This question is similar to my other question Twos For Thought, which arguably was a better pun but I decided to recreate the question, except with threes!

How many threes do you need to get to the number $100$ exactly?

Example: $3^3*3!-33-33+3+\frac{3}{3}$

This example method uses 10 threes and is obviously not ideal. Can you find a way that uses the least amount of threes? The lowest number of threes will win!

Math You Can Use:

  • Addition
  • Subtraction
  • Multiplication
  • Division
  • Exponents
  • Square & Cube Roots
  • Factorials
  • Parenthesis are allowed as long as they are used to order operations

Other Rules For Clarification:

  • The only number you may use are threes; any other numbers must be made with a combination of threes in some way.

  • This problem uses base 10 and the answer should be base 10.

  • Decimal points are allowed (Expressions like .$n$ instead of 0.$n$ is allowed, but .0$n$ instead of 0.0$n$ is not.

  • 2
    $\begingroup$ As Jeremy Stein pointed out, all numbers are 10 in base 10 (themselves). You should say "This problem uses base ten" ;-) $\endgroup$ Commented Jun 22, 2015 at 17:14
  • $\begingroup$ Problem poser, you need to explicitly state that decimal points are permitted. $\endgroup$ Commented Jun 26, 2015 at 16:46
  • $\begingroup$ They are added to the rules $\endgroup$ Commented Jun 26, 2015 at 20:49

4 Answers 4


How about this?


If this counts, then these would also count:

$(33-3)/.3$ and $33/.33$ (credit to Joe Z.)

  • $\begingroup$ Very nice! But is ".3" allowed as an abbreviation for "0.3"? $\endgroup$ Commented Jun 21, 2015 at 19:57
  • $\begingroup$ If you're doing it like that, you can also do $(33-3)/.3$ or $33/.33$. $\endgroup$
    – user88
    Commented Jun 21, 2015 at 20:07
  • $\begingroup$ I suppose this would work, and I'll count the .3 as 0.3 $\endgroup$ Commented Jun 21, 2015 at 23:56
  • 3
    $\begingroup$ @randal'thor But is "3" allowed as an abbreviation of "3.0" (or "03")? $\endgroup$ Commented Jun 22, 2015 at 0:20
  • 1
    $\begingroup$ The problem poser never stated that decimal points were permitted, so I stated that the problem poser needs to state it as such if it is to be allowed. $\endgroup$ Commented Jun 26, 2015 at 16:44

A quick initial attempt gives us five:

$33 \times 3 + \frac 33 = 100$

If we're allowed to put overlines on top of numbers, I can get four:

$33.\overline{3} \times 3 = 100$

The overline means "33.3 repeated", or $33 \frac 13$.

  • $\begingroup$ The overlines are not allowed, but the first one is good. But can you get it smaller? $\endgroup$ Commented Jun 21, 2015 at 19:14
  • $\begingroup$ I'm thinking about it right now. $\endgroup$
    – user88
    Commented Jun 21, 2015 at 19:15
  • $\begingroup$ Do you know of a solution that uses only four? $\endgroup$
    – user88
    Commented Jun 21, 2015 at 19:17
  • $\begingroup$ Not currently. I knew the five threes but I'm not sure about four. It's worth a shot though, I bet there is a way! I'm thinking too! $\endgroup$ Commented Jun 21, 2015 at 19:19
  • $\begingroup$ I think using an overline on top of the 3 is sort of like having an infinite number of 3's haha... $\endgroup$
    – vero
    Commented Jun 21, 2015 at 21:41

If we are allowed to use an infinite number of operations, and we are limited in only the number of 3's that we use, then we can get away with using just one 3.

Since the factorial function increases the number and the square root function decreases the number, applying an infinite number of factorials and square roots on 3 in a certain sequence at some point will result in 100. If we can use any rounding function then this becomes even easier.

Another way to do it would be: 33*3+√√√... ...√√√3 (An infinite number of square roots converges to 1)

And, if what Quark answered is valid, then is 3/0.03 valid? That would be a 2 threes solution.

  • $\begingroup$ I've got to say, this is pretty creative! ;) $\endgroup$ Commented Jun 21, 2015 at 21:41
  • 6
    $\begingroup$ The last line works, but the middle paragraph is very woolly and unrigorous. Is that really true? $\endgroup$ Commented Jun 21, 2015 at 22:09
  • $\begingroup$ Playing around and trying to prove that you can get to 100 i found something interesting: (√(3!))! is almost pi... off by a little less than 0.003. $\endgroup$
    – vero
    Commented Jun 21, 2015 at 23:31
  • $\begingroup$ And (√(4!))! is almost 100. Factorial square root factorial is interesting... $\endgroup$
    – vero
    Commented Jun 21, 2015 at 23:35
  • $\begingroup$ If we can use floor, this works. Otherwise the middle paragraph seems unlikely - if we restrict factorials to natural numbers (which isn't really a restriction so much as "the domain of the factorial" - extensions like the gamma function are not wholly natural, even if they coincide at integers), we can prove that that doesn't happen. The only perfect square factorials are $0!$ and $1!$ and the square root of any other thing is irrational. I would bet that even using the gamma function doesn't suffice. $\endgroup$ Commented Jun 22, 2015 at 0:24

Maybe this one:

33.(3) * 3 = 100

This however does not follow the requirement "Parenthesis are allowed as long as they are used to order operations", in this example it denotes 0.(3) means 0.3333....

The Wikipedia also marks infinite repeating decimals with a line or a dot over the number; here is the image from Wikipedia:

repeating 0,3333

which would allow to not uses parentheses, so this should count (4 threes).

  • $\begingroup$ You can't just use 0.(3) to denote 0.333... $\endgroup$ Commented Jun 26, 2015 at 16:56
  • $\begingroup$ Why not? I provided links showing this is a correct notation $\endgroup$
    – Voitcus
    Commented Jun 26, 2015 at 17:02
  • $\begingroup$ As the problem poser stated "[parentheses] are allowed as long as they are used to order operations." You are not using them to order operations. $\endgroup$ Commented Jun 26, 2015 at 17:18
  • $\begingroup$ Well, yes. I presented also notation with a dot over the number or a line. The decimal point is also not allowed, however it is in the accepted answer. You should rather say that I used the same solution as Joe Z. did (unfortunately, I haven't noticed his answer before, so I accept he was the first). $\endgroup$
    – Voitcus
    Commented Jun 26, 2015 at 19:51
  • $\begingroup$ (BTW I upvoted your comment regarding that decimal point was not allowed) $\endgroup$
    – Voitcus
    Commented Jun 26, 2015 at 19:53

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