Difference between revisions of "2021 AMC 12B Problems/Problem 7"
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<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== | ||
| − | 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) | + | 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> | + | 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> | ||
| + | ~MC | ||
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| + | ==Video Solution (Under 2 min!)== | ||
| + | https://youtu.be/AiWQjjL85ZE | ||
| + | |||
| + | <i>~Education, the Study of Everything </i> | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
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~IceMatrix | ~IceMatrix | ||
| + | ==Video Solution by Interstigation== | ||
| + | https://youtu.be/duZG-jirKRc | ||
| + | |||
| + | ~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.
Contents
Problem
Let 
. What is the ratio of the sum of the odd divisors of 
to the sum of the even divisors of ?
Solution 1
Prime factorize to get 
. For each odd divisor 
of , there exist even divisors 
of , therefore the ratio is 
Solution 2
Prime factorizing , we see . The sum of 's odd divisors are the sum of the factors of without 
, 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 ![\[a = (1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2)\]](http://latex.artofproblemsolving.com/d/4/3/d43ad58db14705adb3a80482071db62edeed9039.png)
and the total sum of divisors is ![\[(1+2+4+8)(1+3+3^2 + 3^3 + 3^4 + 3^5)(1 + 5)(1+7)(1+17+17^2) = 15a.\]](http://latex.artofproblemsolving.com/9/8/8/988f4d1d96ecf8b5b835d9022bec978c99e51fa1.png)
Thus, our ratio is ![\[\frac{a}{15a-a} = \frac{a}{14a} = \boxed{\textbf{(C)} ~1 : 14}.\]](http://latex.artofproblemsolving.com/1/0/c/10cb40e0665d665311e990a490d8a6e57c429ab3.png)
~JustinLee2017
Solution 3
Prime factorizing , we have that there is 
in our factorization. Now, call the sum of the odd divisors 
. We know that if we multiply k by 2, we will have even divisors. So, we can multiply k by 2, 
respectively to get 14k as the sum of the even divisors. Therefore, the answer is ![\[\frac{k}{14k} = \boxed{\textbf{(C)} ~1:14}\]](http://latex.artofproblemsolving.com/8/f/a/8fa9771364fd42c1c41e21e0c43aa4849fbe65a6.png)
~MC
Video Solution (Under 2 min!)
~Education, the Study of Everything
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=qpvS2PVkI8A&t=643s
Video Solution by OmegaLearn (Prime Factorization)
~ 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
~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. 
