Difference between revisions of "1994 AHSME Problems/Problem 1"

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Note that <math>a^x\times a^y=a^{x+y}</math>. So <math>4^4\cdot 4^9=4^{13}</math> and <math>9^4\cdot 9^9=9^{13}</math>. Therefore, <math>4^{13}\cdot 9^{13}=(4\cdot 9)^{13}=\boxed{\textbf{(C)}\ 36^{13}}</math>.
 
Note that <math>a^x\times a^y=a^{x+y}</math>. So <math>4^4\cdot 4^9=4^{13}</math> and <math>9^4\cdot 9^9=9^{13}</math>. Therefore, <math>4^{13}\cdot 9^{13}=(4\cdot 9)^{13}=\boxed{\textbf{(C)}\ 36^{13}}</math>.
  
--Solution by [[User:TheMaskedMagician|TheMaskedMagician]] 23:04, 27 June 2014 (EDT)
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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
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==See Also==
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{{AHSME box|year=1994|before=First Problem|num-a=2}}
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{{MAA Notice}}

Latest revision as of 16:25, 9 January 2021

Problem

$4^4 \cdot 9^4 \cdot 4^9 \cdot 9^9=$

$\textbf{(A)}\ 13^{13} \qquad\textbf{(B)}\ 13^{36} \qquad\textbf{(C)}\ 36^{13} \qquad\textbf{(D)}\ 36^{36} \qquad\textbf{(E)}\ 1296^{26}$

Solution

Note that $a^x\times a^y=a^{x+y}$. So $4^4\cdot 4^9=4^{13}$ and $9^4\cdot 9^9=9^{13}$. Therefore, $4^{13}\cdot 9^{13}=(4\cdot 9)^{13}=\boxed{\textbf{(C)}\ 36^{13}}$.

--Solution by TheMaskedMagician

See Also

1994 AHSME (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AHSME Problems and Solutions

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