32
$\begingroup$

37+32=55

Move $1$ match and make this correct.

Rules: (added 8/23/2018 based on clarifications in comments)

  • You can't break the match.
  • You can't make an inequality sign, just change numbers and/or operators.
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  • 5
    $\begingroup$ Are there any restrictions on what kind of answers are acceptable? I was about to make the same suggestion as El-Guest, but he got there first. $\endgroup$ – F1Krazy Aug 17 '18 at 17:53
  • 10
    $\begingroup$ my sister said //dont break the match, dont make inequality sign, just change numbers or operator// $\endgroup$ – underds Aug 17 '18 at 18:07
  • 4
    $\begingroup$ Maybe greater than sign. Although this is probably cheating too. $\endgroup$ – Bennett Bernardoni Aug 17 '18 at 18:15
  • 9
    $\begingroup$ @underds Would you mind providing a source for this problem, since it is not your own? $\endgroup$ – El-Guest Aug 17 '18 at 18:15
  • 13
    $\begingroup$ Is there actually a known (and valid) solution? $\endgroup$ – jcaron Aug 18 '18 at 23:14

26 Answers 26

66
$\begingroup$

Super easy:

37 - 32 = 5 | 5

Where:

the | is bitwise OR operator

By:

moving the vertical match from the plus sign to the middle of the 55

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  • 3
    $\begingroup$ I have a feeling this is technically the correct answer per the move one match stick requirement. $\endgroup$ – Facebook Aug 18 '18 at 2:52
  • 12
    $\begingroup$ Which is, of course, the best kind of correct. $\endgroup$ – peaceoutside Aug 18 '18 at 4:28
  • 3
    $\begingroup$ I know you mean 37 - 32 = (5 | 5), but I cannot help seeing it as (37 - 32 = 5) | 5, but maybe because that is because I am used to a programming language with a particular convention for operator precedence. I like the interpretation that 5 | 5 means 5 divides 5 (evenly), better. $\endgroup$ – Jeppe Stig Nielsen Aug 18 '18 at 10:52
  • 3
    $\begingroup$ @Jeppe, which language? In C, | has higher precedence than =. $\endgroup$ – Greg Schmit - Reinstate Monica Aug 18 '18 at 21:57
  • 2
    $\begingroup$ @Jeppe I get that; however the symbol actually used is =, so it looks perfectly natural from another fellow programmer :) $\endgroup$ – Greg Schmit - Reinstate Monica Aug 18 '18 at 22:15
63
$\begingroup$

Technically, you could change this to

$37 - 32 \neq 55$

by

moving the cross stick in the plus sign to go diagonally across the equals sign.

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  • 1
    $\begingroup$ Well if that were true you could also rot13(zbir bar sebz gur rdhnyf fvta gheavat svsgl svir vagb avargl svir; guvf jbhyq or gehr fvapr rkcerffvbaf ner arvgure gehr abe snyfr)! $\endgroup$ – PerpetualJ Aug 17 '18 at 18:06
  • 3
    $\begingroup$ That sounds like another variation of the CheaterPants-You-Can't-Do-That solution, I like it! $\endgroup$ – El-Guest Aug 17 '18 at 18:09
  • 2
    $\begingroup$ It looks even better if you rot13(zbir gur yrsg fgvpx va gur bqq-ybbxvat frira gb sbez gur vardhnyvgl fvta.) $\endgroup$ – HKOB Aug 18 '18 at 10:06
  • 3
    $\begingroup$ That way you could solve any match puzzle involving numbers. $\endgroup$ – stackzebra Aug 18 '18 at 15:47
  • $\begingroup$ That's what I thought. $\endgroup$ – prog_SAHIL Aug 19 '18 at 10:52
63
$\begingroup$

