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

Revision as of 03:21, 11 August 2025

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 $1:(2+4+8)=\boxed{\textbf{(C)} ~1 : 14}$

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.

BY SUM OF FACTORS FORMULA (search it up if you don't know): The formula is actually for all factors, but we can just take out the $2^3$ , so we have:

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 $\frac{a}{15a-a} = \frac{a}{14a} = \boxed{\textbf{(C)} ~1 : 14}.$

~JustinLee2017 ~minor edits by SwordAxe

Solution 3

Prime factorizing $N$ , we have that there is $2^3$ in our factorization. Now, call the sum of the odd divisors $k$ . We know that if we multiply k by 2, we will have even divisors. So, we can multiply k by 2, $2^2, 2^3$ respectively to get 14k as the sum of the even divisors. Therefore, the answer is $\frac{k}{14k} = \boxed{\textbf{(C)} ~1:14}$ ~MC

Video Solution (Under 2 min!)

https://youtu.be/AiWQjjL85ZE

~Education, the Study of Everything

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 14: / Line 14: @@
 The sum of <math>N</math>'s odd divisors are the sum of the factors of <math>N</math> without <math>2</math>, and the sum of the even divisors is the sum of the odds subtracted by the total sum of divisors.
-BY SUM OF FACTORS FORMULA (search if you don' know):
+BY SUM OF FACTORS FORMULA (search it up if you don't know):
 The formula is actually for all factors, but we can just take out the <math>2^3</math>, so we have:

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