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Difference between revisions of "2021 AMC 12B Problems/Problem 7"
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{{duplicate|[[2021 AMC 10B Problems#Problem 12|2021 AMC 10B #12]] and [[2021 AMC 12B Problems#Problem 7|2021 AMC 12B #7]]}}
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==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>
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==Solution 1==
 
==Solution 1==
  
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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)\rightarrow\boxed{\textbf{(C)}}</math>
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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==
 
==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} = \frac{1}{14}</cmath> <math>\boxed{C}</math>
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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
 
~JustinLee2017
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==Solution 3==
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Prime factorizing <math>N</math>, we have that there is <math>2^3</math> in our factorization. Now, call the sum of the odd divisors <math>k</math>. We know that if we multiply k by 2, we will have even divisors. So, we can multiply k by 2, <math>2^2, 2^3</math> respectively to get 14k as the sum of the even divisors. Therefore, the answer is <cmath>\frac{k}{14k} = \boxed{\textbf{(C)} ~1:14}</cmath>
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~MC
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==Video Solution (Under 2 min!)==
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https://youtu.be/AiWQjjL85ZE
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 +
<i>~Education, the Study of Everything </i>
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==Video Solution by Punxsutawney Phil==
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https://youtube.com/watch?v=qpvS2PVkI8A&t=643s
  
 
== Video Solution by OmegaLearn (Prime Factorization) ==
 
== Video Solution by OmegaLearn (Prime Factorization) ==
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~ pi_is_3.14
 
~ pi_is_3.14
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==Video Solution by Hawk Math==
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https://www.youtube.com/watch?v=VzwxbsuSQ80
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==Video Solution by TheBeautyofMath==
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https://youtu.be/L1iW94Ue3eI?t=478
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~IceMatrix
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==Video Solution by Interstigation==
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https://youtu.be/duZG-jirKRc
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 +
~Interstigation
  
 
==See Also==
 
==See Also==

Latest revision as of 12:03, 11 November 2024

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

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

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

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