A little of a stretch but

$87 - 32 = 55$

by

moving the upright + matchstick to the left of the 3 in 37

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  • 5
    $\begingroup$ Pun not intended, but great anyway $\endgroup$ – Bennett Bernardoni Aug 17 '18 at 18:26
  • 17
    $\begingroup$ Stretch, as in stretching the matchstick? (+1 from me) Edit: but you got the upvote because of the sweet pun! $\endgroup$ – El-Guest Aug 17 '18 at 18:26
  • 47
    $\begingroup$ You must hold the match closer to your face to see it span the whole number. $\endgroup$ – tyobrien Aug 17 '18 at 22:12
  • $\begingroup$ @tyobrien : indeed, the question doesn't specify you can only move it in two dimensions! $\endgroup$ – vsz Aug 20 '18 at 6:14
  • 1
    $\begingroup$ The funkily-drawn number solution is not unique. Remove the bottom match from the last 5, resulting in a funky 9. Use this match to change the first 5 into a 6, resulting in 37+32=69. $\endgroup$ – David Hammen Aug 21 '18 at 13:01
53
$\begingroup$

Just to prove that this is impossible (without some creativity), I wrote a python script to solve these:

subs = {'1':[],'2':[],'3':[],'4':[],'5':[],'6':['5'],'7':['1'],'8':['6','9','0'],'9':['3','5'],'0':[],'+':['-'],'-':[],'=':['-']}

adds = {'1':['7'],'2':[],'3':['9'],'4':[],'5':['6','9'],'6':['8'],'7':[],'8':[],'9':['8'],'0':['8'],'+':[],'-':['+'],'=':[]}

noop = {'1':[],'2':['3'],'3':['2','5'],'4':[],'5':['3'],'6':['9','0'],'7':[],'8':[],'9':['6','0'],'0':['9','6'],'+':['='],'-':[],'=':['+']}

ts = list("37+32=55")
alt_strings = []

for i, c in enumerate(ts):
    for new_char in noop[c]:
        alt_strings.append(ts[:i]+[new_char]+ts[i+1:])

    for sub_c in subs[c]:
        for i2, c2 in enumerate(ts):
            for add_c in adds[c2]:
                s = [x for x in ts]
                s[i] = sub_c
                s[i2] = add_c
                alt_strings.append(s)

alt_strings = [''.join(x) for x in alt_strings]
print alt_strings

for alt_string in alt_strings:
    split_strings = alt_string.split('=')
    if len(split_strings) != 2:
        continue

    left = eval(split_strings[0])
    right = eval(split_strings[1])
    if left == right:
        print alt_string

The possible combinations I got were:

['27+32=55', '57+32=55', '91+32=55', '31+92=55', '31+32=65', '31+32=95', '31+32=56', '31+32=59', '37=32=55', '97-32=55', '37-92=55', '37-32=65', '37-32=95', '37-32=56', '37-32=59', '37+22=55', '37+52=55', '37+33=55', '37+32+55', '97+32-55', '37+92-55', '37+32-65', '37+32-95', '37+32-56', '37+32-59', '37+32=35', '37+32=53']

And there were no matches (hehe).

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  • 7
    $\begingroup$ As with the above (Bennett's comment)...+1 for the excellent pun (and the cool programming solution too, I guess $\endgroup$ – El-Guest Aug 17 '18 at 20:39
  • $\begingroup$ @tyobrien how would you make the first number a 67? $\endgroup$ – ale10ander Aug 17 '18 at 22:24
  • 1
    $\begingroup$ Noop 3 can also turn into a 9, btw, if you don't have a segment on the bottom $\endgroup$ – No don't shown my real name Aug 18 '18 at 0:27
  • 2
    $\begingroup$ and 7 into 4. and how does 7->1 (I can see 11)? $\endgroup$ – JMP Aug 18 '18 at 4:06
  • 5
    $\begingroup$ You can also remove the left vertical match from 7 to keep it as 7. $\endgroup$ – Heimdall Aug 18 '18 at 9:28
27
$\begingroup$

37+32>55

There you go,

just change the equals sign to a more than sign!

