2005 AMC 12A Problems/Problem 11
Problem
How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?
Solution
Solution 1
Let the digits be so that . In order for this to be an integer, and have to have the same parity. There are possibilities for , and for . depends on the value of both and and is unique for each . Thus our answer is .
Solution 2
Thus, the three digits form an arithmetic sequence.
- If the numbers are all the same, then there are possible three-digit numbers.
- If the numbers are different, then we count the number of strictly increasing arithmetic sequences between and and multiply by 2 for the decreasing ones:
Common difference | Sequences possible | Number of sequences |
1 | 8 | |
2 | 6 | |
3 | 4 | |
4 | 2 |
This gives us . However, the question asks for three-digit numbers, so we have to ignore the four sequences starting with . Thus our answer is .
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 10 |
Followed by Problem 12 |
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 12 Problems and Solutions |
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