calculation puzzle - clock values using just "9" (and some operators)

Sometimes a clock is posted on Facebook and the likes where digits are substituted by mathematical expressions. A friend of mine asked me what is the smallest numbers of "nine" which let one express numbers from 1 to 12, using the four basic operations, square root, factorial, decimal point and the notation .(9) = 1. I managed to come up with a solution using 21 nines; is it possible to get 20?

EDIT: the notation .(9) means $.{\bar 9}$ (I forgot that puzzling allows MathJaX). For example, 11 may be expressed as 99/9, while 12 is $9 + \sqrt(9)$. So you must have twelve expressions, one for each number from 1 to 12, and using in total the least number of 9.

Answer

How many numbers can you represent using only one 9?

• $9$ itself


• The four basic operations require at least two numbers, so they're out


• $\sqrt 9 = 3$


• $9! = 362880$ and $\sqrt 9! = 6$ and $\sqrt{9!} = 602.395...$


• Decimal point requires at least two numbers


• $.{\bar 9} = 1$ and $\sqrt{.{\bar 9}} = 1$ and $.{\bar 9}! = 1$

Therefore the only numbers on a clock face that can be represented with only one 9 are 1, 3, 6 and 9. All the rest require at least two nines. That gives us a minimum of $4 + 2 \times 8 = 20$ nines.

1. $.{\bar 9}$



2. $.{\bar 9} + .{\bar 9}$


3. $\sqrt 9$


4. $\sqrt 9 + .{\bar 9}$


5. $\sqrt 9! - .{\bar 9}$


6. $\sqrt 9!$


7. $\sqrt 9! + .{\bar 9}$


8. $9 - .{\bar 9}$


9. $9$


10. $9 + .{\bar 9}$


11. ???



12. $9 + \sqrt 9$

Is it possible to get 11 from only 2 nines?

• The square of 11 is 121. If we can get 121 using two nines, we could take its square root.


• 11 or 121 is not a factorial of any integer, so that's out. ($5! = 120$ which is very close but not enough.)


• The basic operations won't help (none of numbers 1, 3, 6, 9, 362880 and 602,395... add up to 11 or 121, they're too small or too big for subtraction or division, and 11 is a prime and $121 = 11 \times 11$ so multiplication won't do it.)

Based on this it looks like 21 is the minimum (4 numbers using 1 nine, 7 numbers using 2 nines and one number using 3 nines.)

But if we relax the requirements a bit and allow any mathematical notation, as long as it doesn't use any other numbers than 9? Then you could do:

$$\lceil \sqrt{ (\sqrt 9! - .{\bar 9})!}\rceil$$

That is, the square root of the factorial of 5 (=120), rounded up.