Is it possible to make 1 with any 3 numbers, all the same, such as using 3, 3 and 3 or 6, 6 and 6? Can you think of a proof which will work no matter which number it is? Trig functions allowed.


8 Answers 8


It is possible.

$x^{x-x} = 1$ for all $x \in \mathbb{C}$

See it in action here.

  • 1
    $\begingroup$ Yep. In fact, it works for all complex numbers. $\endgroup$
    – Zerris
    Feb 28, 2016 at 7:27
  • 8
    $\begingroup$ @Zerris: Except 0, depending on how you define $0^0$. $\endgroup$
    – Deusovi
    Feb 28, 2016 at 7:39
  • 4
    $\begingroup$ 0 will not work. $\endgroup$
    – Oray
    Feb 28, 2016 at 8:53
  • 4
    $\begingroup$ actually 0^0 =1 $\endgroup$
    – Mohammad
    Feb 28, 2016 at 18:21
  • 3
    $\begingroup$ @Mhmd There are mathematicians who say 0^0 is undefined. $\endgroup$
    – nhgrif
    Feb 29, 2016 at 3:01

There are certainly many ways. Here are a few ones.

$$\cos\left(x\left(x-x\right)\right) = \cos\left(0\right) = 1$$


$$\sec\left(x\left(x-x\right)\right) = \sec\left(0\right) = 1$$


$$\left(x\left(x-x\right)\right)! = 0! = 1$$

These all rely on the fact that $x\left(x-x\right)$ is 0 for every $x\in\mathbb C$.

Additionally (and this could be done with only one copy of any complex number $x$),


Where $\left|x\right|$ is the modulus of $x$ and $\lfloor r\rfloor$ is the value of the floor function at $r$, where $r\in\left[0,1\right]$ for any $x$. The same works if $\sin$ is replaced with $\cos$ or $\lfloor r\rfloor$ with $\lceil r\rceil$.

Finally, if we are allowed to cheat and use set theoretical constructs, then


will work for whatever (number) $x$: the cardinality of a set containing only one element is 1. This is true even if $x$ is not a complex quantity.

  • $\begingroup$ Even if x is not a number $\endgroup$ Feb 29, 2016 at 17:28

For $x \neq 0$:


Alternatively, the following works for all $x$:


  • $\begingroup$ Nice. Easily the simplest yet, but I mark Zerris' one because his is so ingenious (and as far as I'm concerned, his is also first) $\endgroup$ Feb 28, 2016 at 12:55
  • 1
    $\begingroup$ If you replace x with x! (or any other function that doesn't return zero), you will be able to work with all numbers, AFAIK. $\endgroup$ Feb 29, 2016 at 11:16
  • $\begingroup$ @BojidarMarinov $x!$ is undefined for some $x$. Something like $\cos(\sin(x))$ should work. $\endgroup$
    – wythagoras
    Mar 2, 2016 at 13:40

Here's one that works for 0:

$x!^{x - x} \forall x \in \{0, 1, 2, 3, ...\}$ (adapted from Zerris' answer)

  • 2
    $\begingroup$ To who ever down voted - I originally made this a comment and was asked to make at an answer. $\endgroup$
    – Paul Evans
    Feb 28, 2016 at 20:20
  • $\begingroup$ This is the best right now. I'd prefer gamma function since it is defined for real numbers, but it's a slight change $\endgroup$ Feb 29, 2016 at 17:29
  • $\begingroup$ @LuisMasuelli Yes, $\Gamma(x)$ is good on reals that way but $\Gamma(0) = \infty$! Also true for all negative integers. $\endgroup$
    – Paul Evans
    Feb 29, 2016 at 17:44

From a more computer science angle (depending on how your language handles boolean to integer conversions):

(int)(A == (A & A))
Means: ConvertToInteger(Is A equal to (A bitwise-anded with A)?)
Because: Bitwise-anding A with A always gives A, and that A always equals A, so the result is True, which typically evaluates to 1.

Probably only works for integers, though technically it would work for floats if you had a language that allowed it.

