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...")
 
Line 5: Line 5:
  
 
==Solution==
 
==Solution==
 +
 +
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)

Revision as of 22:04, 27 June 2014

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 23:04, 27 June 2014 (EDT)

Invalid username
Login to AoPS