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$\begin{array}{ccc}6&7&4\\ 4&11&14\\ 13&2&11\\ 6&8&1\\ 15&3&?\end{array}$

Own attempt at building a puzzle, find the missing number.

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Here's the closest I get

13

Reasoning

if we sum the numbers in an X Formation, they turn out to be equal. Like: \begin{array}{cccccc}6&&7&&4\\& + &&& \\4&&11&&14\\&&& + & \\13&&2&&11 & =28\\\\6&&8&&1\\\\15&&3&&?\end{array}- \begin{array}{cccccc}&6&&7&&4\\&&&&+& \\&4&&11&&14\\&&+&&&\\28 = &13&&2&&11\\\\&6&&8&&1\\\\&15&&3&&?\end{array} and if we use the same rule on the lower X formation like this \begin{array}{cccccc}&6&&7&&4\\\\&4&&11&&14\\\\&13&&2&&11&\\&&&&+&\\&6&&8&&1\\&&+&&&\\34 =&15&&3&&?\end{array}- \begin{array}{cccccc}6&&7&&4\\\\4&&11&&14\\\\13&&2&&11&\\&+&&&\\6&&8&&1\\&&&+&\\15&&3&&?& = 34 \end{array} so "?" should be 13 to maintain this rule

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    $\begingroup$ Doesn't seem to hold in general: for the X in the center $4+2+1 \neq 6+2+14$ $\endgroup$ – elias Jul 12 at 10:27
  • $\begingroup$ Yeah, I'm aware but this question didn't get any attempt so i think it's either too hard or too broad (the solution is not going to satisfy many). I still wanted to give it a try this way $\endgroup$ – Emre Ünsal Jul 12 at 10:45

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