2007 AMC 12A Problem 16

The three distinct digits form an increasing arithmetic progression. Counting by common difference: $8$ with difference $1,$ $6$ with difference $2,$ $4$ with difference $3,$ and $2$ with difference $4,$ for $20$ sets. Of these, $4$ sets contain $0$ (namely $\{0,1,2\},$ $\{0,2,4\},$ $\{0,3,6\},$ $\{0,4,8\}$); each yields $2\cdot 2!=4$ valid numbers since $0$ cannot lead. The other $16$ sets each yield $3!=6$ numbers. The total is $4\cdot 4+16\cdot 6=112.$ Thus, the correct answer is C.