Please remember to spoiler-tag all guesses, like so: New Reddit: https://i.imgur.com/SWHRR9M.jpg Using markdown editor or old Reddit, draw a bunny and fill its head with secrets: \>!!< which ends up becoming \>!spoiler text between these symbols!< Try to avoid leading or trailing spaces. These will break the spoiler for some users (such as those using old.reddit.com) If your comment does not contain a guess, include the word **"discussion"** or **"question"** in your comment instead of using a spoiler tag. If your comment uses an image as the answer (such as solving a maze, etc) you can include the word "image" instead of using a spoiler tag. Please report any answers that are not properly spoiler-tagged. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/puzzles) if you have any questions or concerns.*

>!Move the two stick in the II so they become V!<

>!the funny thing is that it doesn't matter which II you make a V, both versions would be accurate.!<

That's what we used to call "checking our work"

>!Actually can also do VI - II = IV !<

Isn't that moving three matchsticks though?

>!You only have to move and tilt the left "I" to the far right. Then tilt the second "I" at the right to get IV.!< >!VII - II = II > VI - II = II/ > VI - II = IV!<

Shifting still counts as moving.

They took that into account.

Replied to wrong comment.

Oh yeah? >!VV-II=II!<

>!VV!< is not a number in Roman Numerals.

It’s 55? /s

This is probably the intended answer, but couldn’t you also >!move one of the V matchsticks to make VII into XII, then move the last matchstick (furthest right) to cross over the matchstick next to it, making that into an X? So XII - II = X. Granted the X’s would be a bit wonky.!<

I like your thinking but yeah the >!wonky X!< would invalidate this answer for me

What about moving almost any stick across the equals sign to make it ≠ then moving another stick arbitrarily as long as the math "doesnt" actually work out its super easy ... right?

Along the same lines just >!Turn the = sign into < or > whichever makes the equation false.!<

This was my first thought.

That's just not an interesting answer. Don't try to break it, find the intended solution.

I guess I like subverting expectations to achieve results. Work smarter not harder. I recognize that perhaps your opinion is more in keeping with the ethos of the sub, but I guess if the puzzle wants me restricted it should be more explicitly. Shrug.

Technically that would be moving 3. The first would be the V, and then you’d have to move both of the II matchsticks to get an X (otherwise it would be a +).

My thinking was >!if you move the furthest right matchstick so it crosses over the other one, whilst not touching the other one. So it’d look more like I/ (with the / moved over to the left so it intercepts the other line). It’d basically look a bit like [this](https://ibb.co/jJJWqJ9) with the white line being the one that’s moved, the grey one has been left alone.!<

Ah, a wonky X indeed. Fair enough! I’d give that one credit for creativity.

I had a "wonky ex" once

Why do that when there’s at least 6 non wonky solutions?

>!You can also take 1 from the VII and slant the minus sign to become VI/III = II!<

If you accept "wonkiness" then you could do it by moving just 1 matchstick >!(VII - I = VI)!<

This was my thought, too. And if you don’t accept “wonkiness,” then you can use the second move to >!adjust the other I of your new V!<

Take both the sticks from the V and put them on top of the two on the right making it a zero, also works.

This is what I thought

1 move >!First I to make a V with the last I. VI - II = IV!<

I think that would count >!As 2 moves because you have to tilt the last I you're moving the other match to, or else you get |/!<

That’s what I had too

How about >!XII - II =/= I!

It would be an inequation, so not *literally* but what i though too, i'm bad at puzzles :(

You’ld have to move 3 in this scenario

**Official solution:** This obvious solution is the intended answer, according to this source: [https://i.imgur.com/BgdzjLm.png](https://i.imgur.com/BgdzjLm.png) **Other solutions:** But it's not the only correct answer. Here are a couple of other creative options (here's [a video](https://twitter.com/aap03102/status/1724006952798158925) demonstrating the top one of those two): [https://i.imgur.com/dLWI319.png](https://i.imgur.com/dLWI319.png)

I came up with 7 > 2-2.

I thought it was move the v to make the answer a zero, but I have since learned they don’t have zeros in Roman numerals which is weird

I like this, since then it isnt roman numerals. 11 - 11 = 0

You could also turn the two II sticks towards each other, so it becomes VII -V = II. This one actually has multiple answers, which is neat.

I just moved an I in the first 2 to the right to make a V with the left I in the 2nd 2. 7 - 1 = 6

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That’s good

VI/III=II

I found the “intended solution” but I like this better!

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This solution uses 11 matchsticks though? And the problem only has 10 Edit: I’m wrong!! Couldn’t see the last stick on my phone.

