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Fill in the numbers 1 through 9 in the 9 empty cells (each number appears exactly once) to make the row and column expressions equal to the number at the end of the row or column.
Regular math operations rules apply; multiplication and division before addition and subtraction.
A solid B means it is a blank space in the table- no numbers go there.
Answers should explain a sequence of logical steps to arrive at the solution.
My relative sent it to me, I don't know where they got it.

B B B B B B
+ - 12
× B + B + B
+ + 12
- B × B - B
+ + 11
23 B 37 B 1 B
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Here is a solution inspired by Beastly Gerbil's observation in a comment that:

If we label the numbers A through I in reading order, then (A+B-C)+(D+E+F)+(G+H+I) is 12+12+11=35 by the conditions on the rows; on the other hand, A+B+C+D+E+F+G+H+I=45, the sum of all 9 values. Subtracting the first equation from the second, we get 2C=10, or C=5.

From here, we can finish with almost no casework:

The first row now tells us A+B-5=12, or A+B=17, so A and B are 8 and 9 in some order. Setting B=8 means that E×H=37-8=29 in the second column, which is impossible, so A=8 and B=9. Then E×H=28, which means E and H are 4 and 7 in some order.

From the first column, 8×D-G=23. D must be at least 3 for 8×D to be bigger than 23, but if D is 4 or more, then G=8×D-23 is too big. So D=3 and G=1.

If E=4 and H=7, then G+H+I=11 means I=4, which was already used. So E=7 and H=4. G+H+I=11 gives I=6 and D+E+F=12 gives F=2, so we have the complete solution: A=8, B=9, C=5, D=3, E=7, F=2, G=1, H=4, I=6.

We didn't even use the condition in the third column, though we can check that it's satisfied at the end.

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5
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The number $9$

  • cannot be in the bottom row (only $9,1,1$ can sum to 11),

  • cannot be in the middle row (only $9,2,1$ can sum to 12, and $1$ cannot be in any of the three positions in this row),

  • cannot be in the top right (that would mean two digits summing to 21 in the top row),

  • so it must be in the top left or top centre.

Option 1:

$9$ is in the top left, so the middle left is $3$ and the bottom left is $4$.
The top centre and top right are two digits differing by three, so they must be either $5,2$ or $8,5$, so $5$ must be in the top row.
The middle centre and middle right are two numbers adding to nine, so they must be either $1,8$ or $2,7$ (in some order).
The bottom centre and bottom right are two numbers summing to seven, so they must be $1,6$ in some order.
Therefore the middle row has $2,7$ in some order and the top row has $8,5$. But $8$ at the top centre gives a contradiction.

Option 2:

$9$ is in the top centre, so the middle centre and bottom centre are $4,7$ (in some order).
The top left and top right are two numbers differing by three, so they must be either $8,5$ or $6,3$ or $5,2$. There's no possibilities for either $5$ or $6$ in the top left, so the top left is $8$ and the top right is $5$.
So the middle left is $3$ and the bottom left is $1$. Then the middle right is $2$ and the bottom right is $6$.
Checking again the centre column, we have $7$ in the middle centre giving a sum of 12, and $4$ in the bottom centre giving a sum of 11.

So the full solution is

8 + 9 - 5 = 12
× + +
3 + 7 + 2 = 12
- × -
1 + 4 + 6 = 11

23 37 1

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  • 5
    $\begingroup$ You can place the 5 top right really easily, 1 to 9 sums to 45, and the sums on the right add up to 35. As top right is the only thing 'subtracted' it must be a 5 $\endgroup$ – Beastly Gerbil Jan 20 at 22:28
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    $\begingroup$ @BeastlyGerbil Nice spot! And of course that immediately gives the rest of the top row, and then the rest as in my final spoilertag here. I just kinda brute-forced the possibilities in my head :-) $\endgroup$ – Rand al'Thor Jan 20 at 22:29
  • $\begingroup$ The 5 was the only thing I managed to work out, I feel like brute force is usually quicker with things like this :P $\endgroup$ – Beastly Gerbil Jan 20 at 22:30
  • $\begingroup$ @BeastlyGerbil Your trick with the 5 gives a very quick solution! I've written up an answer that continues it. $\endgroup$ – Misha Lavrov Jan 21 at 0:12
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Here's another way to solve this puzzle. We can actually proceed one digit at a time, and the middle column product trick (see below) seems way too neat to be a coincidence.

Step 1: Where can the 1 go?

  • It can't be in the rightmost column (it's too small to be subtracted, and if it's added, the other two digits in the column would have to be equal to each other)
  • It can't a part of either product (can't reach the high totals), and
  • It can't be in the top row's addition (can't reach 12 in the row).

So the

1 is in the bottom left corner.

   _ + _ - _ = 12
   x   +   +
   _ + _ + _ = 12
   -   x   -
   1 + _ + _ = 11
   =   =   =
  23  37   1
 

Step 2: The three rows contain only additions and a single subtraction. The row totals add up to 35, and the numbers 1-9 add up to 45, so the subtracted number

must be a 5.

   _ + _ - 5 = 12
   x   +   +
   _ + _ + _ = 12
   -   x   -
   1 + _ + _ = 11
   =   =   =
  23  37   1
 

Step 3: The product in the middle column is now very limited. We know it must be a number between 28 and 36. But

  • 29 and 31 are prime
  • 30 and 35 would need a 5 in the product
  • 32 would need a 5 in the addition
  • 33 would need an 11 in the product
  • 34 would need a 17 in the product, and finally
  • 36 would need a 1 in the addition.

Therefore, the product must be 28, which places the

9 at the top of the middle column, and the other two digits in the column are 4 and 7 in some order.

   _ + 9   - 5 = 12
   x   +     +
   _ + 4/7 + _ = 12
   -   x     -
   1 + 7/4 + _ = 11
   =   =     =
  23  37     1
 

The left column is now easily solved (there are equations with only one number missing) so we know that the remaining two digits at the rightmost column are 2 and 6, and that column's equation fixes the order of the rest of the digits:

   8 + 9 - 5 = 12
   x   +   +
   3 + 7 + 2 = 12
   -   x   -
   1 + 4 + 6 = 11
   =   =   =
  23  37   1
 

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  • $\begingroup$ Can you justify "Step 1: Where can the 1 go?... can't be in the top row's addition (can't reach 12 in the row)". The top row could be 8+3+1 or 7+4+1 or 6+5+1 or various others in some order, so there's some small step missing. It seems far easier to try to narrow down the 9, like others are doing. $\endgroup$ – smci Jan 21 at 21:45
  • $\begingroup$ @smci those are not the operators on the first row. $\endgroup$ – Bass Jan 21 at 21:46
  • $\begingroup$ Ok then you meant "top row's expression" not "addition". $\endgroup$ – smci Jan 21 at 21:47
  • $\begingroup$ @smci No, I didn't. There is an addition on the top row, and that is exactly the part of the top row's expression that cannot contain the 1. Hope this clears up your confusion. $\endgroup$ – Bass Jan 21 at 22:05
  • $\begingroup$ What you wrote is at best very confusing: without inspecting the table, we wouldn't know "the top row's addition" refers to only the first two digits X,Y in "X+Y-Z" expression, not "X+Y+Z". Like I said, too many extra steps required vs brute-force locating the '9'. $\endgroup$ – smci Jan 21 at 22:10

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