Difference between revisions of "2021 AMC 12B Problems/Problem 7"
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<math>\boxed{\textbf{(C)} ~1 : 14}</math> | <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> | + | 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> |
==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}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 20:08, 11 February 2021
Problem
Let 
. What is the ratio of the sum of the odd divisors of 
to the sum of the even divisors of ?
Solution
Prime factorize to get 
. For each odd divisor 
of , there exist even divisors 
of , therefore the ratio is 
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
