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2007 AMC 12A Problems/Problem 16

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Problems

How many three-digit numbers are composed of three distinct digits such that one digit is the average of the other two?

\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256

Solution

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 two digit numbers, and hence our numbers cannot start with zero. There are numbers which start with zero, so our answer is 120 - 8 = 112 \Longrightarrow \mathrm{(C)}.

See also

similar problem

2007 AMC 12A (Problems)
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
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's NEW Intermediate Counting & Probability by David Patrick.
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