Correct Option

The given equation is 136 + 5B7 = 7A3, where A and B are integers (single digits from 0-9).

First, analyze the condition that 7A3 is exactly divisible by 3. For a number to be divisible by 3, the sum of its digits must be divisible by 3. Therefore, (7 + A + 3) must be divisible by 3, which simplifies to (10 + A) being divisible by 3.

Possible single-digit values for A that satisfy this condition are:
- If A = 2, then 10 + 2 = 12 (divisible by 3).
- If A = 5, then 10 + 5 = 15 (divisible by 3).
- If A = 8, then 10 + 8 = 18 (divisible by 3).

Next, analyze the addition column by column:

136

+ 5B7

-----

7A3

Units column: 6 + 7 = 13. This implies the units digit of the sum is 3, and there is a carry-over of 1 to the tens column.
Hundreds column: 1 + 5 = 6. However, the hundreds digit of the sum is 7. This indicates that there must be a carry-over of 1 from the tens column to the hundreds column.
Tens column: The sum of the digits in the tens column, including the carry-over from the units column, must result in A as the units digit and a carry-over of 1 to the hundreds column.
So, 3 (from 136) + B (from 5B7) + 1 (carry-over from units) = A + 10 (to account for the carry-over of 1 to hundreds).
This simplifies to 4 + B = 10 + A, or B = 6 + A.

Now, substitute the possible values of A (2, 5, 8) into the equation B = 6 + A:

If A = 2: B = 6 + 2 = 8. This is a valid single-digit value for B.
If A = 5: B = 6 + 5 = 11. This is not a valid single-digit value for B, as B must be a single digit (0-9).
If A = 8: B = 6 + 8 = 14. This is not a valid single-digit value for B.

Therefore, the only possible value for B is 8, which occurs when A = 2.

Verification: If B = 8 and A = 2, then 136 + 587 = 723. The sum of digits of 723 is (7 + 2 + 3) = 12, which is divisible by 3. This confirms the solution.

Incorrect Options:

If B = 2:
Using the relation B = 6 + A, we get 2 = 6 + A, which implies A = -4. This is not a valid single-digit integer for A.
Alternatively, 136 + 527 = 663. The sum 663 does not match the hundreds digit of 7A3.
If B = 5:
Using the relation B = 6 + A, we get 5 = 6 + A, which implies A = -1. This is not a valid single-digit integer for A.
Alternatively, 136 + 557 = 693. The sum 693 does not match the hundreds digit of 7A3.
If B = 7:
Using the relation B = 6 + A, we get 7 = 6 + A, which implies A = 1.
If A = 1 and B = 7, the sum would be 136 + 577 = 713.
However, for 713 to be divisible by 3, the sum of its digits (7 + 1 + 3 = 11) must be divisible by 3. Since 11 is not divisible by 3, this option is incorrect.