Difference between revisions of "2002 AMC 10P Problems/Problem 1"
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\begin{align*} | \begin{align*} | ||
| − | \frac{(2^4)^8}{(4^8)^2} | + | \frac{(2^4)^8}{(4^8)^2} \\ |
| − | =\frac{(2^4)^8}{(2^16)^2} | + | =\frac{(2^4)^8}{(2^16)^2} \\ |
| − | =\frac{2^32}{2^32} | + | =\frac{2^32}{2^32} \\ |
| − | =1 | + | =1 \\ |
\end{align*} | \end{align*} | ||
| − | Thus, our answer is < | + | Thus, our answer is <math>\boxed{\textbf{(C) } 1}.</math> |
== See also == | == See also == | ||
{{AMC10 box|year=2002|ab=P|before=First question|num-a=2}} | {{AMC10 box|year=2002|ab=P|before=First question|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 19:15, 14 July 2024
Problem
The ratio 
equals
Solution 1
We can use basic rules of exponentiation to solve this problem.
\begin{align*} \frac{(2^4)^8}{(4^8)^2} \\ =\frac{(2^4)^8}{(2^16)^2} \\ =\frac{2^32}{2^32} \\ =1 \\ \end{align*}
Thus, our answer is 
See also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
