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

(Created page with "==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 consist...")
 
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==Problem==
 
==Problem==
  
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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 the three letters followed by three digits. By how many times is 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>
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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
  
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<cmath>\frac{26^3 \cdot 10^3}{26 \cdot 10^4} = \boxed{\mathrm{(C) \ } \frac{26^2}{10}}</cmath>
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<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==
 
==See Also==
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{{AMC10 box|year=2003|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2003|ab=B|num-b=9|num-a=11}}
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{{MAA Notice}}

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 (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

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