All letters represent unique numbers between 0-9
Basic starting point: If you subtract all the digits from a number, you get a multiple of 9
ABC = 100 x A + 10 x B + C = (99 x A + 9 x B) + A + B + C = 9 (11 x A + B) + A + B + C
Inversely stated, it means
ABC = 9x + A + B + C
ABC + DEF + GHI = (9x + A + B + C) + (9y + D + E + F) + (9z + G + H + I)
ABC + DEF + GHI = 9 (x+y+z) + (A + B + C + D + E + F + G + H + I)
ABC + DEF + GHI + J = 9 (x+y+z) + (A + ... I + J)
123J = (9q + 1 + 2 + 3 + J)
(A + ... + I + J) = (9 + 1) + (8 + 2) + (7 + 3) + (6 + 4) + 5
(A + ... + I + J) = 45 = 9 x 5
ABC + DEF + GHI + J = 9 (x + y + z + 5)
To restate original problem with additional J
123J + J = ABC + DEF + GHI + J = 9 (x + y + z + 5)
123J + J = (9q + (1 + 2 + 3 + J)) + J = 9q + 6 + 2J
To remove the extra 9s, we can modulo the numbers (they are equal, so it should be the same)
(123J + J) % 9 = (ABC + DEF + GHI + J) % 9 = (9 (x + y + z + 5)) % 9 = 0
(123J + J) % 9 = (9q + 6 + 2J) % 9 = (6 + 2J) % 9 = 0
Ending up with a single variable to solve for
(6 + 2J) % 9 = 0
This implies it is a multiple of 2
2 (3 + J) = 9q = 9 x 2 x (m + 1)
J = 9 - 3 + 9 * m
J = 6 + 9 * m
But we know J is < 9, so m = 0. Ending up with J = 6 without question.