Difference between revisions of "2007 AMC 10B Problems/Problem 8"

(Solution)
m (Solution)
Line 9: Line 9:
 
For the average, <math>b,</math> to be an integer, <math>a</math> and <math>c</math> have to either both be odd or both be even. There are <math>\frac{5 \times  
 
For the average, <math>b,</math> to be an integer, <math>a</math> and <math>c</math> have to either both be odd or both be even. There are <math>\frac{5 \times  
 
4}{2 \times 1} = 10</math> ways to choose a set of two even numbers and <math>\frac{5 \times 4}{2 \times 1} = 10</math> ways to choose a set  
 
4}{2 \times 1} = 10</math> ways to choose a set of two even numbers and <math>\frac{5 \times 4}{2 \times 1} = 10</math> ways to choose a set  
 +
 
of two odd numbers. Therefore, the number of five-digit numbers that satisfy these properties is <math>10+10=\boxed{\textbf{(D) }20}</math>
 
of two odd numbers. Therefore, the number of five-digit numbers that satisfy these properties is <math>10+10=\boxed{\textbf{(D) }20}</math>
  

Revision as of 12:57, 4 June 2021

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 $10+10=\boxed{\textbf{(D) }20}$

See Also

2007 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png