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