Discussion

Answer: C

For a number to be divisible by 88, it must be divisible by both 8 and 11.

1. Divisibility by 8: The number formed by the last three digits, 47B, must be divisible by 8. Let's test values for B. If B=2, we have 472. \(472 \div 8 = 59\). So, B=2 is a valid value.

2. Divisibility by 11: The difference between the sums of alternate digits must be 0 or a multiple of 11. The number is 8A51472. Sum of digits at odd places: \(2+4+5+8 = 19\). Sum of digits at even places: \(7+1+A = 8+A\). The difference is \(19 - (8+A) = 11-A\). For this to be a multiple of 11, the only possibility for a single-digit A is when \(11-A=0\), which gives A=11 (not a digit), or when \(11-A=11\), which gives A=0. So, A=0.

We have A=0 and B=2. The value of B-A = 2-0=2.