Make 37+32=55 correct by moving one match

Question

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.

Answer 1

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

Answer 2

A little of a stretch but

$87 - 32 = 55$

by

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

Answer 3

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.

Answer 4

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).

Answer 5

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.

Answer 6

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.

Answer 7

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 $1^55=5$.

Answer 8

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.

Answer 9

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!

Answer 10

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:

Even better:

If you can split the match vertically :)

Answer 11

How about a hexadecimal answer?

17 + 3e = 55

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

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

Answer 12

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:

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.

Answer 13

What about:

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

Answer 14

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.

Answer 15

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.

Answer 16

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.

Answer 17

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.

Answer 18

The solution appears after [re]moving a match:

37 - 32 = 5 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.

Answer 19

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

Answer 20

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.

Answer 21

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.

Answer 22

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!

Answer 23

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.

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

Answer 24

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.

Answer 25

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.

Answer 26

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.