Difference between revisions of "2021 AMC 12A Problems/Problem 1"
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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> | ||
− | ==Video Solution== | + | ==Video Solution (Quick and Easy)== |
+ | https://youtu.be/zxf-CZ97gY0 | ||
+ | |||
+ | ~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 | https://www.youtube.com/watch?v=P5al76DxyHY | ||
+ | |||
+ | == Video Solution by OmegaLearn (Using computation) == | ||
+ | https://youtu.be/90OIMAWAwNg | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/rEWS75W0Q54 | ||
+ | |||
+ | ~IceMatrix | ||
==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
Solution
We evaluate the given expression to get that
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.