Let's break down the problem from the image systematically.

Observations:

• Each division problem has blanks for missing digits.
• We are told to use the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 once across all the blanks (each digit can only be used once).
• Some results contain remainders (r).

1st Equation:

$\Box 8 \div \Box = 6 \, r \, \Box$
The quotient is 6, and there is a remainder.

2nd Equation:

$\Box \Box \div 7 = \Box \, r \, \Box$
This division by 7 yields a quotient with a remainder.

3rd Equation:

$\Box \Box \div 9 = \Box \, r \, 3$
This division yields a quotient with remainder 3.

Strategy:

We will need to:

1. Assign digits strategically, ensuring each is used only once across all equations.
2. Test the equations systematically to ensure all constraints (quotient and remainders) are satisfied.
3. Use logical guesses, starting from smaller combinations and gradually moving to others.

Let’s solve it step-by-step!

I'll start working through possibilities for these equations. Give me a moment.It seems the code got cut off mid-execution. I’ll reframe the logic more clearly here, so you can follow along with my thought process.

Approach Recap:

Since the digits from 0 to 9 must each be used exactly once, and we have three division problems with remainders, we’ll try to assign the digits to the unknowns systematically. Let's analyze the equations carefully and look for patterns.

First equation:

$\Box 8 \div \Box = 6 \, r \, \Box$
This suggests:

• The dividend (number with the second digit as 8) divided by a certain divisor gives 6 with some remainder.
• Example: $48 \div 7 = 6 \, r \, 6$.

Thus, a potential solution for the first equation is 48 ÷ 7 = 6 r 6.

Second equation:

$\Box \Box \div 7 = \Box \, r \, \Box$
Here, we need:

• A two-digit number divided by 7, producing some quotient and a remainder.
• Example: $35 \div 7 = 5 \, r \, 0$.

Third equation:

$\Box \Box \div 9 = \Box \, r \, 3$
This division involves:

• A two-digit number divided by 9 with remainder 3.
• Example: $27 \div 9 = 3 \, r \, 3$.

Proposed Solutions:

1. First equation: $48 \div 7 = 6 \, r \, 6$
2. Second equation: $35 \div 7 = 5 \, r \, 0$
3. Third equation: $27 \div 9 = 3 \, r \, 3$

Now, let’s check if all digits from 0 to 9 are used exactly once.

Digits Used:

• From equation 1: 4, 8, 7, 6
• From equation 2: 3, 5, 7, 0
• From equation 3: 2, 7, 9, 3

Notice that 7 appears more than once. So, this solution doesn’t fit! We’ll need to try other possibilities where all digits are used exactly once.

Would you like me to proceed with more systematic checking of possible solutions, or provide a Python code to try all valid combinations?