-edit

You take the

vertical match from the plus sign

and

use it to burn the first five in 55

to make

37 - 32 = 5

If only the first 5 would be a 6. That would help a lot. Please ask your sister for the solution and double check if it's actually possible. Because I have a hunch that this puzzle is impossible.

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  • 1
    $\begingroup$ You need to move two matches to accomplish this; otherwise you end up with an implied greater than or equal sign which is also not true. $\endgroup$ – PerpetualJ Aug 17 '18 at 18:30
  • 2
    $\begingroup$ Greater than OR equal to, is true too. $\endgroup$ – Alto Aug 17 '18 at 18:36
  • $\begingroup$ Sorry about that, was a typo and it got cut-off. It is also true, but is not the full sign. $\endgroup$ – PerpetualJ Aug 17 '18 at 18:41
  • 2
    $\begingroup$ I figure you haven't read all the comments, but I've mentioned both of these in comments above. Great minds think alike I guess. $\endgroup$ – Bennett Bernardoni Aug 17 '18 at 18:59
  • 3
    $\begingroup$ I like your second solution. It's not the inequality cheat, and you do move one match only, haha. +1 for that. $\endgroup$ – Heimdall Aug 18 '18 at 9:23
23
$\begingroup$

I saw this in "hot network questions" and tried to solve it on paper before I clicked to avoid getting spoiled. So I scribbled down the equation. When I couldn't find a way to solve it, I finally opened the question and saw, that my 7 had one less matchstick (the very left one). So I wondered about the display of numbers we don't get to see. Would a 9 without an underscore be legal for example? It clearly would be a distinct nine, wheter the bottom stick is there or not. You wouldn't return your 80's alarm clock because of such a nine, anyway.

Along this line of thinking I came up with this solution:

37 + 32 = 69 by taking the bottom matchstick from the second five to make the first five a six. Granted, the resulting nine is somewhat weird, but you clearly would not assume another number instead of it. Maybe for arguments sake just now, but not if some hot girl wrote the nines of her phone number in that way. That would probably just be a-okay for you. So just give me the correct flag now. Thanks.

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  • 1
    $\begingroup$ To further my point, here is a list of fonts in which the nines have no "back". This is from dafont.com and I just clicked through the scifi-section. imgur.com/a/8OuHuIB Occasionally (e.g. in font fabricate) you might mistake the five for a nine, but there is no way to confuse a nine with something else. As you can clearIy see: I really want this. $\endgroup$ – Omphaloskopie Aug 18 '18 at 4:56
  • 1
    $\begingroup$ I get the feeling that this is most likely the correct solution, especially after reading the thread on twitter (see comments under OP's post). Since it's a Korean riddle I wouldn't put it past them to use numbers that look slightly differently. $\endgroup$ – Alex Aug 22 '18 at 7:21
20
$\begingroup$

Ok here is one I don't think has appeared yet:

$37-32=1^5 5$

Explanation:

Take the vertical stick in the $+$ turning it into a $-$ and put left-below the first $5$ to get $_15\,5$. Interpret this as $1^55=5$.

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  • 1
    $\begingroup$ This is my new favourite for applying more than simple arithmetic. $\endgroup$ – Ruadhan2300 Aug 21 '18 at 15:37
17
$\begingroup$

Outside the box solution!

Remove the match on the + to make it a -, then eat it. Then crop the picture so that it cuts out the last 5. 37 - 32 = 5, only one match (and the frame of the picture) has moved.

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  • 1
    $\begingroup$ Great solution, although ...you eat matches? $\endgroup$ – El-Guest Aug 17 '18 at 18:23
  • 36
    $\begingroup$ I tell other people to eat matches. I find it improves my quality of life substantially. $\endgroup$ – flashstorm Aug 17 '18 at 18:24
  • 14
    $\begingroup$ Or even better rot13(yvtug gur zngpu naq ohea gur bgure 5) $\endgroup$ – Bennett Bernardoni Aug 17 '18 at 18:30
  • 1
    $\begingroup$ @BennettBernardoni out of all the answers we've put forward, this one in the comments might be my favourite so far $\endgroup$ – El-Guest Aug 17 '18 at 18:32
  • $\begingroup$ Btw: rot13(Vs lbh vagraq gb pebc gur cvpgher naljnl lbh zvtug nf jryy zbir gur zngpu vagb gur ynfg svir vafgrnq bs rngvat vg, perngvat n fvk.) $\endgroup$ – Alex Aug 21 '18 at 12:10
13
$\begingroup$

Take a match from the equal sign and put it anywhere else where it creates a number. For example: 37 + 92 - 55. It's neither true nor false and it's left to the reader to calculate!

