4
$\begingroup$

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 $((6-3)!)! = 6!$.

$\endgroup$
  • 1
    $\begingroup$ 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 $6$. $\endgroup$ – Feeds Apr 18 at 17:44
  • 6
    $\begingroup$ @user477343 or even $$3 \times 4 \times 9 = 108$$ since the upside down $6$ is very much like a $9$. $\endgroup$ – Weather Vane Apr 18 at 18:12
  • 1
    $\begingroup$ @WeatherVane Of course! Huh... funny how I saw what was obscure as opposed to what was actually just plain simple xD $\endgroup$ – Feeds Apr 18 at 18:19
  • 3
    $\begingroup$ @user477343 ah but it took yours to make me think of it. $\endgroup$ – Weather Vane Apr 18 at 18:19
  • 1
    $\begingroup$ 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 Apr 19 at 13:32
11
$\begingroup$

Could this be

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

@Oray found another one, which might possibly be

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

$\endgroup$
  • 1
    $\begingroup$ good finding! mine was different but this seems right too :) $\endgroup$ – Oray Apr 18 at 17:17
  • $\begingroup$ Thank you, @Oray!! $\endgroup$ – El-Guest Apr 18 at 17:26
  • $\begingroup$ @Oray: was this second one the one that you found? $\endgroup$ – El-Guest Apr 18 at 17:29
  • $\begingroup$ no actually :D it was a bit more complicated. $\endgroup$ – Oray Apr 18 at 17:40
14
$\begingroup$

I have found this solution

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

$\endgroup$
2
$\begingroup$

In Excel:

(6^3)/sqrt(4)

or as Word equation:

$\endgroup$
  • 5
    $\begingroup$ 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 Apr 19 at 9:14
1
$\begingroup$

My answer is

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

$\endgroup$
0
$\begingroup$

No frills...

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

$\endgroup$
  • 1
    $\begingroup$ No dice... using 0 is expressly not allowed. $\endgroup$ – Rubio Apr 22 at 22:55
  • $\begingroup$ You're right. Thanks $\endgroup$ – Jon Apr 23 at 15:06
-1
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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 Apr 19 at 13:30
  • 1
    $\begingroup$ 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 Apr 19 at 13:50
  • 2
    $\begingroup$ 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 Apr 19 at 14:37
  • 2
    $\begingroup$ 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 Apr 19 at 15:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.