Difference between revisions of "2000 AMC 12 Problems/Problem 2"
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== Solution == | == Solution == | ||
We can use an elementary exponents rule to solve our problem. | We can use an elementary exponents rule to solve our problem. | ||
| − | We know that <math>a^b | + | We know that <math>a^b\cdot a^c = a^{b+c}</math>. Hence, |
<math> 2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2000+1} = 2000^{2001} \Rightarrow \boxed{\textbf{(A) } 2000^{2001}}</math> | <math> 2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2000+1} = 2000^{2001} \Rightarrow \boxed{\textbf{(A) } 2000^{2001}}</math> | ||
| − | Solution edited by | + | Solution edited by armang32324 and integralarefun |
== Solution 2== | == Solution 2== | ||
Revision as of 13:51, 9 July 2023
- The following problem is from both the 2000 AMC 12 #2 and 2000 AMC 10 #2, so both problems redirect to this page.
Contents
Problem
Solution
We can use an elementary exponents rule to solve our problem.
We know that 
. Hence,
Solution edited by armang32324 and integralarefun
Solution 2
We see that 
Only answer choice 
satisfies this requirement.
-SirAppel
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. 
