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

The correct option is 8.

Let us break down the problem step-by-step to understand why this is the correct answer.

Step 1: Write down the addition operation
We are given that the three-digit number 136 is added to another three-digit number 5B7 (where B represents the tens digit), and their sum is the three-digit number 7A3 (where A represents the tens digit).
This can be written vertically as:
1 3 6
+ 5 B 7
-------
7 A 3

Step 2: Analyze the columns from right to left
First, look at the units (ones) column:
$6+7=13$
This gives a units digit of 3 and a carryover of 1 to the tens column.

Next, look at the tens column, including the carryover of 1:
$1+3+B=4+B$
Since the sum in the hundreds column is $1+5=6$ but the actual hundreds digit in the sum is 7, there must be a carryover of 1 from the tens column to the hundreds column.
Therefore, the sum in the tens column is $10+A$.
So, we can write:
$4+B=10+A$
Simplifying this equation, we get:
$B-A=6$ (or $A=B-6$)

Step 3: Apply the divisibility rule for 3
We are given that the sum 7A3 is exactly divisible by 3.
A number is divisible by 3 if and only if the sum of its digits is divisible by 3.
The sum of the digits of 7A3 is:
$7+A+3=10+A$
For $10+A$ to be divisible by 3, since A is a single-digit integer (from 0 to 9), the possible values for $10+A$ are 12, 15, or 18.
This gives the possible values for A as:
$A=2,5,\text{or}8$

Step 4: Find the value of B
Using our relation $B=A+6$, let us test each possible value of A:
1. If $A=2$, then $B=2+6=8$.
2. If $A=5$, then $B=5+6=11$ (not possible because B must be a single digit).
3. If $A=8$, then $B=8+6=14$ (not possible because B must be a single digit).
Thus, the only possible single-digit value for B is 8.