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. What is the only possible value of B?

Correct Answer: (D) 8

In the addition $136 + 5B7 = 7A3$, the units column gives $6 + 7 = 13$ (carry 1). In the hundreds column, $1 + 5 +$$\text{carry}$ = 7, requiring a carry of 1 from the tens column. Thus $1 + 3 + B = 10 + A$$\implies$ B - A = 6. For $7A3$ to be divisible by 3, the sum of digits $7 + A + 3 = 10 + A$ must be a multiple of 3, so $A$$\in$ \{2, 5, 8\}. Since $B = A + 6$ must be a single digit ($B$$\le$ 9), only $A = 2$ is possible, giving $B = 8$.