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2007 AMC 10B Problems/Problem 8

Revision as of 17:29, 27 May 2011 by Gina (talk | contribs) (Created page with '==Problem 8== On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form <math>bbcac,</m…')
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Problem 8

On the trip home from the meeting where this AMC10 was constructed, the Contest Chair noted that his airport parking receipt had digits of the form $bbcac,$ where $0 \le a < b < c \le 9,$ and $b$ was the average of $a$ and $c.$ How many different five-digit numbers satisfy all these properties?

$\textbf{(A) } 12 \qquad\textbf{(B) } 16 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 20 \qquad\textbf{(E) } 25$

Solution

For the average, $b,$ to be an integer, $a$ and $c$ have to either both be odd or both be even. There are $\frac{5 \times 4}{2 \times 1} = 10$ ways to choose a set of two even numbers and $\frac{5 \times 4}{2 \times 1} = 10$ ways to choose a set of two odd numbers. Therefore, the number of five-digit numbers that satisfy these properties is $10+10=\boxed{\textbf{(D) }20}$

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