I was sent this puzzle by a friend who's recently attended an escape room, but it's really not coming to me.

I wondered if anyone had seen similar puzzles and could suggest an answer?enter image description here


1 Answer 1


Each tree is built by

Performing mathematical operations on two side-by-side numbers, which results in the number above them

Left Tree

x^2+y^2, e.g. 20^2+13^2=569. Missing number is 3

Middle Tree

x/y, e.g. 18/6=3. Missing number is 3.

Right Tree

x^2-y^2, e.g. 9^2-4^2=65. Missing number is 7

  • $\begingroup$ Hi @ Lukas. Are you saying the top of the middle tree = 3? Also with the third tree, the operation conducted on tree 1 doesnt work. Therefore, the operation of 65^2 + 7^2 does not equal 4176 $\endgroup$
    – EB3112
    Dec 19, 2023 at 14:51
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    $\begingroup$ Operations are different for each tree, not row. The entire third tree uses x^2 - y^2 $\endgroup$ Dec 19, 2023 at 15:13
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    $\begingroup$ Fair fair. Thanks for all your feedback @Lukas and @@ Leppy. I've personally never been to an escape room, but I didn't think exponentiation would be expected. $\endgroup$
    – EB3112
    Dec 19, 2023 at 15:19
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    $\begingroup$ @EB3112 If that was a professional escape room that functions as a business for the general public then I would say it's very unexpected, but if it was an escape room made by amateurs for people that they know will enjoy it then everything is possible. Also, it's usual in escape rooms to be in communication (for instance by talkie-walkie) with a host who can give hints if a particular puzzle is too hard; and it's possible they had other hints in the room, like a pocket calculator or a multiplication table. $\endgroup$
    – Stef
    Dec 19, 2023 at 23:14
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    $\begingroup$ I'm pretty sure this is the intended solution, well done for spotting it. But with my mathematicians hat on I feel this puzzle is massively underdetermined unless there are some significant hints added in. One can probably find lots of other solutions that make use of that many free choices. So if this is all there is I would consider this a very badly designed puzzle. $\endgroup$
    – quarague
    Dec 20, 2023 at 7:51

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