8.【Solution】The problem requires forming 5 numbers using seven digits, indicating that three of the numbers are single-digit, and two are two-digit numbers.
It is clear that the box and the dividend are two-digit numbers, while the multiplier and divisor are single-digit numbers.
It is evident that 0 is not suitable as a multiplier and certainly not as a divisor. Therefore, it must be the unit digit of a two-digit number, and thus the unit digit of the dividend.

If the multiplier were 1, the multiplicand would appear twice in the equation regardless of its value. Therefore, the multiplier is not 1. Similarly, the multiplier cannot be 5.

The dividend is the product of three single-digit numbers, one of which is 5, and the other two are not 1 and cannot be 2 (since $2 \times 5 = 10$, which would make the tens digit of the dividend the same as the other multiplier). Therefore, the dividend is at least $3 \times 4 \times 5 = 60$. Since there are no digits greater than 6, the dividend must be 60, and the equation is $3 \times 4 = 12 = 60 \div 5$. Thus, the number in the box is 12.
8. Fill the circles with the numbers $0, 1, 2, 3, 4, 5, 6$ in the grid, with each number appearing exactly once, to form an integer equation consisting of single-digit and two-digit numbers. What number is filled in the grid?
$$
O \times O=\square=O \div O
$$