Difference between revisions of "2003 AMC 10B Problems/Problem 10"
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There are <math>26</math> letters and <math>10</math> digits. There were <math>26 \cdot 10^4</math> old license plates. There are <math>26^3 \cdot 10^3</math> new license plates. The number of license plates increased by | There are <math>26</math> letters and <math>10</math> digits. There were <math>26 \cdot 10^4</math> old license plates. There are <math>26^3 \cdot 10^3</math> new license plates. The number of license plates increased by | ||
| − | <cmath>\frac{26^3 \cdot 10^3}{26 \cdot 10^4} = \boxed{\ | + | <cmath>\frac{26^3 \cdot 10^3}{26 \cdot 10^4} = \boxed{\textbf{(C) \ } \frac{26^2}{10}}</cmath> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2003|ab=B|num-b=9|num-a=11}} | ||
Revision as of 18:02, 26 November 2011
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 9 |
Followed by Problem 11 | |
| 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 | ||
