Difference between revisions of "2021 AMC 12A Problems/Problem 1"
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==Solution== | ==Solution== | ||
We evaluate the given expression to get that <cmath>2^{1+2+3}-(2^1+2^2+2^3)=2^6-(2^1+2^2+2^3)=64-2-4-8=50 \implies \boxed{\text{(B)}}</cmath> | We evaluate the given expression to get that <cmath>2^{1+2+3}-(2^1+2^2+2^3)=2^6-(2^1+2^2+2^3)=64-2-4-8=50 \implies \boxed{\text{(B)}}</cmath> | ||
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| + | ==Video Solution (Quick and Easy)== | ||
| + | https://youtu.be/zxf-CZ97gY0 | ||
| + | |||
| + | ~Education, the Study of Everything | ||
==Video Solution by Aaron He== | ==Video Solution by Aaron He== | ||
https://www.youtube.com/watch?v=xTGDKBthWsw&t=8 | https://www.youtube.com/watch?v=xTGDKBthWsw&t=8 | ||
| + | |||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== | ||
https://www.youtube.com/watch?v=MUHja8TpKGw | https://www.youtube.com/watch?v=MUHja8TpKGw | ||
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==See also== | ==See also== | ||
| − | {{AMC12 box|year=2021|ab=A|before=First | + | {{AMC12 box|year=2021|ab=A|before=First Problem|num-a=2}} |
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 22:28, 22 October 2022
Contents
Problem
What is the value of![\[2^{1+2+3}-(2^1+2^2+2^3)?\]](http://latex.artofproblemsolving.com/2/8/a/28aba0a1e7d8bcd9c7a8c9b2aab526c0329165c5.png)
Solution
We evaluate the given expression to get that ![\[2^{1+2+3}-(2^1+2^2+2^3)=2^6-(2^1+2^2+2^3)=64-2-4-8=50 \implies \boxed{\text{(B)}}\]](http://latex.artofproblemsolving.com/1/d/c/1dc90c6a7f03e6831e8df56990bb7a8e762e2f0b.png)
Video Solution (Quick and Easy)
~Education, the Study of Everything
Video Solution by Aaron He
https://www.youtube.com/watch?v=xTGDKBthWsw&t=8
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=MUHja8TpKGw
Video Solution by Hawk Math
https://www.youtube.com/watch?v=P5al76DxyHY
Video Solution by OmegaLearn (Using computation)
~ pi_is_3.14
Video Solution by TheBeautyofMath
~IceMatrix
See also
| 2021 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by First Problem |
Followed by Problem 2 |
| 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. 
