First, note that 792=23×32×11. So we get that 8∣N⇒8∣7C2⇒8∣10C+6⇒C=1,5,99∣N⇒9∣5+A+B+3+7+C+2⇒A+B+C=1,10,1911∣N⇒11∣5−A+B−3+7−C+2⇒−A+B−C=−11,0 Adding the last two equations, and noting that they sum to 2B, which must be even, we get that B=4,5. Checking values of C we get possible triplets of (0,5,5),(4,5,1), and (6,4,9).