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enter image description here

Since $23$ and $7$ both are primes, so I am try to make $23$ to $14$ by moving only one match stick. But I am unable to do this. Any hints will be appreciated. Here note that the line divided by denominator and numerator is also made with three matchstick.

I am new in puzzling stackexchange. So I am sorry if the the problem is trivial and if it is required to add any tag or change please edit the question.


marked as duplicate by ABcDexter, JonMark Perry, NL628, greenturtle3141, Ankoganit Mar 9 '18 at 4:54

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  • 4
    $\begingroup$ Sometimes the "trick" to these is just using one matchstick to make . Is that against the rules? $\endgroup$ – Dan Russell Mar 8 '18 at 16:04
  • $\begingroup$ I don't know. I saw the problem in school's puzzle book. But I am understood your trick. $\endgroup$ – SAHEB PAL Mar 8 '18 at 16:08
  • 1
    $\begingroup$ Is taking one stick off the second X too simple? $\endgroup$ – Snow Mar 8 '18 at 16:12

Maybe not the answer, but it is so sexy it probably is:

move 1 match from the XXIII (23) and put it on top of the II (2) to make the famous Mathematical coincidence that 22/7 is roughly equal to Pi (π)

  • $\begingroup$ bhavinionline.com/2016/05/… - was about to post this $\endgroup$ – Joe Mar 8 '18 at 16:12
  • $\begingroup$ Upvote for being so sexy $\endgroup$ – Alex Mar 8 '18 at 18:03
  • $\begingroup$ It's not that much of a "coincidence". The difference is one part in 2500. There are 1000 different ways to divide a 2 digit number by a 1 digit number. $\endgroup$ – Acccumulation Mar 8 '18 at 19:51
  • $\begingroup$ This is actually the first thing I thought of, except that OP wants 14. I feel like this is the intended solution, though. $\endgroup$ – Xenocacia Mar 9 '18 at 2:18
  • 2
    $\begingroup$ While this may be the intended answer, it's not a correct answer, as $22/7 = \pi$ is NOT a correct statement, and hence doesn't "correct" the equation at all. $\endgroup$ – AlexanderJ93 Mar 9 '18 at 2:59

Is this cheating?

Possible solution
Take the second match on the RHS, break it in 3 pieces, and create three minus signs.

  • $\begingroup$ I consider this perfectly within the confines of the rules defined and a much better solution than the top voted pi non-equation. $\endgroup$ – Amit Naidu Mar 9 '18 at 3:40

A possibility, depending on how you interpret an arrangement of matchsticks:

The actual arrangement is here: enter image description here

Parts of this layout are, um, ambiguous, to say the least. Here's how I would interpret the layout (without actually moving the other matchsticks): enter image description here

Converting this equation to MathJax, we have:

$$ \frac{10}{5} \times \frac{3}{2} = 3 $$ $$ 2 \times \frac{3}{2} = 3 $$ $$ 3 = 3 $$


If a sloppy-looking and technically-written-wrong answer is allowed, you could

Move one matchstick from the second X in XXIII to join the line of division, producing XIIII/VII=II


The division line looks sloppy (unless you lay the match on top, in 3D space, of one of the existing matches?), the first I in XIIII is slanted, and XIIII is not a technically correct roman numeral (it should instead be written as XIV)

  • $\begingroup$ Nothing says you can't move the one matchstick completely out of the puzzle. Of course, at that rate, nothing says you can't move a matchstick that's not in the puzzle now into it.... $\endgroup$ – RDFozz Mar 9 '18 at 20:38

Make the denominator into XII. Everybody knows $\frac{23}{12}=2$.

  • 2
    $\begingroup$ Quick maths xD. $\endgroup$ – Gustavo Gabriel Mar 8 '18 at 19:29
  • $\begingroup$ Actually thats not true, 24/12 = 2 $\endgroup$ – watchme Mar 8 '18 at 20:00
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    $\begingroup$ I think JonMark Perry forgot to say "...for sufficiently small values of 12 and 2". $\endgroup$ – Rubio Mar 8 '18 at 20:12

ok a bit convoluted but how about....

enter image description here

on the bottom IX = 9 so maybe IXII = 11

rather convoluted and not as good as some of the other answers, but worth a try....

and as pointed out in the comment below the number on the bottom could be |XI|- the modulus of 11....

  • $\begingroup$ IXII = 11 Request Denied. $\endgroup$ – Amit Naidu Mar 9 '18 at 3:47
  • $\begingroup$ Wait, on second thought - instead of reading that as 9+2 as you implied, we could read that as the absolute value of XI. Close, but the left stick is too bent though \XI|. Hmm, unsatisfactory, but provisionally approved pending appeal. $\endgroup$ – Amit Naidu Mar 9 '18 at 3:55
  • $\begingroup$ @AmitNaidu - great idea! I will edit $\endgroup$ – tom Mar 9 '18 at 9:23

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