Match problem: infinite or not?


An intriguing matchstick problem has been running around on Facebook. (I haven’t managed to dig up the original originator). I suggest that the number of possible solutions is  in principle infinite. However, there is a reasonably good logical case to suggest that the maximum number of solutions is at least 9 and most 78. There are four classes of solutions, which have 3, 3-15, 3-15, and 0-45 solutions respectively.


Spoiler alert: these solutions are shown below.





1. The fair class: three solutions

There seem to be three of these.  One stick can be moved to create either the equations 5+4=9, 0+4=4, or 8-4=4.



2. The inequivalance class. At least 3, at most 15 solutions

A match can be moved to change the equals sign into an inequivalence sign (≠), giving 6-4≠4, 5+4≠4, 6-4≠4.   As far as I can determine, these are the only three solutions containing numbers.


However, nothing in the problem statement requires us to stick to numbers.  We can consider the characters to be variable names, giving us for example h-4≠4.


The variables do not even have to be legible Latin or other alphabetic characters. 6+whatever≠4 is a perfectly valid equation, as long as whatever≠-2.


The number of possible solutions is however limited, as there are only 15 matches that could be moved — any of the matches in the numbers, or the vertical bar in the plus sign.

Note that in principle the horizontal bar in the minus sign could be moved as well. The pipe operator | is in fact a valid symbol, but it does not have a generally accepted arithmetic definition, and thus I feel it should be left out of this class.

3: The identity class. At least 3, at most 15 solutions

The equals sign can be transformed into an  identity sign (≡) by moving one match. As in class 3, there are at least three and at most 15 solutions. By definition, all of these solutions are true: if we say that 6+4≡4, then 6+4 is in fact 4 because we say so.


4: The variable class. At most 45 solutions.

A hat tip goes to Jarmo Korteniemi for pointing out this class of solutions. If we consider the numbers to be variables, then in the below case h+4=9 is true when h=5.


However, the number of solutions is infinite, since we can now place the removed matchstick in any position whatsoever, rather than having to place it over the equals sign. Hence, h+H=4 is true whenever it is true.


If we do not need to stick to known characters, then we can place the match absolutely anywhere we wish. h+whatever=4 is true for whatever values of whatever we want. Thus, the number of possible equations would be infinite.


There is, however, a case to be made that the number of solutions is in fact finite. In the solution above, we removed a match from the first number to create the variable h. We can then add that match to either the second number (“4”) or the last number (“9”). We can thus in fact only produce two logically distinct equations: h+whatever=4, or h+4=whatever. In addition, if we place the match over the first number itself, we get a third equation whatever+4=4.


Thus, for any match that is moved, we can only create 3 logically distinct equations. As in the previous classes, there are 15 movable matches and hence 45 logically different outcomes. None of these is numerical in the sense meant in classes 1-3.

5: Rejected: The non-equation class.

If we remove the top bar from the equals sign, then we get a statement of the type 8-4-4, or 6+4-4, or 6-9-4. However, after consideration, I believe this class has to be dropped, as is it impossible to determine the validity of the new statements; i.e. they are not equations.  Thus, these are not valid solutions to the question.



Additions after first publication

Sammeli Heikkinen and Jarmo Korteniemi independently noted that 6+4≥4 would also be a valid solution. I am reserving judgment on this, as I have never seen the ≥ operator written in the way it would be displayed.


Also, Olli Välke has noted that the first solution for Class 1 is not logical, as the 9 should have an extra match in order to be rotationally symmetrical with the 6. I find the point compelling, but am reserving judgment on this also, as life is not always logical.



What is the lesson learned?

There does not seem to be one. Nor is it clear whether there was any point for this exercise. It was done because it was there.

See Weirdmath page for similar blog posts.



3 thoughts on “Match problem: infinite or not?”

  1. That should have read “You could also use symbols for larger than, smaller than, larger or equal than, or smaller or equal than…”.

  2. Class 2 is broader than what you describe. You could also use , ≤ or ≥, if you allow some distortion of the symbol in question.

    Also note that the job was to “fix the equation”, which may be done in many ways. The class 2 fixes all result in non-equations. If you allow them, you should also allow the resulting math problems of class 4.

    To me the lesson was: Oh man, one can go this far outside of the box!

    1. Very valid points. Larger than would indeed be a valid solution for Class 1. The Class 4 question is a bit trickier. I think there needs to be a class 4b which allows changes to the operators. I’m pretty sure its size would be zero, as I don’t see how to make sure the operators are valid.

      Indeed this problem is a good example of the continuum between out-of-the-box and lunacy. I wonder where we are on the continuum.

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