Difference between revisions of "2021 Fall AMC 12B Problems/Problem 4"
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==Video Solution by Interstigation== | ==Video Solution by Interstigation== | ||
https://youtu.be/p9_RH4s-kBA?t=429 | https://youtu.be/p9_RH4s-kBA?t=429 | ||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2021 Fall|ab=B|num-a=5|num-b=3}} | ||
+ | {{AMC12 box|year=2021 Fall|ab=B|num-a=5|num-b=3}} | ||
+ | {{MAA Notice}} |
Revision as of 01:20, 24 November 2021
- The following problem is from both the 2021 Fall AMC 10B #5 and 2021 Fall AMC 12B #4, so both problems redirect to this page.
Problem
Let . Which of the following is equal to
Solution 1
We have Therefore,
~kingofpineapplz
Solution 2
The requested value is Thus, the answer is
~NH14
Video Solution by Interstigation
https://youtu.be/p9_RH4s-kBA?t=429
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 |
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
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.