Difference between revisions of "2003 AMC 10B Problems/Problem 10"
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<cmath>\frac{26^3 \cdot 10^3}{26 \cdot 10^4} = \boxed{\textbf{(C) \ } \frac{26^2}{10}}</cmath> | <cmath>\frac{26^3 \cdot 10^3}{26 \cdot 10^4} = \boxed{\textbf{(C) \ } \frac{26^2}{10}}</cmath> | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2003|ab=B|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Revision as of 00:21, 5 January 2019
Problem
Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of the three letters followed by three digits. By how many times is the number of possible license plates increased?
Solution
There are 
letters and 
digits. There were 
old license plates. There are 
new license plates. The number of license plates increased by
![\[\frac{26^3 \cdot 10^3}{26 \cdot 10^4} = \boxed{\textbf{(C) \ } \frac{26^2}{10}}\]](http://latex.artofproblemsolving.com/b/c/e/bce4220f1b7d53e103e486a9453b3f8368d28bf8.png)
See Also
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
