1994 AHSME Problems/Problem 1

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)

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1994 AHSME Problems/Problem 1

Problem

Solution