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

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--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
 
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
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{{AHSME box|year=1994|before=First Problem|num-a=2}}
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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 (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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