# Please find the missing number in the question [closed]

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

## closed as off-topic by F1Krazy, Rupert Morrish, Glorfindel, boboquack, SteveVJan 18 at 0:45

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• 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! – Greg Jan 17 at 18:45
• A friend send it to me. – Mursaleen Salroo Jan 17 at 19:16
• The problem is undertermined. You can make the answer X by using the formula: 9(A + 2B + 5C - 15) + X*(-A+B-C+3). – Florian F Jan 19 at 12:55

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.