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  • 4
    $\begingroup$ In fact, as the resulting expression is non-0 , many programming languages would say it is indeed true. $\endgroup$ – RJFalconer Aug 20 '18 at 12:29
13
$\begingroup$

The extra match on the 7 seems suspicious, so I assume the solution is:

Removing the extra match from 7, breaking it in half and making it 69 in the other end:
enter image description here

Even better:

If you can split the match vertically :)

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  • $\begingroup$ From the comments: no splitting of the match! $\endgroup$ – UKMonkey Aug 22 '18 at 12:06
10
$\begingroup$

How about a hexadecimal answer?

17 + 3e = 55

Of course, it makes a weird looking one...

_  _      _  _      _  _
 || | _|_ _||_| -- |_ |_
_|  |  |  _||_  --  _| _|

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  • $\begingroup$ Make sure to hide your answer in spoiler quotes >!, but $(+1)$ anyway :) $\endgroup$ – Mr Pie Aug 19 '18 at 13:36
  • 2
    $\begingroup$ I tried to hide it, but no amount of >! and <pre><code> tags would allow the code block to remain lined up in the spoiler quote. Is there a good way of doing that? $\endgroup$ – BM- Aug 20 '18 at 1:40
  • 1
    $\begingroup$ There you go @BM-: note the `\`s which stop italicisation. $\endgroup$ – boboquack Aug 20 '18 at 22:15
  • $\begingroup$ Of course! Thank you, I hadn't though of that. $\endgroup$ – BM- Aug 20 '18 at 23:11
7
$\begingroup$

My contribution to the answer pile:

Take the upper left match on the 7 so that it's still a 7. Put the match on top of the equals sign, so that it and the upper part of the equals sign form an angle. Like this: enter image description here

The result looks a bit sloppy, but it can be read as

$37+32 \geq 55$

The inequality is technically correct, which is the best kind of correct.

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  • $\begingroup$ *Technically*, the equation is not technically correct, it is the (rot13) *vardhnyvgl* which is correct. $\endgroup$ – boboquack Aug 20 '18 at 22:19
  • $\begingroup$ @boboquack Thanks for pointing that out, I edited my answer $\endgroup$ – MikeQ Aug 20 '18 at 23:50
6
$\begingroup$

Here's a boolean logic solution:

Take one of the $=$ matches, break it in half, and put it on the right side of the remaining = match to make a right facing arrow $\rightarrow$. Then, the LHS is $69$, interpreted as a boolean is True, and the RHS is $55$, interpreted as a boolean is True. Then, the statment True $\rightarrow$ True is True. Thus, the statement becomes True.

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6
$\begingroup$

I think the following is technically a solution to the problem:

Make use of the fact that those are not arbitrary sticks, but matches. So, take the vertical match from the plus (making it a minus), move it quickly over the side of the match box (so it starts burning), and then move it in turn to all the matches making up the left digits (so they all catch fire and burn away, without moving). Then put that match anywhere out of the way to finish its burning without affecting the rest of the matches.

The remaining matches form the equation 7-2=5.

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  • $\begingroup$ Or use your idea to burn the first 5: 37 - 32 = 5. $\endgroup$ – DrSheldon Aug 21 '18 at 5:06
6
$\begingroup$

Along the lines of El-Guest and Alto:

37 + 3P = 55 where P = 6.

