Difference between revisions of "2007 AMC 12A Problems/Problem 16"
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Thus the total is <math>96 + 16 = 112 \Longrightarrow \mathrm{(C)}</math>. (by scrabbler94) | Thus the total is <math>96 + 16 = 112 \Longrightarrow \mathrm{(C)}</math>. (by scrabbler94) | ||
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+ | == See Also == | ||
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+ | == See also == | ||
+ | {{AMC12 box|year=2007|ab=A|num-b=15|num-a=17}} | ||
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+ | {{MAA Notice}} |
Revision as of 15:16, 19 December 2020
Problems
How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?
Solution 1
We can find the number of increasing arithmetic sequences of length 3 possible from 0 to 9, and then find all the possible permutations of these sequences.
Common difference | Sequences possible | Number of sequences |
1 | 8 | |
2 | 6 | |
3 | 4 | |
4 | 2 |
This gives us a total of sequences. There are to permute these, for a total of .
However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are numbers which start with zero, so our answer is .
Solution 2
Observe that, if the smallest and largest digit have the same parity, this uniquely determines the middle digit. If the smallest digit is not zero, then any choice of the smallest and largest digit gives possible 3-digit numbers; otherwise, possible 3-digit numbers. Hence we can do simple casework on whether 0 is in the number or not.
Case 1: 0 is not in the number. Then there are ways to choose two nonzero digits of the same parity, and each choice generates 3-digit numbers, giving numbers.
Case 2: 0 is in the number. Then there are ways to choose the largest digit (2, 4, 6, or 8), and each choice generates 3-digit numbers, giving numbers.
Thus the total is . (by scrabbler94)
See Also
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.