Looking at the last digit: A × 8 must end in 4. The possibilities for A are 3 (3×8=24) or 8 (8×8=64). If A=3, the problem is 633 × 8 = 5064. Here B=6. A and B are distinct (3 and 6). This works. If A=8, the problem is 688 × 8 = 5504. Here B=0. A and B are distinct (8 and 0). Let's recheck the problem source... Assuming 633x8=5064, B is 6.