Difference between revisions of "2002 AMC 10P Problems/Problem 1"

Latest revision as of 14:55, 13 August 2024

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

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

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 $\boxed{\textbf{(C) } 1}.$

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

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

2002 AMC 10P (Problems • Answer Key • Resources)
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

@@ Line 17: / Line 17: @@
 We can use basic rules of exponentiation to solve this problem.
-<math>\frac{(2^4)^8}{(4^8)^2}</math>
+<math>\frac{(2^4)^8}{(4^8)^2}
+=\frac{(2^4)^8}{(2^{16})^2}
+=\frac{2^{32}}{2^{32}}
+=1</math>
-=\frac{(2^4)^8}{(2^16)^2}
+Thus, our answer is <math>\boxed{\textbf{(C) } 1}.</math>
+== Solution 2==
-=\frac{2^32}{2^32}
+We can rearrange the exponents on the bottom to solve this problem:
-=1<math>
-Thus, our answer is </math>\boxed{\textbf{(C) } 1}.$
+<math>\frac{(2^4)^8}{(4^8)^2}
+=\frac{(2^4)^8}{(4^{2})^8}
+=\frac{16^{8}}{16^{8}}
+=\boxed{\textbf{(C) } 1}</math>
 == See also ==
 {{AMC10 box|year=2002|ab=P|before=First question|num-a=2}}
 {{MAA Notice}}