Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 10B Problems/Problem 10"
(→‎See Also)
Line 11: Line 11:
 
<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>
  
−
==See Also==
+
I MA DA ONE
−
 
 
−
{{AMC10 box|year=2003|ab=B|num-b=9|num-a=11}}
 
−
{{MAA Notice}}
 

Revision as of 11:22, 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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10B_Problems/Problem_10&oldid=97953"