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Can somebody please solve this? My daughter's school teacher gave her this puzzle to solve at home. But to me it seems a little out of order, and that's why I am asking here for help.

it is a difficult to solve math problem please help required

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    $\begingroup$ 1) Where did you get this from? Attribution is required for all outside puzzles. 2) Is order of operations preserved? $\endgroup$ – Avi Oct 16 at 20:44
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    $\begingroup$ this looks like algebra homework. You could easily put letters in the appropriate blanks, and render it down to simple (if involved) algebra. $\endgroup$ – Ben Barden Oct 16 at 20:53
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    $\begingroup$ My daughters school teacher gave her this puzzle to solve at home but to me also it seems a little out of order thats why i asked for help here.. $\endgroup$ – Rizwan Asghar Oct 16 at 20:56
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    $\begingroup$ Hi @RizwanAsghar, welcome to Puzzling SE! Take the tour if you haven't already! I edited in the information you provided in your comment above, since there was a close vote for not providing attribution for your question. In the future, please ensure you provide sources on all problems and puzzles you did not create yourself. Thanks! $\endgroup$ – PiIsNot3 Oct 16 at 22:58
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I think the answer is as folows

enter image description here

Reasoning

Usually, in these types of puzzles (when there are nine slots) there is an extra restriction of filling in each digit from 1-9 exactly once. With that in mind, the middle column must automatically have both 2 and 5. Then, for the top row to work, we must have the 2 at the top and the 5 at the bottom and, assuming we perform row operations left to right, it must be that we have 3 + 2 x 4 = (3 + 2) x 4 = 20 in the top row.
From there, the middle row must contain 1 and 6 and if 1 is on the left then 4 must be in the lower left corner. But, since we've already used 4, it must be that 6 is on the left and 1 on the right. The rest of the grid then follows quickly.

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  • $\begingroup$ 3 + 2 x 4 = 11 unless the teacher wants the students to ignore order of operations. $\endgroup$ – Engineer Toast Oct 17 at 12:20
  • $\begingroup$ @EngineerToast An annoying convention of these puzzles is that they fly in the face of BODMAS. $\endgroup$ – Paul Evans Oct 17 at 13:25
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Assuming the order of operation is preserved (this is significant only for the first line) and all numbers must be integers

there are 13 different solutions (however, all of them but one involve at least one negative number), as follows (numbers are from left to right, from top to bottom):


 5       1       15      -2      -3      3       10      -8
 -5      -1      -25     -2      -3      -7      -10     32
 10      2       5       -2      -3      8       5       2
 -10     -2      -15     -2      -3      -12     -5      22
 25      5       -1      -2      -3      23      2       8
 -25     -5      -9      -2      -3      -27     -2      16
 50      10      -3      -2      -3      48      1       10
 0       1       20      3       2       3       10      -8
 50      10      -3      3       2       53      1       15
 -4      -1      -24     -3      -2      -7      -10     32
 -50     -10     -7      -3      -2      -53     -1      15
 2       2       9       6       1       8       5       2
 -6      -2      -13     -6      -1      -12     -5      22
 

So, the only solution with all numbers positive is

2 + 2 x 9 = 20, 6 x 8 x 1 = 48, 8 x 5 - 2 = 38, 2 + 6 - 8 = 0, 2 x 8 x 5 = 40, 9 - 1 + 2 = 10.

Explanation:

Let A to H be the 8 unknown numbers. We get the following system of equations: $A+BC=20$, $8DE=48$ (or $DE=6$), $FG-H=38$, $A+D-F=0$, $8BG=80$ (or $BG=10$), $C-E+H=10$. Now it's clear that $D$ and $E$ must be either 1 and 6, or 2 and 3 in any order (or the opposite numbers -1 and -6, or -2 and -3). The same can be said about $B$ and $G$ (1 and 10 or 2 and 5, etc.). So we get $8\times8=64$ different possibilities for $B,D,E,G$, so all of them can be quickly bruteforced. Note that $$F=A+D=20-BC+D=20-B(10+E-H)+D=20-B(10+E-FG+38)+D=20-B(E-FG+48)+D.$$
After simplification, we get $F=(-\frac{20+D}{B}+E+48)/(G-\frac1B)$. So if $F$ is integer, we can quickly compute $A$, $C$ and $H$, which must be all integers.
I've written simple Python script which does the job (giving all 13 solutions): Try it online!

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    $\begingroup$ Beautiful! I had the middle column and the bottom row right, but couldn't figure out the top corners and the two squares below those corners. $\endgroup$ – Avi Oct 17 at 13:38
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Assuming order of operations is not perserved (multiplication and addition occur from left to right, top to bottom in the column/row), the solution is:

 1 + 1  x 10|20
 +   x    - |
 3 x 8  x 2 |48
 -   x    + |
 4 x 10 - 2 |38
 -----------+
 0   80   10
 

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    $\begingroup$ Assuming order of operations is preserved, I'm working on it $\endgroup$ – Avi Oct 16 at 20:59
  • $\begingroup$ Thanks a million... You guyz rock.. $\endgroup$ – Rizwan Asghar Oct 16 at 21:03
  • $\begingroup$ puzzling.se users are too smart to be true. this is skynet. $\endgroup$ – George Menoutis Oct 16 at 21:19
  • $\begingroup$ I failed :( Plus, it's not algebraically solvable - 6 equations 8 unknowns $\endgroup$ – Avi Oct 16 at 21:29
  • $\begingroup$ wouldn't less equations than unknowns actually enable infinitely many solutions? $\endgroup$ – George Menoutis Oct 16 at 22:47

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