>>14296389Justification:
1. F + M = T
2. T + M (+ 1?) = F
Therefore, either F + 2*M = F (without carry) or F + 2*M + 1 = F (with carry)
So either 2M which is even or 2M + 1 which is odd
A + 10*x = A
10*x is even, therefore F + 2*M = F, 10/2 = 5, therefore M = 5 and there is no carry for last digit. Therefore, T = F + 5. Since F is not 6, F in (3, 4). There is carry for the penultimate digit, therefore N = B + B + 1 = T + 1. Since N is not 0, therefore F = 3, therefore T = 8, therefore B = 4, therefore N = 9, therefore there is no carry for the antepenultimate digit. Since C + C = 4, 4/2=2 or 5 + 2=7, and C!= 2, then C = 7, and therefore H = 1.
QED
