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?
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The most restrictive equations are □ × □ = □ and (equivalently) □ ÷ □ = □ .
There are only the following possibilities where we use distinct digits:
2 × 3 = 6
2 × 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 × 2 = 8 , 6 ÷ 2 = 3
□ + 6 = □ + ÷ 4 × 2 = 8 = = □ 3
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.
1 + 6 = 7 + ÷ 4 × 2 = 8 = = 5 3
Case 1: 3 × 2 = 6 , 8 ÷ 2 = 4
□ + 8 = □ + ÷ 3 × 2 = 6 = = □ 4
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.