I know that's not the answer, however; from my testing I have found that the only (to my knowledge) possible (true) combinations are:

37 + 22 = 59, 27 + 32 = 59, 57 - 22 = 35, 97 - 32 = 65, 34 + 22 = 56

But all of those require more than one match. I'll keep at it, if I solve it I'll update. All in all, I have found 210 total numeric combinations. None of which are achieved with a single move. I have written a loop in C# that goes through multiple arrays of all possible number combinations to confirm this. I may be missing something, but mathematically; this seems quite impossible, aside from the Cheater-Pants, you can't do that solutions that El-Guest and several others (including myself) have posed.

Also, not sure as to why I got a down vote as my answer evaluates true, and can be achieved with a single match:

37 + 3P = 55 evaluates true when P = 6; this breaks down to 37 + 18 = 55.

If you down vote, please explain why the down vote is justified.

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  • 1
    $\begingroup$ I like this answer, and thanks for the kind mention! :P (+1) from me! $\endgroup$ – El-Guest Aug 20 '18 at 17:04
  • 1
    $\begingroup$ I liked your reference, so no problem on the mention! Your answer was the first I thought of, then several other answers that fell into the same category. I only answered because my solution had not been recommended yet, and I hadn't found any possible numeric combinations! $\endgroup$ – PerpetualJ Aug 20 '18 at 17:08
6
$\begingroup$

What about:

Match picture

You can either interpret that as a zero or a five that's been slashed out.

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6
$\begingroup$

My answer, similar to @alto's, is a little bit more elegant.

Take the leftmost downstroke from the 7 and place it at an angle above and and touching the top bar of the = sign to form 'a greater-than-or-equal' (or less-than-or-equal") sign.

BTW, this brute force & ignorance approach (in Python 3.x.) shows that it is not possible without modifying the + or = chararacter:

'''  Dictionary of all possible "matchstick' substitutions.  Note that
     some "matchstick" characters can be transformed into variants of 
     themselves by adding or removing one stroke/matchstick, e.g. 7 (by removal of 
     leftmost downstroke); 6 and 9 (by adding a horizontal top/bottom stroke).    

SubstitutionList = {
                    '+':['+','-'',='], '-':['+','-'',='],     '=':['+','-'',='],
                    '1':['1'],         '2':['2','3','6'],     '3':['2','3',5','9'], 
                    '4':['4','9'],     '5':['3','5','6','9'], '6':['6','9','0'],
                    '7':['4','7','9'], '8':['8'],             '9':['9','6','0'], 
                    '0':['0','9','6'], 
'''     
''' Only these values are needed (after allowing for character variations) '''

SubstitutionList = { '2':['2','3','6'],     '3':['2','3','5','9'],
                     '5':['3','5','6','9'], '7':['4','7','9'], 
                     '+':['+','-','='],     '-':['+','-','='],
                     '=':['+','-','='] }

TestEquation = '37+32=55'    
PossibleSolutions = []   # None yet

''' For clarity the code that extracts values from "TestEquation" has been omitted  '''

for I in SubstitutionList.get( '3' )  : 
    for J in SubstitutionList.get( '7' ) :
        for K in SubstitutionList.get( "+" ) : 
            for L in SubstitutionList.get( "3" ) :
                for M in SubstitutionList.get( "2" ) :
                    for N in SubstitutionList.get( "=" ) :
                        for O in SubstitutionList.get( "5" ) :
                            for P in SubstitutionList.get( "5" ) :                                    
                                Equation = I+J+K+L+M+N+O+P    # concatinate the letters into a string
                                FirstNumber  = int( Equation[   : 2 ] )
                                SecondNumber = int( Equation[ 3 : 5 ] )
                                ThirdNumber  = int( Equation[ 6 :   ] )
                                if ( (Equation[2] == '=' ) and (Equation[5] == '+' ) ) :                                        
                                    if ( FirstNumber  == (SecondNumber + ThirdNumber) ) :
                                        PossibleSolutions += [ Equation ]                                   
                                elif ( (Equation[2] == '=' ) and (Equation[5] == '-' ) ) : 
                                    if ( FirstNumber  == (SecondNumber - ThirdNumber)  ) :
                                        PossibleSolutions += [ Equation ]
                                elif ( (Equation[2] == '+' ) and (Equation[5] == '=' ) ) :                                         
                                     if ( (FirstNumber + SecondNumber) == ThirdNumber  ) :
                                        PossibleSolutions += [ Equation ]
                                elif ( (Equation[2] == '-' ) and (Equation[5] == '=' ) ) :                                         
                                     if ( (FirstNumber - SecondNumber) == ThirdNumber ) :
                                        PossibleSolutions += [ Equation ]

