 \cdot 5 \cdot 7 \cdot 17^2 \ \text{(an odd number)}.$$

Step 3: add up the divisors.

6. The odd divisors of $N$ are exactly the divisors of $M$. Call their sum $S$.
7. Each even divisor of $N$ is one of these odd divisors times $2$, $4$ or $8$, and each such product is a different divisor. So the even divisors add up to
$$2S + 4S + 8S = 14S.$$
8. The ratio of odd to even is $S : 14S = 1 : 14$.

The answer is $\boxed{\textbf{(C)}\ 1 : 14}$.

Only the number of 2s in $N$ matters. Every divisor is "an odd part times a power of 2," so the divisors split into groups that are $1$, $2$, $4$ or $8$ times the odd ones. The big odd part $M$ never needs to be worked out. Next time a problem compares odd and even divisors, pull out the powers of 2 first.