In the units column, A x 8 must end in a 4. This means A can be 3 ($3 \times 8 = 24$) or A can be 8 ($8 \times 8 = 64$). Case 1: If A=3. The multiplication is $633 \times 8$. $3 \times 8 = 24$ (4, carry 2). $3 \times 8 + 2 = 26$ (B=6, carry 2). $6 \times 8 + 2 = 50$. This gives 5064. So A=3 and B=6. This is a valid solution. Case 2: If A=8. The multiplication is $688 \times 8$. $8 \times 8 = 64$ (4, carry 6). $8 \times 8 + 6 = 70$ (B=0, carry 7). $6 \times 8 + 7 = 55$. This gives 5504. This doesn't match the 50B4 pattern. So, the only solution is A=3, B=6.