$Given$
The six Interrelated Equations A, B, C, D, E, F
All equal to 3.
Allowed Digits.. 1 to 9.
All the missing 9 are distinct...9?s...9 Different Digits
Fillin all the digits that satisfy all these Relations.
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$\begingroup$ On the ordering of operations, do we assume multiplication and division are performed before addition and subtraction? Also, is the bottom row and rightmost column "(?-?)+?" or "?-(?+?)" $\endgroup$ – hexomino Jun 13 at 9:30
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$\begingroup$ Is there a unique solution and are all the values unique? $\endgroup$ – Bee Jun 13 at 9:31
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$\begingroup$ Order of operations ..A, B, C ...3 Horizontal. D, E, F...3 Vertical...Bottom right 4 squares...blank..like other black cells. $\endgroup$ – Uvc Jun 13 at 9:36
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$\begingroup$ Based on what I could figure out, it is unique $\endgroup$ – Uvc Jun 13 at 9:38
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1$\begingroup$ Order of operations as given would be..3+3=6,,,6x2=12....follow the operations in the order given in both horizontal and vertical sets. $\endgroup$ – Uvc Jun 13 at 9:50
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I've found a solution, which is unique given the constraints.
1 * 9 - 6 = 3 * - - 7 + 8 / 5 = 3 - * + 4 - 3 + 2 = 3 = = = 3 3 3
Proof
The digit in the bottom middle must be either $1$ or $3$ since the product of the second column is $3$.
If it is $1$ then the other entries in the bottom row are either $2,2$ or $1,3$ but we are not allowed to repeat digits. Hence, it must be $3$.
This means the other digits in the bottom row will be $2,4$ or $1,5$.
Now, suppose $1$ is on the left in the bottom row. This means the first product on the left-hand column is $4$ which cannot be constructed without repeating digits.
Alternatively, suppose $5$ is on the left in the bottom. Then the other digits in the left-hand column must be $2$ and $4$ since $1$ has already been used. This means the last entry in the second row is at least $6$ which is not permissible since the sum cannot be greater than $17$.
Hence, the other two entries in the bottom row must be $2$ and $4$.
Now suppose $2$ is on the left. This means that the other entries in the left-hand column are $1$ and $5$ which forces the last entry in the second row to be at least $6$, not allowed. Hence the bottom row must be $4,3,2$.
From there, we know that the other entries in the left-hand column are $1$ and $7$.
This means the last entry in the second row must be $5$ since it is the only option available that could possibly work. Hence the sum of the first two digits in the second row is $15$ which makes the first entry $7$ and the second $8$.
This means the top left-hand corner must be $1$ and we are left with $9$ and $6$ to complete the grid.
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$\begingroup$ This morning, my mind is not sharp..forgot to mention that there are no repeating digits...will edit to reflect that.. $\endgroup$ – Uvc Jun 13 at 9:55
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Here's a solution:
1 * 3 - 0 = 3 * - - 3 + 0 / 1 = 3 - * + 0 - 1 + 4 = 3 = = = 3 3 3
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$\begingroup$ I forgot to mention..that only 1 to 9 are allowed...will edit it now..sorry about that $\endgroup$ – Uvc Jun 13 at 9:46
