Find A in the addition, when B=1
A+A+AB A
5
BA can be written as 10B +A, as B is in the tens digit and A in the ones digit.
3A=10×B+A
⇒2A=10×B
⇒B=A5
The above solution satisfies this relation.
as B=1, A will be equal to 5
5+5+51 5
That is, A= 5 and B= 1