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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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{{AMC10 box|year=2003|ab=B|num-b=9|num-a=11}}
 
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Revision as of 10:22, 26 September 2018

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?

$\textbf{(A) } \frac{26}{10} \qquad\textbf{(B) } \frac{26^2}{10^2} \qquad\textbf{(C) } \frac{26^2}{10} \qquad\textbf{(D) } \frac{26^3}{10^3} \qquad\textbf{(E) } \frac{26^3}{10^2}$

Solution

There are $26$ letters and $10$ digits. There were $26 \cdot 10^4$ old license plates. There are $26^3 \cdot 10^3$ new license plates. The number of license plates increased by

\[\frac{26^3 \cdot 10^3}{26 \cdot 10^4} = \boxed{\textbf{(C) \ } \frac{26^2}{10}}\]

I MA DA ONE

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