Difference between revisions of "2019 AMC 10A Problems/Problem 1"
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==Solution== | ==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 <math>\boxed{\textbf{(C) }2}</math> | + | Adding these up gives <math>\boxed{\textbf{(C) }2}.</math> |
== See Also == | == See Also == | ||
Revision as of 17:12, 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/6/5/f/65f791e3b5ae823aba59f11775813e81b9eb1db6.png)
The second part is ![\[(1^{1})^{9}=1^{9}=1.\]](http://latex.artofproblemsolving.com/2/8/6/286780308ed4829e01fd188354d093d804d65865.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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