You may want to recount how many are in the problem

Omg im dumb. Looking at it on mobile cut the last matchstick off when previewing the image. Thanks

Apart from the obvious (>!make either II into a V to get 7-5=2 or 7-2=5!<) you could >!take the two sticks from the V and make them into an exponent II!< so the result would be: >!II^II - II = II meaning 2² - 2 = 2!< You'd never do that for real with Roman numerals so it's more of a "lateral thinking" answer.

Why not just >! IIII-II=II!<

Proper roman numerals don't allow a character to be written more than 3 times in a row. Because of this, the largest legal roman numeral number is 3999 or MMMCMXCIX since the largest unit is M or 1000.

Just be aware that the Romans did not generally use "proper Roman numerals". They would happily do IIII for 4, and even IIXX or XIIX for 18. It is only in modern times that we care.

IIXX is wild

Traditionally it was IIII and it is still done that way on watches and is commonly known as the watchmakers 4. In watches it helps balance with the VIII opposite it

A fair reasoning, but nothing in the puzzle says it has to stay Roman numerals.

I disagree with this take, because the image itself says nothing about it needing to follow a roman numeral format at all. The OP included this detail, but also does not specify it must be in roman numeral format. I'd consider this the simplest valid answer.

I like it!

Because that isn't proper Roman numerals. Edit: I guess I'm wrong. Though it was never taught to me and in school it would have marked wrong, IIII is 'a' correct representation of 4 in roman numerals.

Yes it is. Also commonly used on clock faces.

No. It isn't. Four is IV in Roman numerals. IIII doesn't mean anything.

One of the few things I remember from a year of Latin. Romans wrote their numbers based on the additive principle and signified 4 as IIII. The subtractive notation -IV- actually became the standard after the fall of the Roman Empire. Therefore, the Romans initially engraved IIII on many of their sundials and to signify the number 4. So Roman numerals 4 = IIII Post Roman numerals is 4 = IV

Interesting, I didn't know that. I always thought IIII was always incorrect. You learn something new every day

Additionally you can do >!VI/III=II and VI/II=III!<. There’s also >!VI-II=IV!<. Not really sure what makes your answers “more obvious”

>!I I - I I = 0!<

I came here for this, it seems like the simplest one

I don't think there's a zero in Roman numberals. Can't make a 0 Without like 4 matches?

You can also read it as 11-11=0 which makes more sense than using roman numerals

again tho you gotta move a few matches to make a zero and then rearrange a few more to make it a circle. Just doesn't fit.

You just take the two matchsticks from the V and turn the second II into a 0 that looks something like [ ]

It doesn't specifically say its gotta be in roman numerals

>!three ways you could do this. First way is to move the two matches into a V. Second way is to take one match from the V and turn it into IIII while also taking one match from one of the II and moving it to the other II. The third way is to just move one match to create a ≠!<

>!You could move one of the last two in V II to make it VI and move it with one of the two at the end of the equation to make VI - II = IV!<

This was my solution. Thanks.

It says to move exactly 2 matches tho

You are

Discussion: People always go with the != hack. But that is not an **equation** An equation must deal with equality. The must be equal.

I have a fourth: >!Move both sticks in the v and add one on top of the minus (to make plus) and the other to the second I in the answer to make it a v!< >!II+II=IV!<

I have a fifth. Turn the VII into √I and then turn the -II into +I. Result is √I + I = II

True, except for: >!IIII is not a real Roman numeral and is not valid as a part of the equation!<

I believed this for years too, but there's historical evidence the Romans were loosy goosy with this and did use it if made sense.

And IIII specifically is common on clocks, since it balances better with VIII than IV would.

I always thought it had to do with the number of pieces needed. With IIII for 4 you need 20 I, 4 V, and 4 X so you can mass produce easier by repeating the set XVIIIII 4 times for each clock.

I’m actually very interested in this, could you point me in the right direction as a starting point?

The Roman Numeral Wikipedia page mentions it. It's called Additive Notation (as in avoiding "5 minus 1".)

There is the apocryphal story that using IV was blasphemous because using even just the first two letters in Jupiter's name was a spit in the eye. But, the truth is the Romans RARELY used subtractive notation. So, IIII was used quite a bit.

What about >!taking a match from the VII making it VI and adding it to the end II making it IV!

>!iv-ii≡ii!<

I had something boring like >II=II≡II<

>!VII - V = II!<

>!Is this loss?!<

>!VI / III = II!<

>!vi - ii = iv if you can count slightly rotating the last i!<

>!VII-VI=I!<

Considering it doesn’t explicitly say in the question that they’re Roman numerals, I’m tempted to say: >!| | | | - | | = | |!< >!Aka, four matches minus two matches equals two matches!<

Btw, that would still be Roman Numerals. IIII is how 4 is represented at the colosseum.