OneCharacterMoves = []   # match changed position inside one character, e.g. change "3' to '2', "6" to "9" etc. 
TwoCharacterMoves = []   # match moved from one character to another
for ValidEquation in PossibleSolutions :   # valid solutions change one or two characters
    DifferenceCount = 0  
    for i, _ in enumerate( ValidEquation ) :
         if ValidEquation[ i ] != TestEquation[ i ] :
            DifferenceCount += 1                
    if ( DifferenceCount == 1 ) :  
        OneCharacterMoves += [ ValidEquation ]
    elif ( DifferenceCount == 2 ) :
        TwoCharacterMoves += [ ValidEquation ]

print( 'Original :                ', TestEquation )
print( 'Involving One Character:  ', OneCharacterMoves )        
print( 'Involving Two Characters: ', TwoCharacterMoves )

The program produces :

Original :                 37+32=55
Involving One Character:   []
Involving Two Characters:  ['27+32=59', '37+22=59', '37+32=69']

Inspecting the values shows that none of them can be produced by moving one matchstick.

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  • 1
    $\begingroup$ It could well be argued, that the substitution-list is incomplete. Check out my answer or this font dafont.com/de/space-age.font?fpp=100&text=1234567890 and run it again to be amazed! $\endgroup$ – Omphaloskopie Aug 18 '18 at 4:02
  • $\begingroup$ I'm assuming we are limited to 7 segment "characters' - like the numbers on digital clocks., The code was inspired by @John Aaron. And of course I may have missed some "subtitutions" . Neat font by the way... $\endgroup$ – user1459519 Aug 18 '18 at 4:17
  • $\begingroup$ yeah, sure. my nine would be a AFGC in that regard. en.wikipedia.org/wiki/… Wikipedia basically says "uncommon", I say "weird", but I found eight fonts that use this style of nine and in no one it could be confused with something other... I am totally going to build a picture! $\endgroup$ – Omphaloskopie Aug 18 '18 at 4:25
  • $\begingroup$ Changing two matches you can also get 97-32=65 $\endgroup$ – user Aug 19 '18 at 0:05
5
$\begingroup$

Here is a solution exploiting that the algebraic operations are not specified in the question.

Take the vertical part of the plus sign and place it horizontally above the equality sign to obtain 37 "minus" 32 "is defined as" 55.

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  • $\begingroup$ That's not the symbol for "is defined as" ... ? $\endgroup$ – Rand al'Thor Aug 19 '18 at 17:01
  • 1
    $\begingroup$ it is used as such, see en.wikipedia.org/wiki/Triple_bar $\endgroup$ – HRSE Aug 20 '18 at 2:59
  • $\begingroup$ Nice. The brute-force general-case approach. $\endgroup$ – Ruadhan2300 Aug 21 '18 at 15:39
5
$\begingroup$

The solution appears after [re]moving a match:

37 - 32 = 5 solution a clue is that in second 5 match heads are aligned so that they burn each other properly until 5 is fully burnt, so perhaps we are supposed to ignite it with the match we took away.

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  • $\begingroup$ This is a good observation: the first 5 is drawn differently to the other numbers in that its heads are together, whereas the others are apart. Could be onto something $\endgroup$ – Tas Aug 22 '18 at 2:16
4
$\begingroup$

It gives a negative value but I would like to share it anyways:

37-92 = 55

How?

