How many three-digit numbers are composed of three distinct digits such that one digit is the [average](https://artofproblemsolving.com/wiki/index.php/Average) of the other two? 
$\mathrm{(A)}\ 96\qquad \mathrm{(B)}\ 104\qquad \mathrm{(C)}\ 112\qquad \mathrm{(D)}\ 120\qquad \mathrm{(E)}\ 256$
We can find the number of increasing [arithmetic sequences](https://artofproblemsolving.com/wiki/index.php/Arithmetic_sequence) 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
$012, \ldots, 789$
8


2
$024, \ldots, 579$
6


3
$036, \ldots, 369$
4


4
$048, \ldots, 159$
2

This gives us a total of $2 + 4 + 6 + 8 = 20$ sequences. There are $3! = 6$ to permute these, for a total of $120$.
However, we note that the conditions of the problem require three-digit numbers, and hence our numbers cannot start with zero. There are $2! \cdot 4 = 8$ numbers which start with zero, so our answer is $120 - 8 = 112 \Longrightarrow \mathrm{(C)}$.