2003 CEMC Pascal Problems/Problem 5

Problem

The value of $\frac{2^8}{8^2}$ is

$\text{ (A) }\ \frac{1}{16} \qquad\text{ (B) }\ 8 \qquad\text{ (C) }\ 4 \qquad\text{ (D) }\ \frac{1}{4} \qquad\text{ (E) }\ 2$

$\frac{2^8}{8^2} = \frac{256}{64} = \boxed {\textbf {(C) } 4}$

~anabel.disher

We can use the fact that $(a^x)^y = a^{xy}$ and $\frac{a^x}{a^y} = a^{x - y}$ :

$\frac{2^8}{8^2} = \frac{2^8}{(2^3)^2} = \frac{2^8}{2^{3 \times 2}} = \frac{2^8}{2^6}$

$=2^{8 - 6} = 2^2 = \boxed {\textbf {(C) } 4}$

~anabel.disher