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Create an equation using only the following numbers and mathematical symbols: $$4,2,1,2,4,+,=$$

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    $\begingroup$ Welcome to Puzzling.SE! Is this a puzzle you created yourself? If not, you need to provide attribution to the original, otherwise this could be plagiarism and your question could be closed. $\endgroup$
    – F1Krazy
    May 28, 2020 at 9:46

7 Answers 7

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$4^2 = 12 +4$ (if exponentation is allowed)

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  • $\begingroup$ (+1) You ninja’d me by a minute! $\endgroup$
    – Culver Kwan
    May 28, 2020 at 10:27
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    $\begingroup$ OP specifies "the following numbers", as distinct from "the following numerals". This doesn't seem to follow that rule. $\endgroup$
    – Jeremy Nottingham
    May 29, 2020 at 12:22
  • $\begingroup$ This would require rot13(pbapngrangvba, v.r. bs 1 naq 2 gb trg 12 ), not just rot13(rkcbaragvngvba) $\endgroup$
    – Earlien
    May 31, 2020 at 13:01
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Solution 1: $$2^{2^{1^4}} = +4$$ Another possible solution: $$^24 = + ^24^1$$ (See Knuth's up-arrow notation) Another solution: $$4+2=12_4$$ i.e. $$6 = 12 \text{ (base 4)}$$

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    $\begingroup$ Could you expand/clarify the up-arrow solution? Maybe just linking to Tetration would be good enough. IIUC, ²4 means 4↑↑2 means a very big number — but wouldn't "24¹ = +24" require less explanation? $\endgroup$
    – Quuxplusone
    May 28, 2020 at 18:19
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    $\begingroup$ $^24^1 = (4^1)\uparrow\uparrow 2$ $\endgroup$
    – John Brookfields
    May 28, 2020 at 18:42
  • $\begingroup$ What does "but wouldn't "24¹ = +24" require less explanation?" mean? @Quuxplusone $\endgroup$
    – John Brookfields
    May 28, 2020 at 18:43
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    $\begingroup$ I mean, your equation is of the form "x = +x¹", where x happens to be "²4"; but you could have used plain old "24" instead and then you wouldn't have needed to explain anything about tetration notation at all. Other solutions in the same vein are "24¹ = +24", "42¹ = +42" and "2₄¹ = +2₄" (base-4 notation). It just seemed like you picked the absolutely most obscure one! $\endgroup$
    – Quuxplusone
    May 28, 2020 at 18:53
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    $\begingroup$ $2^{1^{4+4}}=2$ $\endgroup$
    – the default.
    May 29, 2020 at 7:55
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Almost similar to the other answers, but a non-trivial change nonetheless:

$2^4=12+4$

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Suppose we permit ourselves just one freedom of interpretation: that "numbers" in the question actually means "digits", which means that a formula like 42+1=24 is potential solution. We do not allow any other freedom of interpretation: no exponentiation or anything else.

We must necessarily choose a base for the digits; there is no concept of digit without reference to a base. If we choose the digits to be decimal then none of the possible formulas happen to be true. However why should we restrict to that base? Since the digits are no higher than four, any of the bases 5 and above are possible. It so happens that under the base 6 interpretation, there is the following solution: $$4+14=22$$. No solution appears for base 5, or bases above 6.

We can make the some informal remarks and arguments about this:

Formulas are formed by inserting the two operators into permutations of the five digits. The operators are binary, requiring argument material on both sides, and so can only be inserted into the digit string in limited ways. The only possible formulas are of the form nnn op n op n or nn op nn op n. The former cannot work in any base, because nnn + n = n cannot be true (three digit number plus one digit number cannot equal one digit number, zero not being available) and nnn = n + n similarly cannot be true. If a solution exists it must be of the form nn + nn = n or nn = nn + n. Of these, the former can, again, be squarely ruled out leaving only nn = nn + n: the possibility that in some base, adding a two-digit figure and one digit will leave a result expressible as a two-digit figure using the remaining digits.

