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This question already has an answer here:

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.

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 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
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    $\begingroup$ Is taking one stick off the second X too simple? $\endgroup$ – Snow Mar 8 '18 at 16:12
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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)

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  • $\begingroup$ 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
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    $\begingroup$ 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
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Is this cheating?

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

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  • $\begingroup$ 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
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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:

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

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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

But,

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)

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

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    $\begingroup$ 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
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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....

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  • $\begingroup$ 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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