$\LARGE \begin{array} { l l l l l } \square & + & \square & = & \square \\ + & & \times & & \\ \square & \div & \square & = & \square \\ \hspace{1mm} {\small ||} & & \hspace{1mm} {\small ||} & & \\ \square & & \square \\ \end{array}$

Fill in the 8 boxes with distinct digits, such that all 4 equations are true.

Clearly, we cannot use the digit 0. What other digit is not used?

1
3
5
7
9

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The most restrictive equations are $\square \times \square = \square$ and (equivalently) $\square \div \square = \square$ .

There are only the following possibilities where we use distinct digits:

$2 \times 3 = 6$

$2 \times 4 = 8$ .

Hence, this tells us that the middle square must be the number that appears twice in these equations, which is 2. We have 2 possible ways to fill in the squares. Let's consider them seperately:

Case 1:$4 \times 2 = 8, 6 \div 2 = 3$$\begin{array} { l l l l l } \square & + & 4 & = & \square \\ + & & \times & & \\ 6 & \div & 2 & = & 3 \\ = & & = & & \\ \square & & 8 \\ \end{array}$

We have the numbers $1, 5, 7, 9$ remaining. By considering the vertical summation, $5, 7, 9$ are too larger for the top left square. If we place 1 in, we see that the other numbers we use are 5 and 7, which are allowed. In this case, the number that isn't used is 9.

$\begin{array} { l l l l l } 1 & + & 4 & = & 5 \\ + & & \times & & \\ 6 & \div & 2 & = & 3 \\ = & & = & & \\ 7 & & 8 \\ \end{array}$

Case 1:$3 \times 2 = 6, 8 \div 2 = 4$$\begin{array} { l l l l l } \square & + & 3 & = & \square \\ + & & \times & & \\ 8 & \div & 2 & = & 4 \\ = & & = & & \\ \square & & 6 \\ \end{array}$

We have the numbers $1, 5, 7, 9$ remaining. By considering the vertical summation, $5, 7, 9$ are too larger for the top left square. If we place 1 in, we see that the other numbers we use are 9 and 4. However, 4 would be duplicated and so this case doesn't lead to a solution.