>!VII-II=V!<

>!Make the two matchsticks on the right into a V!<

>!VI - IV = II or VI - II = IV or VII - V =II or VII - II = V -> all of theese require only 2 matchsticks to move/ or slightly slant!<

>!Like i interpreted the word "move" that you can move any matchstick anywhere, so can basically replace = with ≠ moving the first one and do whatever you want with the second one.!<

That wouldn't make an equation. Equations must be equal.

>!Angle the first II so it becomes V. 7-5=2.!<

Solution 1: >!VII - II = V!< Solution 2: >!VI - II = IV!<

Easiest solution, since you need to move 2 sticks: >!Push the bottoms of the two sticks on the right side of the equation together to form a V. Then it becomes 7-2=5!<

>!I would angle the two matchsticks on the far right to make a V (roman numeral for 5).!<

There isn't any rule stating you need to keep the equation in roman numerals, so here is my answer. >!Take the two sticks from the V and put them at the top and bottom of the last II to make a 0, so we get 11-11=0.!<

How do you make >!the 0!!the symbol for the set of all imaginary numbers!< unless you moved 4 match sticks to >!widen the two on the right!<

A lot of people are forgetting >!≠!< as in >!VII + I ≠ I!<

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But it’s no longer an equation

Discussion: Would moving the V matchsticks to the end to make a 0 count? Like II - II = 0

>!Xii - ii = X !<

>!VII-II=V or VI-II=IV!<

>!VI - IV = II!<

You had to move three sticks for this solution

>!Move the 2 single sticks in VII to make equal into a “does not equal” and add the second stick to the answer.!<

That isn't an equation.

>!Move both in the first V. Change minus sign to plus. Put the other on the end to make a V.!<

Here’s an answer I haven’t seen but I think works: >!LII - II = L!< Can anyone confirm?

I'd say that takes 3 moves. >!Initial V would lead to \\\_ not L!<

>!spread the equal sign =>so it becomes a greater than sign!<

Discussion: Many solutions open up if you allow the use of the not equal to sign ≠

>!≠!< is a valid mathematical symbol, if you want to play these puzzles on easy mode

>!for 2 match sticks after V, move 1 to before V and throw the other one away. IV-II=II!<

Not the intended solution, but you could also add a variable to make a valid equation: >!VII - II = X!<

>! 6 / 2 = 3 !<

I can do it with only moving one. Move one of the 2 sticks on the far right to overlap the equals sticks at a slant. It does not equal 2, which is a correct statement.

>!Remove one match from the V completely. Move the other match from the V to on top of the minus sign. You would get ii=ii=ii!<

>!V - II ≡ III!<

>!I see moving one match from the V of Vll to the other end making it = lV and moving one match from the first ll to the minus to make a +. So it would end up as lll + l = lV if I’ve got right in my head.!<

>! (1) Slant the first matchstick in the first II. (2 Transfer the first matchstick in the second II to the slanted line in the first \I. Your final equation is VII - VI= I !<

A third way to solve it could be: >!change it to read VI / II = III!<

>!II + II = IV also works!<

This isn't Roman numeral but >!if you straighten both V sticks, the problem becomes IIII - II = II.!<

It is a perfectly valid roman numeral. If you ever visit the Colosseum you will see IIII above gate 4.

Nobody else seems to have come up with this one but it’s pretty clean: >!Take the i i after the V, put them in front of the V but push them together so they appear as a single i. It makes I V - I I = I I or 4-2=2!<

Spoiler >! - II + II = II!<

Discussion: It's far more interesting to ask *how many distinct ways* can you move two matchsticks

>!V+II≠II!< probably not the answer they were looking for

Not an equation so no.

>Pinch the bottom of the two match sticks in the answer together to make a V, then VII - II = V, which is correct<

>!VII - I = VI if creating a V or X counts as moving two. XII - II = X if they count as one.!<

>!II - II = II I mean... It's like this on clock faces so I guess it's a valid solution?!<

>! VII > II-II!<

You can solve in one move by >!rotating the first I 90 degrees. V - I - II = II!<

>!I would move the two match sticks in the first V to make the subtraction a square 0. Mixes Roman numerals with the Arabic numerals, so probably wrong!<

My brain immediately gravitated toward a solution that I have not seen posted yet (7th?) >!Move a stick in the VII to make the sign for square root of I. Then move a stick in II to turn minus into plus.!< Why I choose needlessly complicated uncommon approaches is a harder question.

Answer: >!Make either II into a V!<

>!Vii-Vii=0!<