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> | ||
| − | ==Solution== | + | ==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)\rightarrow\boxed{\textbf{(C)}}</math> | 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> | ||
| − | ==Solution== | + | ==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> | 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> | ||
Revision as of 01:48, 12 February 2021
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/b/1/0/b10e907736316fa2340951e0cc06d45d23d8993a.png)
. Thus, our ratio is ![\[\frac{a}{15a-a} = \frac{a}{14a} = \frac{1}{14}\]](http://latex.artofproblemsolving.com/3/b/f/3bfd715a81b9409bec2d8865929266ae3964d50a.png)
~JustinLee2017
Video Solution by OmegaLearn (Prime Factorization)
~ pi_is_3.14
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. 
