# Clock values using only the digit 3

What is the minimal number of 3s required to create a clock where each of the hours is replaced by a mathemetical expression using only the digit 3 and standard mathematical operators?

A friend mentioned this puzzle to me, but didn't remember the answer. The puzzle is inspired by this clock. I tried searching for similar puzzles, but only came across this one, which has has rules I don't really like. I prefer this puzzle in a simpler form, only allowing rational numbers.

I'm interested in both the minimal number of 3s per expression and the minimal number of 3s in total.

My best solution so far uses thirty 3s, but I feel like it can be improved. It would be nice to be able to formulate the puzzle such that you are only allowed to use 3 three times per expression.

My solution:

$1 = 3\div3$
$2 = 3!\div3$
$3 = 3$
$4 = 3 + (3\div3)$
$5 = 3! - (3\div3)$
$6 = 3!$
$7 = 3! + (3\div3)$
$8 = 3! + (3!\div3)$
$9 = 3\times3$
$10 = (3\times3) + (3\div3)$
$11 = 33\div3$ or with one extra 3: $((3!\times3!) - 3)\div3$
$12 = (3!\times3!)\div3$

Summary:

Create a clock face using only the digit 3 and standard mathematical operations

Allowed:

• The digit 3
• Square root
• Factorial
• Concatenation (33 counting as two 3s)
• It can't be further 'optimized' with the operations you allow. – DrunkWolf Oct 20 '15 at 10:17
• If you allow 33/3, then you should be asking for solutions with the digit or character 3, not the number 3. – user1717828 Oct 20 '15 at 10:26
• If you allow '33' (concatenation), how about decimal points and the recurring-decimal dot? You could then write $10 = 3 \times 3.\dot{3}$. This reduces the number of instances of '3' by one. – Lawrence Oct 20 '15 at 10:54
• One optimisation: $12 = 3! + 3!$. – Lawrence Oct 20 '15 at 10:55
• @Rovin It seems that square root is pointless unless expanded with round/floor/ceil. – dmg Oct 20 '15 at 12:39