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

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{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #2]] and [[2000 AMC 10 Problems|2000 AMC 10 #2]]}}
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== Problem ==
 
== Problem ==
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<math>\displaystyle 2000(2000^{2000}) =</math>
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<math>2000(2000^{2000}) = ?</math>
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<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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<math> \mathrm{(A) \ 2000^{2001} } \qquad \mathrm{(B) \ 4000^{2000} } \qquad \mathrm{(C) \ 2000^{4000} } \qquad \mathrm{(D) \ 4,000,000^{2000} } \qquad \mathrm{(E) \ 2000^{4,000,000} }  </math>
 
 
== Solution ==
 
== Solution ==
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<math> 2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2001} \Rightarrow A</math>
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We can use an elementary exponents rule to solve our problem.
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We know that <math>a^b\cdot a^c = a^{b+c}</math>. Hence,
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<math> 2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2000+1} = 2000^{2001} \Rightarrow \boxed{\textbf{(A) } 2000^{2001}}</math>
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Solution edited by armang32324 and integralarefun
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== Solution 2==
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We see that <math>a(a^{2000})=a^{2001}.</math> Only answer choice <math>\boxed{\textbf{(A)}}</math> satisfies this requirement.
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-SirAppel
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== See also ==
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{{AMC12 box|year=2000|num-b=1|num-a=3}}
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{{AMC10 box|year=2000|num-b=1|num-a=3}}
  
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== See Also ==
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[[Category:Introductory Algebra Problems]]
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*[[2000 AMC 12 Problems]]
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{{MAA Notice}}

Latest revision as of 14:51, 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.

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 $a^b\cdot a^c = a^{b+c}$. Hence, $2000(2000^{2000}) = (2000^{1})(2000^{2000}) = 2000^{2000+1} = 2000^{2001} \Rightarrow \boxed{\textbf{(A) } 2000^{2001}}$

Solution edited by armang32324 and integralarefun

Solution 2

We see that $a(a^{2000})=a^{2001}.$ Only answer choice $\boxed{\textbf{(A)}}$ 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. AMC logo.png

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