Difference between revisions of "2007 AMC 10B Problems/Problem 8"
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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 | ||
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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 11:57, 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 
where 
and 
was the average of 
and 
How many different five-digit numbers satisfy all these properties?
Solution
For the average, 
to be an integer, and 
have to either both be odd or both be even. There are 
ways to choose a set of two even numbers and 
ways to choose a set
of two odd numbers. Therefore, the number of five-digit numbers that satisfy these properties is 
See Also
| 2007 AMC 10B (Problems • Answer Key • Resources) | ||
| 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. 
