Difference between revisions of "2000 AMC 12 Problems/Problem 2"
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<math>2000(2000^{2000}) =</math> | <math>2000(2000^{2000}) =</math> | ||
− | <math> \ | + | <math> \textbf{(A)} \ 2000^{2001} \qquad \textbf{(B)} \ 4000^{2000} \qquad \textbf{(C)} \ 2000^{4000} \qquad \textbf{(D)} \ 4,000,000^{2000} \qquad \textbf{(E)} \ 2000^{4,000,000} </math> |
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== Solution == | == Solution == | ||
<math> 2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2001} \Rightarrow \boxed{A}</math> | <math> 2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2001} \Rightarrow \boxed{A}</math> |
Revision as of 20:19, 23 November 2020
- The following problem is from both the 2000 AMC 12 #2 and 2000 AMC 10 #2, so both problems redirect to this page.
Problem
Solution
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
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 |
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.