Since \(24 = 8 \times 3\), the number must be divisible by both 8 and 3.

For divisibility by 8, the last three digits \(76B\) must be divisible by 8. Testing values, \(760\) and \(768\) are divisible by 8, so \(B = 0\) or \(B = 8\).

For divisibility by 3, the digit sum \(3 + 2 + A + 7 + 7 + 6 + B = 25 + A + B\) must be a multiple of 3.

Take \(B = 0\): then \(25 + A\) must be a multiple of 3, and the smallest valid digit is \(A = 2\) (giving 27), so \(A + B = 2\).

Take \(B = 8\): then \(33 + A\) must be a multiple of 3, and the smallest is \(A = 0\), giving \(A + B = 8\).

The minimum of these is \(A + B = 2\).

Hence, the minimum value of \((A + B)\) is 2.