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Find the missing number, and show how you found it.

enter image description here

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    $\begingroup$ Hi Mursaleen. What is the source of this puzzle? Please include it in the question. Also, please include the picture, not just a link to the picture. Thanks and happy puzzling! $\endgroup$ – eye_am_groot Jan 17 '19 at 18:45
  • $\begingroup$ A friend send it to me. $\endgroup$ – Mursaleen Salroo Jan 17 '19 at 19:16
  • $\begingroup$ The problem is undertermined. You can make the answer X by using the formula: 9(A + 2B + 5C - 15) + X*(-A+B-C+3). $\endgroup$ – Florian F Jan 19 '19 at 12:55
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Let $A$ be the top number, $B$ be the left number, $C$ be the bottom number and $D$ be the right number. The pattern is:

$$2(A+B+C+D+AC+BD)-(A-B)+(D-C)$$

I don't have time to simplify it now, but I guess it looks less random in the way I have put it, and perhaps looks better to the eye. I might look back at this puzzle later in the day. But anyway, let's continue to finish off this answer.

So for the first one, we have:

$A=7$
$B=5$
$C=1$
$D=2$

So using our formula, we should get $63$:

$$2\times\big(7+5+1+2+(7\times 1)+(5\times 2)\big)-(7-5)+(2-1)=63$$

For the second one, we have:

$A=5$
$B=4$
$C=2$
$D=3$

So using our formula, we should get $72$:

$$2\times\big(5+4+2+3+(5\times 2)+(4\times 3)\big)-(5-4)+(3-2)=72$$

It works both times, so let us apply it to the third and final case. We have:

$A=4$
$B=3$
$C=1$
$D=2$

So using our formula, we get:

$$2\times\big(4+3+1+2+(4\times 1)+(3\times 2)\big)-(4-3)+(2-1)=40$$

However, none of the answers seem to be this... even though it works. Is it an error, or there might be a different formula. I am leaving this answer here as it is too long for a comment, and I can perhaps give others ideas to find an answer.

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