7

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.

Xandawesome
  • 1,274
  • 9
  • 21

4 Answers4

9

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 $33 \frac 13$.

8

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.

vero
  • 738
  • 4
  • 12
  • I've got to say, this is pretty creative! ;) – Xandawesome Jun 21 '15 at 21:41
  • 6
    The last line works, but the middle paragraph is very woolly and unrigorous. Is that really true? – Rand al'Thor Jun 21 '15 at 22:09
  • 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. – vero Jun 21 '15 at 23:31
  • And (√(4!))! is almost 100. Factorial square root factorial is interesting... – vero Jun 21 '15 at 23:35
  • 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. – Milo Brandt Jun 22 '15 at 00:24
  • @Meelo I'm not an expert in this field by any means but even if the square root is irrational, wouldn't it be possible for a result to become infinitely close to 100? Maybe after applying 9001 trillion trillion factorials to 3, it becomes a number that can be square rooted to 100.000000001. And if that's the case, then maybe that decimal can always be moved to the right with more operations. – Quark Jun 22 '15 at 00:59
  • @Quark Well, I'm relatively confident that one can get arbitrarily close to $100$ (though I can't prove it), but a convergent sequence cannot, in any case, be built by successively adding a single operation - so we need to group multiple operations into one to make a sensible limit. If we restrict factorials to integers, it is even worse: there is no sequence of operations with $100$ as a convergent subsequence. Once we start taking square roots we may never take another factorial, so the sequence passes as close to $100$ as it will, then shrinks to near $10$ and keeps shrinking. – Milo Brandt Jun 22 '15 at 01:05
  • factorial does work on all real numbers, or at least on every single calculator that I have gotten my hands on. I think the definition of the factorial is the gamma function... its up to the OP. – vero Jun 22 '15 at 02:34
  • @Meelo I don't think you can get a concise way to express that limit, though, without using the number "100" itself which would not be allowed. –  Jun 22 '15 at 16:10
  • "3/0.03" is not valid, because it uses digits other than threes. – Olive Stemforn Jun 26 '15 at 16:53
7

How about this?

$(3*3)\over{(.3*.3)}$

If this counts, then these would also count:

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

Quark
  • 6,077
  • 1
  • 20
  • 46
0

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).

Voitcus
  • 445
  • 3
  • 6
  • You can't just use 0.(3) to denote 0.333... – Olive Stemforn Jun 26 '15 at 16:56
  • Why not? I provided links showing this is a correct notation – Voitcus Jun 26 '15 at 17:02
  • 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. – Olive Stemforn Jun 26 '15 at 17:18
  • 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). – Voitcus Jun 26 '15 at 19:51
  • (BTW I upvoted your comment regarding that decimal point was not allowed) – Voitcus Jun 26 '15 at 19:53