Difference between revisions of "2000 AMC 12 Problems/Problem 2"

(Solution)
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== Solution ==
 
== Solution ==
 +
We can use an elementary exponents rule to solve our problem.
 +
We know that <math>a^b * 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: armang32324
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2000|num-b=1|num-a=3}}
 
{{AMC12 box|year=2000|num-b=1|num-a=3}}

Revision as of 00:27, 7 April 2023

The following problem is from both the 2000 AMC 12 #2 and 2000 AMC 10 #2, so both problems redirect to this page.

Problem

$2000(2000^{2000}) = ?$

$\textbf{(A)} \ 2000^{2001}  \qquad \textbf{(B)} \ 4000^{2000}  \qquad \textbf{(C)} \ 2000^{4000}  \qquad \textbf{(D)} \ 4,000,000^{2000}  \qquad \textbf{(E)} \ 2000^{4,000,000}$

Solution

We can use an elementary exponents rule to solve our problem. We know that $a^b * a^c = a^(b+c)$. Hence, $2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2000+1} = 2000^{2001} \Rightarrow \boxed{\textbf{(A) } 2000^{2001}}$

Solution edited by: armang32324

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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