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Assemble a formula using the numbers $3$, $4$ and $6$ in any order to make 108.

You may use the operations;

  • $x + y$

  • $x - y$

  • $x \times y$

  • $x \div y$

  • $x!$

  • $\sqrt{x}$

  • $\sqrt[\leftroot{-2}\uproot{2}x]{y}$

  • $x^y$

  • Brackets to clarify order of operations "(",")"
  • Concatenate two or more of the three digits you start with (concatenation of numbers from calculations is not permitted)

as long as all operands are either $3$, $4$ and $6$.

Note that double, triple, etc. factorials (n-druple-factorials) are not allowed though factorials of factorials are fine, such as $((6-3)!)! = 6!$.

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    $\begingroup$ If I could, I would have done something like $$3\times \style{display: inline-block; transform: rotate(180deg)}{4}\times 6=108$$ since the upside down $4$ looks a bit like a $6$. $\endgroup$
    – Mr Pie
    Commented Apr 18, 2019 at 17:44
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    $\begingroup$ @user477343 or even $$3 \times 4 \times 9 = 108$$ since the upside down $6$ is very much like a $9$. $\endgroup$
    – Weather Vane
    Commented Apr 18, 2019 at 18:12
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    $\begingroup$ @WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD $\endgroup$
    – Mr Pie
    Commented Apr 18, 2019 at 18:19
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    $\begingroup$ @user477343 ah but it took yours to make me think of it. $\endgroup$
    – Weather Vane
    Commented Apr 18, 2019 at 18:19
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    $\begingroup$ Based on Vilx's comment, and the fact someone's now posted an answer that uses 6 twice and doesn't use 4, I'm voting to close until it's clarified whether a) you have to use all three numbers, and b) you have to use them only once. $\endgroup$
    – F1Krazy
    Commented Apr 19, 2019 at 13:32

6 Answers 6

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Could this be

$\frac{6^3}{\sqrt{4}} = \frac{216}{2} = 108$?

@Oray found another one, which might possibly be

$6^{\sqrt{4}} \times 3 = 6^2 \times 3 = 36 \times 3 = 108$.

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    $\begingroup$ good finding! mine was different but this seems right too :) $\endgroup$
    – Oray
    Commented Apr 18, 2019 at 17:17
  • $\begingroup$ @Oray: was this second one the one that you found? $\endgroup$
    – El-Guest
    Commented Apr 18, 2019 at 17:29
  • $\begingroup$ no actually :D it was a bit more complicated. $\endgroup$
    – Oray
    Commented Apr 18, 2019 at 17:40
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I have found this solution

$6 \times (4! - 3!) = 6 \times (24 - 6) = 6 \times 18 = 108$

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In Excel:

(6^3)/sqrt(4)

or as Word equation:

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    $\begingroup$ Hi and welcome to puzzling SE! Good, but your answer is same as the accepted answer above. Please avoid posting similar answers, and also have a look at the tour page to familiarise yourselves with the workings of this site. Happy Puzzling! $\endgroup$
    – Eagle
    Commented Apr 19, 2019 at 9:14
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My answer is

6!/3! - (4 x 3) = 108

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No frills...

((6 x 6) x 3) + (4 x 0)

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    $\begingroup$ No dice... using 0 is expressly not allowed. $\endgroup$
    – Rubio ♦
    Commented Apr 22, 2019 at 22:55
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The most simple answer would appear to be

3x6x6=108

Or this, for those who think all three numbers need to be used

(6!/3√4)-6√4=108

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    $\begingroup$ Welcome to Puzzling.SE! First of all, you should hide your answer using a spoiler tag >!, to avoid spoiling the solution for anyone who wants to have a go at the puzzle themselves. Secondly, it's unclear, but it seems like you have to use all three numbers, not just 3 and 6. $\endgroup$
    – F1Krazy
    Commented Apr 19, 2019 at 13:30
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    $\begingroup$ Sorry I should have used a spoiler tag will try to remember in future. Nothing was said about having to use 4 or only using each number once, my answer to the question as it was defined is correct. Someone has down-voted presumably because my answer doesn't fit their interpretation of the problem but that is an issue with how the question was worded not with my answer. I posted this trivial solution primarily to highlight the fact that the question people seem to be answering is not the one that was asked. $\endgroup$
    – fluffy eragon
    Commented Apr 19, 2019 at 13:50
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    $\begingroup$ It does say “using the numbers 3, 4 and 6” which seems clear enough to me to require all three. (It doesn’t say only one of each. I assume that was intended, but it’s certainly not stated.). $\endgroup$
    – Rubio ♦
    Commented Apr 19, 2019 at 14:37
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    $\begingroup$ It also says "as long as all operands are either 3, 4 and 6". It does not say "using all of the numbers 3, 4 and 6 at least once". If that was the intention of the op then they phrased it very badly. Perhaps my answer should have been 3x6x6+4-4=108 $\endgroup$
    – fluffy eragon
    Commented Apr 19, 2019 at 15:07

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