2002 AMC 10P Problems/Problem 1

Problem

The ratio $\frac{(2^4)^8}{(4^8)^2}$ equals

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

Solution 1

We can use basic rules of exponentiation to solve this problem.

$\frac{(2^4)^8}{(4^8)^2}  =\frac{(2^4)^8}{(2^{16})^2}  =\frac{2^{32}}{2^{32}}  =1$

Thus, our answer is $\boxed{\textbf{(C) } 1}.$

Solution 2

We can rearrange the exponents on the bottom to solve this problem:

$\frac{(2^4)^8}{(4^8)^2}  =\frac{(2^4)^8}{(4^{2})^8}  =\frac{16^{8}}{16^{8}}  =\boxed{\textbf{(C) } 1}$

See also

2002 AMC 10P (ProblemsAnswer KeyResources)
Preceded by
First question
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png