Difference between revisions of "2000 AMC 12 Problems/Problem 2"
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<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 armang32324 and integralarefun nigga | + | Solution edited by armang32324 and integralarefun |
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== Solution 2== | == Solution 2== | ||
Revision as of 20:36, 22 December 2024
- 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
2. nigga
Solution 2
We see that 
Only answer choice 
satisfies this requirement.
-SirAppel
Solution 3
Say you somehow forgot the basic rules of exponents, you could just deduce 
is a line of 2000 2000's being multiplied, and if we multiplied by this line by an additional 2000, we would then have 2000's lined up multiplying each other.
- itsj
Video Solution (Daily Dose of Math)
https://www.youtube.com/watch?v=h0QtF9J0oPs
~Thesmartgreekmathdude
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. 
