2021 AMC 10B Problem 12

Factor: $34 = 2 \cdot 17$, $63 = 3^2 \cdot 7$, $270 = 2 \cdot 3^3 \cdot 5$. So
$$N = 2^3 \cdot 3^5 \cdot 5 \cdot 7 \cdot 17^2.$$

Let $M = 3^5 \cdot 5 \cdot 7 \cdot 17^2$ be the odd part of $N$, and let $S$ be the sum of all divisors of $M$. The odd divisors of $N$ are exactly the divisors of $M$, so their sum is $S$.

Every even divisor of $N$ is $2^k d$ with $k \in \{1, 2, 3\}$ and $d$ a divisor of $M$, and each such pair gives a different even divisor. Grouping by $k$, the sum of the even divisors is
$$2S + 4S + 8S = 14S.$$

The ratio of odd sum to even sum is $S : 14S = 1 : 14$.

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

The divisor sum factors over primes: $\sigma(N) = (1 + 2 + 4 + 8)\,\sigma(M)$. The "$1$" term collects the odd divisors and the rest collects the even ones, so odd $:$ even is $1 : (2 + 4 + 8)$ regardless of what $M$ is. Only the exponent of $2$ matters; the exact value of $N$ is a distraction.

• Comparing odd divisors to all divisors (ratio 1 : 15) instead of odd to even, or miscounting the power of 2 in N.
• Believing the ratio depends on the odd primes and starting to compute $\sigma(M)$ numerically.