# Difference between revisions of "2021 AMC 12B Problems/Problem 7"

The following problem is from both the 2021 AMC 10B #12 and 2021 AMC 12B #7, so both problems redirect to this page.

## Problem

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 $N$?

$\textbf{(A)} ~1 : 16 \qquad\textbf{(B)} ~1 : 15 \qquad\textbf{(C)} ~1 : 14 \qquad\textbf{(D)} ~1 : 8 \qquad\textbf{(E)} ~1 : 3$

## Solution 1

Prime factorize $N$ to get $N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}$. For each odd divisor $n$ of $N$, there exist even divisors $2n, 4n, 8n$ of $N$, therefore the ratio is $1:(2+4+8)\rightarrow\boxed{\textbf{(C)}}$

## Solution 2

Prime factorizing $N$, we see $N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}$. 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 $$a = (1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2)$$ and the total sum of divisors is $$(1+2+4+8)(1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2) = 15a$$. Thus, our ratio is $$\frac{a}{15a-a} = \frac{a}{14a} = \frac{1}{14}$$ $\boxed{C}$

~JustinLee2017

~ pi_is_3.14

~IceMatrix

~Interstigation