Furthermore,

No solution adds 1 to an operand. 1 is the only odd digit we have, so if 1 is an operand by itself, the other two numbers are even. But adding 1 to an even number produces an odd number.

Furthermore,

No solution adds 2. Given XY=ZW+2, we need XY and ZW to both be even, or both be odd. Our only tool for making an odd number is the digit 1, and we have only one, so we must use it only as a left digit: 1A=BC+2 or BC=1A+2. The former is eliminated due to 1A being necessarily smaller than BC since B must be 2 or 4. The latter, BC=1A+2, seems viable, but only if B is not 4. That's because 4C is so large that it cannot be reached by adding 2 to 1A, in any base: if adding 2 to A produces a carry, at most that will bump the 1 to a 2. Since B is not 4, it must be 2, and so C and A are therefore 4: there is only one permutation which is exactly 24=14+2. But this can only hope to be true in a base in which 4+2 produces a carry into the next digit place. The only two viable bases with that property are 5 and 6, and the equation is false in both. Since no solution adds 1 and no solution adds 2, all solutions must work by adding 4.

Finally,

We are left with the only viable solution pattern XY=1Z+4, where X, Y and Z are permutations of 224. The list of possible formulas is rather short, so let us write it: 22=14+4, 24=12+4 and 42=12+4. The first is our base 6 solution. The second can only possibly work in base 5 and 6, and is false. In all higher bases, 12+4 is 16. The third permutation is rubbish. Thus $22 = 14 + 4$ in base 6 is unique.

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  • $\begingroup$ Doesn't $4+4+2=12$ (base 8) also work? $\endgroup$
    – JS1
    Jun 1, 2020 at 7:57
  • $\begingroup$ @JS1 That solution requires a freedom of interpretation not being taken in this answer, namely that the $+$ operator may appear more than once. $\endgroup$
    – Kaz
    Jun 1, 2020 at 8:02
  • $\begingroup$ Ah yes, I didn't realize that the + should be used only once. $\endgroup$
    – JS1
    Jun 1, 2020 at 23:29
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Is your solution this?

$4^2=12+4$

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Here's an answer that uses no "trickery" besides exponentiation (not even multi-digit numbers):

$2 + 2^{1^4} = 4$

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$$+ = 42124$$

That is, we are defining a variable

"+"

to be equal to the value

42124. The question never said "+" couldn't be used as a variable name!

Some have suggested that this doesn't count as an "equation". The mathematical definition of an equation is essentially "any statement with an equals sign" (more formally, any mathematical statement that states the equivalence of two expressions). The items on both sides of my equation are expressions and I am asserting that they are equal, hence this is an equation.

For reference, a mathematics.SE discussion on the topic of "what counts as an equation" can be found here: https://math.stackexchange.com/questions/2738360/what-exactly-is-an-equation

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    $\begingroup$ It should be an equation, not an assignment. $\endgroup$
    – Eric Duminil
    May 29, 2020 at 19:25
  • $\begingroup$ @Eric: you don't consider $x=2$ to be an equation? $\endgroup$
    – JDL
    Jun 1, 2020 at 7:20
  • $\begingroup$ @JDL I do not consider this an equation because why isn't 42124+= an equation as well? It has an equality mark! I also assert that 42124+ equals the empty expression. $\endgroup$
    – the default.
    Jun 1, 2020 at 11:29
  • $\begingroup$ I'm not sure, mathematically, that there is such a thing as "the empty expression" (I know such a thing exists in many computer programming languages, but this is more of a maths puzzle than a CS one) $\endgroup$
    – JDL
    Jun 1, 2020 at 11:30
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    $\begingroup$ Mathematically, I'm fairly sure there is no such thing as a variable assignment (where, of course, the variable is named $+$) $\endgroup$
    – the default.
    Jun 1, 2020 at 12:55

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