Difference between revisions of "2021 Fall AMC 10B Problems/Problem 3"
(→See Also) |
Cellsecret (talk | contribs) (→Solution) |
||
| Line 9: | Line 9: | ||
~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] | ~[[User:Aops-g5-gethsemanea2|Aops-g5-gethsemanea2]] | ||
| + | ==Video Solution by Interstigation== | ||
| + | https://youtu.be/p9_RH4s-kBA?t=160 | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=B|num-a=4|num-b=2}} | {{AMC10 box|year=2021 Fall|ab=B|num-a=4|num-b=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 02:18, 24 November 2021
Problem
The expression 
is equal to the fraction 
in which 
and 
are positive integers whose greatest common divisor is 
. What is 
Solution
We write the given expression as a single fraction: ![\[\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}\]](http://latex.artofproblemsolving.com/e/a/2/ea2a8b57c3444b4d6d105cf4b1bb576bc94c0724.png)
by cross multiplication. Then by factoring the numerator, we get ![\[\frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}.\]](http://latex.artofproblemsolving.com/d/c/0/dc055fe35682fe8e51fe5e3752c645442fb00ed0.png)
The question is asking for the numerator, so our answer is 
giving answer choice 
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=160
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 2 |
Followed by Problem 4 | |
| 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. 
