# All the Signs point to the Second Prime…Figure out the Missing Digits

$$Given$$

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.

• 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 "?-(?+?)" – hexomino Jun 13 at 9:30
• Is there a unique solution and are all the values unique? – Bee Jun 13 at 9:31
• Order of operations ..A, B, C ...3 Horizontal. D, E, F...3 Vertical...Bottom right 4 squares...blank..like other black cells. – Uvc Jun 13 at 9:36
• Based on what I could figure out, it is unique – Uvc Jun 13 at 9:38
• 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. – Uvc Jun 13 at 9:50

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.

• This morning, my mind is not sharp..forgot to mention that there are no repeating digits...will edit to reflect that.. – Uvc Jun 13 at 9:55
• Got it !!...sorry about all those edits – Uvc Jun 13 at 10:04
• @Uvc No, that's fine. Nice puzzle! – hexomino Jun 13 at 10:10

Here's a solution:

 1 * 3 - 0  = 3
*   -   -
3 + 0 / 1  = 3
-   *   +
0 - 1 + 4  = 3
=   =   =
3   3   3

• I forgot to mention..that only 1 to 9 are allowed...will edit it now..sorry about that – Uvc Jun 13 at 9:46