Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2021 AMC 12B Problems/Problem 7"
(→‎Solution 2)
 
(13 intermediate revisions by 11 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2021 AMC 10B Problems#Problem 12|2021 AMC 10B #12]] and [[2021 AMC 12B Problems#Problem 7|2021 AMC 12B #7]]}}
 +
 
==Problem==
 
==Problem==
 
Let <math>N = 34 \cdot 34 \cdot 63 \cdot 270</math>. What is the ratio of the sum of the odd divisors of <math>N</math> to the sum of the even divisors of <math>N</math>?
 
Let <math>N = 34 \cdot 34 \cdot 63 \cdot 270</math>. What is the ratio of the sum of the odd divisors of <math>N</math> to the sum of the even divisors of <math>N</math>?
  
 
<math>\textbf{(A)} ~1 : 16 \qquad\textbf{(B)} ~1 : 15 \qquad\textbf{(C)} ~1 : 14 \qquad\textbf{(D)} ~1 : 8 \qquad\textbf{(E)} ~1 : 3</math>
 
<math>\textbf{(A)} ~1 : 16 \qquad\textbf{(B)} ~1 : 15 \qquad\textbf{(C)} ~1 : 14 \qquad\textbf{(D)} ~1 : 8 \qquad\textbf{(E)} ~1 : 3</math>
−
==Solution==
 
−
<math>\boxed{\textbf{(C)} ~1 : 14}</math>
 
  
−
Prime factorize <math>N</math> to get <math>N=2^{3}3^{5}5\cdot 7\cdot 17^{2}</math>. For each odd divisor <math>n</math> of <math>N</math>, there exist even divisors <math>2n, 4n, 8n</math> of <math>N</math>, therefore the ratio is <math>1:(2+4+8)\rightarrow\boxed{\textbf{(C)}}</math>
+
==Solution 1==
 +
 
 +
Prime factorize <math>N</math> to get <math>N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}</math>. For each odd divisor <math>n</math> of <math>N</math>, there exist even divisors <math>2n, 4n, 8n</math> of <math>N</math>, therefore the ratio is <math>1:(2+4+8)=\boxed{\textbf{(C)} ~1 : 14}</math>
 +
 
 +
==Solution 2==
 +
Prime factorizing <math>N</math>, we see <math>N=2^{3} \cdot 3^{5} \cdot 5\cdot 7\cdot 17^{2}</math>. 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. The sum of odd divisors is given by <cmath>a = (1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2)</cmath> and the total sum of divisors is <cmath>(1+2+4+8)(1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2) = 15a.</cmath> Thus, our ratio is <cmath>\frac{a}{15a-a} = \frac{a}{14a} = \boxed{\textbf{(C)} ~1 : 14}.</cmath>
 +
 
 +
~JustinLee2017
 +
 
 +
 
 +
==Video Solution (Under 2 min!)==
 +
https://youtu.be/AiWQjjL85ZE
 +
 
 +
<i>~Education, the Study of Everything </i>
 +
 
 +
==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
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2021|ab=B|num-b=6|num-a=8}}
 
{{AMC12 box|year=2021|ab=B|num-b=6|num-a=8}}
 +
{{AMC10 box|year=2021|ab=B|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 23:07, 18 July 2023

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. 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


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

See Also

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_AMC_12B_Problems/Problem_7&oldid=195884"