# May the fours be with you!

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)).

• May I suggest a name change one the question? 'May the fours be with you' seems fitting. – Tim Couwelier Oct 3 '15 at 11:58
• @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. – mathmandan Oct 3 '15 at 16:15
• @mathmandan: Anachor changed the image :-) – Stewie Griffin Oct 3 '15 at 17:29
• Oh, I see. I neglected to check the edit history. – mathmandan Oct 3 '15 at 17:47

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]
• $8=4 + \sqrt4 +\sqrt4$
• $9={\sqrt4 \over .4}+4$
• $10=4 + 4 +\sqrt4$
• $11=(4+.4)/.4$
• $12=4 + 4 +4$
• @Anachor Getting rid of factorials? $\frac{\sqrt{4}}{.4} + \sqrt{4} = 7$ – Somo145 Oct 3 '15 at 12:41
• @Somo145 Nice one, don't know why I didn't think of that. – Rohcana Oct 3 '15 at 12:44
• 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. – Stewie Griffin Oct 3 '15 at 13:18
• @StewieGriffin Impressive! Can you get rid of the radix point for 7 and 11 as well? – Lawrence Oct 3 '15 at 13:32
• @StewieGriffin Nice one without the radix, I don't think I would have thought of that. – Rohcana Oct 3 '15 at 14:53