Each of A, B and C is a positive integer in arithmetic sequence, with A < B < C such that the last four digits in the base ten expansion of A*B*C is 2016.

Determine the three smallest values of A+B+C

(In reply to

computer solution by Charlie)

Trying to solve it without software, I've assumed possibility of the 1st triplet to produce P=A*B*C=2016.

Since the last digit of A could be 1,3,6, or 8 only and B could be easily determined to fit P, I've tested the cases of A below 10.

The 1st trial produced 1,32,63 and very quickly I have reached the conclusion that the next 2 triplets must produce products bigger than 10000.

As usual - the computer wins.