Correct the equation by moving $1$ match stick [duplicate]

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.

• Sometimes the "trick" to these is just using one matchstick to make ≠. Is that against the rules? Mar 8 '18 at 16:04
• I don't know. I saw the problem in school's puzzle book. But I am understood your trick. Mar 8 '18 at 16:08
• Is taking one stick off the second X too simple?
– user30599
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)

• bhavinionline.com/2016/05/… - was about to post this
– Joe
Mar 8 '18 at 16:12
• Upvote for being so sexy
– Alex
Mar 8 '18 at 18:03
• 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. Mar 8 '18 at 19:51
• This is actually the first thing I thought of, except that OP wants 14. I feel like this is the intended solution, though. Mar 9 '18 at 2:18
• 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. Mar 9 '18 at 2:59

Is this cheating?

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

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

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

The actual arrangement is 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):

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

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)

• 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.... Mar 9 '18 at 20:38

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

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

ok a bit convoluted but how about....

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....

• IXII = 11 Request Denied. Mar 9 '18 at 3:47
• 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. Mar 9 '18 at 3:55
• @AmitNaidu - great idea! I will edit
– tom
Mar 9 '18 at 9:23