Solved each of the following Cryptarithms:
Here, in unit’s place,
A + B = 9 …(1)
Now by this condition we know that sum of 2 digits can be greater than 18.
Hence, there is no need to carry one in ten’s place.
Now, for ten’s place,
2 + A = 0
Which means A = -2 which is never possible
Hence, 2 + A > 9
So, we carry one in hundred’s place and hence subtract 10 in ten’s place.
∴ Solving for ten’s place,
2 + A – 10 = 0
∴ A = 8
Now, substituting in 1,
A + B = 9
B = 9 – 8
∴ B = 1
Hence, A = 8 and B = 1