Difference between revisions of "2019 AMC 10A Problems/Problem 1"
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| + | ==Problem 1== | ||
| + | What is the value of <cmath>2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?</cmath> | ||
| + | <math>\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4</math> | ||
| + | |||
| + | ==Solution== | ||
The first part can be rewritten as <cmath>2^{0^{1}}=2^{0}=1</cmath> | The first part can be rewritten as <cmath>2^{0^{1}}=2^{0}=1</cmath> | ||
The second part is <cmath>(1^{1})^{9}=1^{9}=1</cmath> | The second part is <cmath>(1^{1})^{9}=1^{9}=1</cmath> | ||
| − | Adding these up gives (C) | + | Adding these up gives <math>\textbf{(C) }2</math> |
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{AMC10 box|year=2019|ab=A|before=First Problem|num-a=2}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:11, 9 February 2019
Problem 1
What is the value of ![\[2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9?\]](http://latex.artofproblemsolving.com/6/7/6/67606c2b8141058216ca18ee5e71821458b74b01.png)
Solution
The first part can be rewritten as ![\[2^{0^{1}}=2^{0}=1\]](http://latex.artofproblemsolving.com/7/e/2/7e2e77e5a74459fb40e84717608fa12cf0c6732b.png)
The second part is ![\[(1^{1})^{9}=1^{9}=1\]](http://latex.artofproblemsolving.com/c/a/3/ca33597adbfe49478e6dd72ab9163f8b88cdbd4a.png)
Adding these up gives 
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