Just remove the vertical match from the "+" sign and add onto 32, now it is 92.

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  • 2
    $\begingroup$ Where does the - in front of 55 come from? $\endgroup$ – puck Aug 18 '18 at 10:30
  • $\begingroup$ @puck it’s not a match.. it’s negative value (-) $\endgroup$ – CR241 Aug 18 '18 at 21:25
  • $\begingroup$ the result is -55 not 55. To make the term match you need a minus sign in front of 55. $\endgroup$ – puck Aug 19 '18 at 6:52
  • $\begingroup$ I like it even if it's not correct. :-) $\endgroup$ – Alex Aug 21 '18 at 12:16
4
$\begingroup$

By moving one match to

turn the second 5 into 3 and then interpreting the suspicious "extra" match in 7 as 1

we get

317+32=53

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3
$\begingroup$

Move the lower match of the plus and the top match moves too! Make $37\times32=1184=550\times2+84$, and the picture has been (unfairly in my opinion) cropped so that you can't see the last bit!

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  • 1
    $\begingroup$ Are you claiming that "0×2+84" is missing from the right part of the image? That's quite a bold claim. $\endgroup$ – xhienne Aug 19 '18 at 20:24
  • $\begingroup$ JonMark Perry; do you have any evidence of this? Perhaps an online sample, or the original image itself? $\endgroup$ – PerpetualJ Aug 20 '18 at 18:47
  • $\begingroup$ it later struck me the cropped portion could be anything, $\times21+29$, for example, so no direct proof yet... $\endgroup$ – JMP Aug 21 '18 at 1:24
3
$\begingroup$

Remove the lower right matchstick to give 3 to the power of pi. Break the stick a bit and the extra matchstick on the equals to give an approximately equals symbol.

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  • 1
    $\begingroup$ That's two operations (removing one matchstick, breaking another), but the question requires just one. $\endgroup$ – Rand al'Thor Aug 19 '18 at 17:13
  • $\begingroup$ No it does not seem to mention anything about number of operations. Only one stick is bent. But then a comment says no match breaking. $\endgroup$ – Sentinel Aug 19 '18 at 19:31
3
$\begingroup$

To combine user477343's comment with CR241's answer,

Remove the vertical match from the +, break it in half, put one half on the upper left of the first 3 to make it a 9, and the other half in front of the first 5 as a minus sign. Illustration

I know the way I constructed this pattern doesn't match the original picture, but I'm sure I nailed the right pattern.

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  • $\begingroup$ OP specifically says, albeit in comments "don't break the match" $\endgroup$ – Tas Aug 22 '18 at 2:14
  • $\begingroup$ @Tas Yes, so it does. sigh $\endgroup$ – Post169 Aug 22 '18 at 2:18
  • $\begingroup$ Oh, kind of you to include me in your answer! $(+1)$ :P $\endgroup$ – Mr Pie Jun 24 at 10:01
3
$\begingroup$

The equation is actually

$$3^1 7 + 32 = 55$$

So,

$$3^1 \times 7 + 32 = 53$$ ...just turn the $55$ into a $53$ by moving one match.


Edit: Sorry, I didn't notice that Kamil posted the same answer earlier.

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  • $\begingroup$ Excellent - this looks to be it! $\endgroup$ – Tom Aug 30 '18 at 12:14
  • $\begingroup$ @Tom couldn't agree more! Unfortunately, however, this is a duplicate of KamilMaciorowski's answer :\ $\endgroup$ – Mr Pie Aug 31 '18 at 4:57
  • $\begingroup$ @user477343 - good spot! Kamil posted this earlier. These solutions resolve the suspicious 1 before the 7 and give a valid equation. Nice :) $\endgroup$ – Tom Aug 31 '18 at 5:06
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I can only think of

$ 37 - 32 ≡ 55 \pmod{5}$

I have

moved the vertical bar of the $+$ sign moved to the $=$ sign to get a triple bar symbol. This turns the expression into a modular arithmetic expression.

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