ABC x C = DBC
(100A + 10B + C) C = (100D + 10B + C)
100AC + 10BC + C² = 100D + 10B + C
Since they are equal, they must have the same units digit. The units digit of C² must be equal to C.
C 0 1 2 3 4 5 6 7 8 9
C² 0 1 4 9 6 5 6 9 4 1
C can be 0, 1, 5, or 6.
100AC will have a maximum value of 900 since we have 100D at the right side of the equation and D would only be a digit (meaning it ranges from 0 to 9).
From this part it becomes a bit tricky.
100AC + 10BC + C² = 100D + 10B + C
If C = 0
0 = 100D + 10B
A digit cannot be equal to 0 so C≠0.
If C = 1
100A + 10B + 1 = 100D + 10B + 1
100A = 100D
Since two letters represent different digits C≠1.
If C = 5
500A + 50B + 25 = 100D + 10B + 5
500A - 100D + 40B + 20 = 0
25A - 5D + 2B + 1 = 0
5(5A - D) = - (2B + 1)
Since B is positive this would mean that D>5A
A = 1 since A cannot be 0 and D ranges from 0 to 9
5(5 - D) = -2B - 1
25 - 5D = -2B - 1
26 = 5D - 2B
Since 26 and -2B are even, D is even. Since D is greater than 5 the only possible values of D would be 6 and 8. The corresponding values of B would be 2 and 7.
A = 1, B = 2, C = 5, D = 6
A = 1, B = 7, C = 5, D = 8
If C = 6
600A+ 60B + 36 = 100D + 10B + 6
600A - 100D + 50B + 30 = 0
60A - 10D + 5B + 3 = 0
5(12A - 2D + B) = -3
Since they do not have equal signs, 12A - 2D + B is negative meaning 12A + B < 2D. This would mean A=1 (because A≠0), we cannot have a higher value of A since that would make 12A greater than 18 which means that 2D cannot be higher than 12A + B (since D ranges from 0 to 9).
5(12 - 2D + B) = -3
60 - 10D + 5B = -3
63 = 10D - 5B
63 = 5 (2D - B)
63 is not a multiple of 5 so C≠6.
Final Answer:
A = 1, B = 2, C = 5, D = 6
A = 1, B = 7, C = 5, D = 8
