2007 AMC 10B Problems/Problem 8
Contents
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
Solution 2
Case : The numbers are separated by .
We this case with and . Following this logic, the last set we can get is and . We have sets of numbers in this case.
Case : The numbers are separated by .
This case starts with and . It ends with and . There are sets of numbers in this case.
Case : The numbers start with and . It ends with and . This case has sets of numbers.
It's pretty clear that there's a pattern: sets, sets, sets. The amount of sets per case decreases by , so it's obvious Case has sets. The total amount of possible five-digit numbers 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.