Difference between revisions of "2019 AMC 10A Problems/Problem 1"

(Video Solution 2)
(Video Solution by T2L Academy)
 
(7 intermediate revisions by the same user not shown)
Line 5: Line 5:
  
 
== Solution ==  
 
== Solution ==  
<math>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9</math>
+
<math>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9=  1+1 = \boxed{\textbf{(C) } 2}</math>.
 
 
<math>=  1+1 = \boxed{2}</math> which corresponds to <math>\boxed{\text{C}}</math>.
 
 
 
== Video Solution 1 ==
 
  
 +
==Video Solution by Education, the Study of Everything==
 
https://youtu.be/K8je0WYBHFc
 
https://youtu.be/K8je0WYBHFc
  
Education, The Study Of Everything
+
~Education, The Study Of Everything
 
 
  
== Video Solution 2==
+
== Video Solution by WhyMath==
 
https://youtu.be/Ad8WKcwZcTA
 
https://youtu.be/Ad8WKcwZcTA
  
 
~savannahsolver
 
~savannahsolver
  
== Video Solution 3==
+
== Video Solution by T2L Academy==
https://www.youtube.com/watch?v=OhCy9c2RTFo&list=PLbhMrFqoXXwnSCZShTPinX9ZHWjzIlQr_&index=1
+
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

Problem

What is the value of \[2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?\]

$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$

Solution

$2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9=  1+1 = \boxed{\textbf{(C) } 2}$.

Video Solution by Education, the Study of Everything

https://youtu.be/K8je0WYBHFc

~Education, The Study Of Everything

Video Solution by WhyMath

https://youtu.be/Ad8WKcwZcTA

~savannahsolver

Video Solution by T2L Academy

https://youtu.be/OhCy9c2RTFo?si=UPgBHbW5Bn0yxP1s

See Also

2019 AMC 10A (ProblemsAnswer KeyResources)
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. AMC logo.png