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

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-SirAppel
 
-SirAppel
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==Solution 3==
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Say you somehow forgot the basic rules of exponents, you could just deduce <math>2000^{2000}</math> is a line of 2000 2000's being multiplied, and if we multiplied by this line by an additional 2000, we would then have <math>\boxed{\textbf{(A)}}</math> 2000's lined up multiplying each other.
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- itsj
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== Video Solution (Daily Dose of Math) ==
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https://www.youtube.com/watch?v=h0QtF9J0oPs
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~Thesmartgreekmathdude
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2000|num-b=1|num-a=3}}
 
{{AMC12 box|year=2000|num-b=1|num-a=3}}

Latest revision as of 19:38, 27 October 2024

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

Solution 3

Say you somehow forgot the basic rules of exponents, you could just deduce $2000^{2000}$ is a line of 2000 2000's being multiplied, and if we multiplied by this line by an additional 2000, we would then have $\boxed{\textbf{(A)}}$ 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. AMC logo.png

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