Difference between revisions of "2021 Fall AMC 12B Problems/Problem 12"
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<cmath>f(768)=(1+\frac{1}{2}+\ldots+\frac{1}{256})(1+\frac{1}{3})=\frac{511}{192}</cmath> | <cmath>f(768)=(1+\frac{1}{2}+\ldots+\frac{1}{256})(1+\frac{1}{3})=\frac{511}{192}</cmath> | ||
<cmath>f(384)=(1+\frac{1}{2}+\ldots+\frac{1}{128})(1+\frac{1}{3})=\frac{510}{192}</cmath> | <cmath>f(384)=(1+\frac{1}{2}+\ldots+\frac{1}{128})(1+\frac{1}{3})=\frac{510}{192}</cmath> | ||
| − | so the difference is <math>\boxed{(B) \frac{1}{192}}</math> | + | so the difference is <math>\boxed{\textbf{(B)}\ \frac{1}{192}}</math> |
~lopkiloinm | ~lopkiloinm | ||
Revision as of 02:20, 24 November 2021
Problem
For 
a positive integer, let 
be the quotient obtained when the sum of all positive divisors
of n is divided by n. For example,
![\[f(14)=(1+2+7+14)\div 14=\frac{12}{7}\]](http://latex.artofproblemsolving.com/a/3/0/a30498646d780aea6d03d5130640089c4775112f.png)
What is 
Solution 1
The prime factorization of 
is 
and the prime factorization of 
is 
so
![\[f(768)=(1+\frac{1}{2}+\ldots+\frac{1}{256})(1+\frac{1}{3})=\frac{511}{192}\]](http://latex.artofproblemsolving.com/e/5/c/e5c18015481892cb263aa4715af8908e1fd60b87.png)
![\[f(384)=(1+\frac{1}{2}+\ldots+\frac{1}{128})(1+\frac{1}{3})=\frac{510}{192}\]](http://latex.artofproblemsolving.com/6/4/a/64acddaee95d9825e1506ba4cf483f39f2d526c8.png)
so the difference is 
~lopkiloinm
See Also
| 2021 Fall AMC 12B (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 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
