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

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


Here's the closest I get



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

  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.