Difference between revisions of "2003 AMC 10B Problems/Problem 10"

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==Problem==
 
==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 three letters followed by three digits. By how many times hascq  the number of possible license plates increased?
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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 three letters followed by three digits. By how many times has the number of possible license plates increased?
  
 
<math>\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} </math>
 
<math>\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} </math>

Latest revision as of 19:53, 16 December 2023

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 three letters followed by three digits. By how many times has 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

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


See Also

2003 AMC 10B (ProblemsAnswer KeyResources)
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

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