Number 136 is added to 5B7 and the sum obtained is 7A3, where A and B are integers. It is given that 7A3 is exactly divisible by 3. The only possible value of B is

Answer is (d).

It is given 136 + 5B7 = 7A3.

Add the unit’s numbers to get 6 + 7 = 13. So, carry over 1.

=> 1 + 3 + B = 1A => 1 + 3 + B = 10 + A => B – A = 6.

Which means if A = 0, B = 6; if A = 1, B = 7, if A = 2, B = 8 and if A = 3, B = 9.

But given 7A3 is completely divisible by 3. So, as per rules of divisibility, 7 + A + 3 = 10 + A should also be completely divisible by 3.

So the possible values of A are 2, 5 and 8. (12 / 15 / 18 divisible by 3) Out of these, only 2 satisfies both the conditions so A = 2, so B = 8. Ans.(d)