In the given problem, the multiplication of
A with
A itself gives a number whose ones digit is
A again. This happens only when
A=1,5,6.
If A=1 then the multiplication will be 31×1=31. However, here the tens digit is given as B and hundred digit is 2, therefore, A=1 is not possible.
Similarly, if A=5 then the multiplication will be 35×5=175. However, here the tens digit is given as B and hundred digit is 2, therefore, A=5 is not possible.
Now, if A=6 then the multiplication will be 36×6=216. However, here the tens digit is given as B and hundred digit is 2, therefore, A=6 and B=1 are possible.
Hence, the letter A=6 and B=1.