Difference between revisions of "2019 AMC 10A Problems/Problem 1"
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== Solution == | == Solution == | ||
− | <math>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9 | + | <math>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9= 1+1 = \boxed{\textbf{(C) } 2}</math>. |
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==Video Solution by Education, the Study of Everything== | ==Video Solution by Education, the Study of Everything== | ||
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~savannahsolver | ~savannahsolver | ||
− | == Video Solution | + | == Video Solution by T2L Academy== |
− | https:// | + | https://youtu.be/OhCy9c2RTFo?si=UPgBHbW5Bn0yxP1s |
== See Also == | == See Also == | ||
{{AMC10 box|year=2019|ab=A|before=First Problem|num-a=2}} | {{AMC10 box|year=2019|ab=A|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:55, 16 July 2024
Contents
Problem
What is the value of
Solution
.
Video Solution by Education, the Study of Everything
~Education, The Study Of Everything
Video Solution by WhyMath
~savannahsolver
Video Solution by T2L Academy
https://youtu.be/OhCy9c2RTFo?si=UPgBHbW5Bn0yxP1s
See Also
2019 AMC 10A (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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.