  • $\begingroup$ In this spirit, use the truth value of $a \neq a$ to get 0. Now raise $a$ to get the 0 to get 1. $\endgroup$
    – wythagoras
    Mar 2, 2016 at 13:43

This should work too:



This works, too, but it's in the category of trick question not mathematics

$1 - 1 + 1 = 1$

also, for $x \ne 0$ this works:

$x^0 - x^0 + x^0 = 1$

working the problem almost suggests a possible quadratic form, if $a$ and $b$ are each $1$, or as above if each evaluate to $1$, and if $x$ is also $1$, as long as $c$ is negative:

$ax^2 + bx - 1 = 1$

this works even for $x = 0$, but I still want the answer to be funny somehow not functional:

$1^x - 1^x + 1^x = 1$

here's a humorous answer, where $x = 6$:

why is $6$ afraid of $7$? because $((x + 1) - (x + 2)) ^ {(x + 3)}$.

($6$ is afraid of $7$ because $7, 8, 9$)

evaluated: because $(7 - 8) ^ 9$. or: because $(-1) ^ 9 = -1$

$-1$ to the power of any number is still $1$, positive if even exponent and negative if odd -- is $-1$ to the power of $0$ a positive $1$? such obscure theoretical math, is there even a right answer or did the first person ever to write a book about math just decide which sign she wanted to attribute to the unusual $(-1) ^ 0$ case?

Another way to phrase the punchline to the joke could be:

"because $-1$ is the loneliest number."

since $(-1) ^ 9 = -1$ you get the "because $7, 8, 9$" answer as the lazy student's "obvious" funny answer that appears to satisfy the problem, but the "right" funny answer expected by the puzzle could be to finish working the problem, arriving at the number $-1$, which would be most clearly suggested as the intended solution like this:

"Why is $x$ afraid of $7$? Because $((x + 1) - (x + 2)) ^ {(x + 3)}$ is the loneliest number."

solve for $x$. what is the loneliest number?

most students will know to find an $x$ that results in "$1$ is the loneliest number" so when they see $x = 6$ it will cause confusion and humor on the way to the intended "right" funny answer that says being negative AND being alone is why $6$ is afraid of $7$.

and as a bonus, the math teacher gets to joke about how NEGATIVE $1$ is surely a lonelier number than POSITIVE $1$, just ask any lonely math teacher!

This is fun stuff for puzzles. Thanks for posing the question!

  • 1
    $\begingroup$ Didn't I say any number? If I didn't, I meant any number. $\endgroup$ Feb 28, 2016 at 9:12
  • $\begingroup$ 1 is a number. your question was clear, but it did suggest two parts -- first provide a solution that works in at least one instance, such as 3,3,3 or 1,1,1, and then try to make it work for ALL inputs and show a proof. $\endgroup$
    – Anonymous
    Feb 28, 2016 at 9:17
  • $\begingroup$ The second answer's nice - it makes use of 1 - 1 + 1 nicely. You could've been hinting at the ^0 with that, if you were I apologise. $\endgroup$ Feb 28, 2016 at 11:54
  • 1
    $\begingroup$ when you said "x^0 - x^0 + x^0 = 1"...aren't you using two numbers, 0 and x? same with 1^x - 1^x + 1^x = 1 $\endgroup$
    – manshu
    Feb 28, 2016 at 13:52
  • $\begingroup$ that's one way to interpret the question, but is it not valid also to view it as a fill-in-the-blank problem that asks the puzzle solver to supply three numbers of the same value to solve? $\endgroup$
    – Anonymous
    Feb 28, 2016 at 20:26

It is possible and simple.

$(\frac11 )\times1 =1$

$1\times1\times1 = 1$

$x^{(x-x)} =1$ (except $x$ not $0$)

  • $\begingroup$ why down voted to my answer? any reason, please elaborate. $\endgroup$
    – mbdAli
    Feb 29, 2016 at 14:38
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    $\begingroup$ I didn't downvote you. However: The first two parts are unnecessary, the last (only useful) part is already an accepted answer, so that's probably why you got downvoted. Look at the answers posted before posting your answer, to avoid such situations, especially if it already has an accepted answer ;) $\endgroup$ Feb 29, 2016 at 14:42
  • $\begingroup$ Thanks @PiotrPytlik . I did not know that. i am new to this forum. $\endgroup$
    – mbdAli
    Feb 29, 2016 at 14:45
  • $\begingroup$ upvoted. although repeated, this one is good $\endgroup$ Feb 29, 2016 at 17:31

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