4
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![enter image description here

In the picture above, all the numbers 1-12 are given, using only three nines. Now, I want to make a similar watch, but using fours instead of nines.

Can you make a list of the numbers 1-12 using exactly three fours (for each number). So, for instance, you can't do 4+4 to get 8.

A few rules:

  • Combining two 4s to make 44 is not legal
  • Repeating / recurring decimal representation is not legal (as used to get seven in the example watch).
  • Flooring / rounding is not legal
  • The answer must be exact.
  • Parentheses are of course OK

You are free to use other functions, but it's possible to do this using only elementary functions and the factorial ! (no other "Special functions" (as they are called in the wiki-article)).

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  • 1
    $\begingroup$ May I suggest a name change one the question? 'May the fours be with you' seems fitting. $\endgroup$ Commented Oct 3, 2015 at 11:58
  • $\begingroup$ @Winther it appears to me that the factorial symbol in the picture actually is outside of the square root sign: $\sqrt{9}!$ vs $\sqrt{9!}$. Of course the difference is extremely difficult to see, and this ambiguity could have been avoided by a simple use of parentheses. $\endgroup$
    – mathmandan
    Commented Oct 3, 2015 at 16:15
  • $\begingroup$ @mathmandan: Anachor changed the image :-) $\endgroup$ Commented Oct 3, 2015 at 17:29
  • $\begingroup$ Oh, I see. I neglected to check the edit history. $\endgroup$
    – mathmandan
    Commented Oct 3, 2015 at 17:47

1 Answer 1

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Using only 5 ops, $+,\ -,\ \times,\ \div,\ \sqrt{}$, I can get all (although it might be debatable whether the radix point counts as one)

  • $1=\sqrt4 -{ 4 \over 4}$
  • $2={4+4 \over 4} $
  • $3=4 - { 4 \over 4}$
  • $4=4+4-4$
  • $5=4+{4 \over 4}$
  • $6=4 + 4 -\sqrt4 $
  • $7={\sqrt4 \over .4}+\sqrt4={4!+4 \over 4}$ [Solution without !, thanks to Somo145]
  • $8=4 + \sqrt4 +\sqrt4$
  • $9=(4-.4)/.4$
  • $10=4 + 4 +\sqrt4$
  • $11=(4+.4)/.4={4! - \sqrt4 \over \sqrt 4}$
  • $12=4 + 4 +4$

In fact, you can get rid of $-$ and $\times$ and do it with three $\big(+,\div,\sqrt{} \big)$:

  • $1={\sqrt4 + \sqrt4 \over 4}$
  • $2={4+4 \over 4} $
  • $3=\sqrt4 + { 4 \over 4}$
  • $4={4+4 \over \sqrt4}$
  • $5=4+{4 \over 4}$
  • $6=\sqrt4 + \sqrt4 +\sqrt4 $
  • $7={\sqrt4 \over .4}+\sqrt4 $ [Somo145]
  • $8=4 + \sqrt4 +\sqrt4$
  • $9={\sqrt4 \over .4}+4 $
  • $10=4 + 4 +\sqrt4$
  • $11=(4+.4)/.4$
  • $12=4 + 4 +4$
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  • $\begingroup$ @Anachor Getting rid of factorials? $\frac{\sqrt{4}}{.4} + \sqrt{4} = 7$ $\endgroup$
    – Somo145
    Commented Oct 3, 2015 at 12:41
  • $\begingroup$ @Somo145 Nice one, don't know why I didn't think of that. $\endgroup$
    – Rohcana
    Commented Oct 3, 2015 at 12:44
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    $\begingroup$ Well done! =) My answer for the tricky number 9: cot(acos(sqrt(4)/4))^-4 = 9. Not exactly pretty, but the radix point is avoided, and only elementary functions are used. $\endgroup$ Commented Oct 3, 2015 at 13:18
  • $\begingroup$ @StewieGriffin Impressive! Can you get rid of the radix point for 7 and 11 as well? $\endgroup$
    – Lawrence
    Commented Oct 3, 2015 at 13:32
  • $\begingroup$ @StewieGriffin Nice one without the radix, I don't think I would have thought of that. $\endgroup$
    – Rohcana
    Commented Oct 3, 2015 at 14:53

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