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

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=qpvS2PVkI8A&t=643s

Video Solution by OmegaLearn (Prime Factorization)

https://youtu.be/U3msAYWeMbI

~ pi_is_3.14

Video Solution by Hawk Math

https://www.youtube.com/watch?v=VzwxbsuSQ80

Video Solution by TheBeautyofMath

https://youtu.be/L1iW94Ue3eI?t=478

~IceMatrix

Video Solution by Interstigation

https://youtu.be/duZG-jirKRc

~Interstigation

2021 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

2021 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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