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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
  • Addition, subtraction, multiplication, division
  • Square root
  • Factorial
  • Concatenation (33 counting as two 3s)
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    $\begingroup$ It can't be further 'optimized' with the operations you allow. $\endgroup$
    – DrunkWolf
    Oct 20, 2015 at 10:17
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    $\begingroup$ If you allow 33/3, then you should be asking for solutions with the digit or character 3, not the number 3. $\endgroup$ Oct 20, 2015 at 10:26
  • $\begingroup$ 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. $\endgroup$
    – Lawrence
    Oct 20, 2015 at 10:54
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    $\begingroup$ One optimisation: $12 = 3! + 3!$. $\endgroup$
    – Lawrence
    Oct 20, 2015 at 10:55
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    $\begingroup$ @Rovin It seems that square root is pointless unless expanded with round/floor/ceil. $\endgroup$
    – dmg
    Oct 20, 2015 at 12:39

1 Answer 1

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I've scratched my head at this for an hour and I could only find one improvement:

12 = 3! + 3!

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  • $\begingroup$ This was pointed out by @Lawrence. See comments on the question. $\endgroup$
    – Mad Max
    Oct 20, 2015 at 19:02
  • $\begingroup$ Oh wow, I didn't see that and I even replied to his earlier comment. Oh well, it took me forever to find it and I suppose that's what happens when people post answers as comments. $\endgroup$ Oct 20, 2015 at 23:31

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