To solve the problem of finding the sum A + B where the number \(9A7B\) is divisible by 55, we need to understand the divisibility rule for 55. A number is divisible by 55 if it is divisible by both 5 and 11.

Divisibility by 5: A number is divisible by 5 if its last digit is 0 or 5. Therefore, \(B\) should be 0 or 5. Let's consider \(B = 5\) as this satisfies divisibility by 5.

Divisibility by 11: A number is divisible by 11 if the difference between the sum of its digits at odd positions and the sum of its digits at even positions is a multiple of 11.

For the number 9A7B:

• Odd-positioned digits: 9 and 7
• Even-positioned digits: A and 5

To satisfy the equation, we need:

\(11 - A = 0 \Rightarrow A = 11\)

However, \(A\) must be a singular digit. Adjust \(B\) so \(11 - A = \text{multiple of } 11\), such that it keeps within a valid digit scope.

Using \(B = 5\) previously determined:

For \(A = 0\), the expression becomes: \(16 - (0 + 5) = 11\), satisfying the divisibility rule for 11.\)

Thus, \(A = 0\) and \(B = 5\).

Therefore, \(A + B = 0 + 5 = 5\).

Conclusion: The sum A + B is \(5\).