Find the missing number, and show how you found it.
closed as off-topic by F1Krazy, Rupert Morrish, Glorfindel, boboquack, SteveV Jan 18 at 0:45
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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:
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:
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:
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:
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.