Difference between revisions of "2002 AMC 10P Problems/Problem 1"
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We can use basic rules of exponentiation to solve this problem. | We can use basic rules of exponentiation to solve this problem. | ||
| − | <math>\frac{(2^4)^8}{(4^8)^2} | + | <math>\frac{(2^4)^8}{(4^8)^2} |
=\frac{(2^4)^8}{(2^16)^2} | =\frac{(2^4)^8}{(2^16)^2} | ||
| Line 23: | Line 23: | ||
=\frac{2^32}{2^32} | =\frac{2^32}{2^32} | ||
| − | =1<math> | + | =1</math> |
| − | 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 18:12, 14 July 2024
Problem
The ratio 
equals
Solution 1
We can use basic rules of exponentiation to solve this problem.
$\frac{(2^4)^8}{(4^8)^2}
=\frac{(2^4)^8}{(2^16)^2}
=\frac{2^32}{2^32}
=1$ (Error compiling LaTeX. Unknown error_msg)
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
