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

To find the value of B, follow these steps:

1. Analyze the addition: 136 + 5B7 = 7A3. Looking at the units place: 6 + 7 = 13. We write 3 and carry over 1 to the tens place. Looking at the tens place: 1 (carry) + 3 + B = A. This gives us the equation: 4 + B = A. Looking at the hundreds place: 1 + 5 = 6. Since the result is 7A3, there must be a carry over of 1 from the tens place to the hundreds place. This means A must be at least 10. Specifically, 4 + B must result in a value where 1 is carried over, so 1 + 5 + 1 (carry) = 7.

2. Use the divisibility rule for 3: A number is divisible by 3 if the sum of its digits is divisible by 3. For 7A3, the sum is 7 + A + 3 = 10 + A. The possible values for A (where 10 + A is divisible by 3 and A is a single digit from the carry calculation) are 2, 5, or 8.

3. Determine the carry: As established, for the hundreds digit to become 7 (1 + 5 + carry), the tens place calculation (4 + B) must produce a carry. This means 4 + B must be 10 or greater. If A = 2, then 4 + B = 12. This gives B = 8. If A = 5, then 4 + B = 15. This gives B = 11 (not possible as B must be a single digit). If A = 8, then 4 + B = 18. This gives B = 14 (not possible).

Therefore, the only possible value for B is 8. This results in A = 2, making the sum 723, which is divisible by 3.

Correct Answer: D (8)