Difference between revisions of "1994 AHSME Problems/Problem 1"
(Created page with "==Problem== <math>4^4 \cdot 9^4 \cdot 4^9 \cdot 9^9=</math> <math> \textbf{(A)}\ 13^{13} \qquad\textbf{(B)}\ 13^{36} \qquad\textbf{(C)}\ 36^{13} \qquad\textbf{(D)}\ 36^{36} \qqu...") |
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==Solution== | ==Solution== | ||
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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>. | ||
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| + | --Solution by [[User:TheMaskedMagician|TheMaskedMagician]] 23:04, 27 June 2014 (EDT) | ||
Revision as of 23:04, 27 June 2014
Problem
Solution
Note that 
. So 
and 
. Therefore, 
.
--Solution by TheMaskedMagician 23:04, 27 June 2014 (EDT)
