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Given to me, by a friend:

How would you make 20, using two threes?

  • You may use any basic operation. And others, such as square roots (the symbol), factorials, etc. Any operation is allowed.
  • Just two threes, though. You can't raise a three to any digit other than three, etc.

Any out of the box, non-mathematical solutions are encouraged.

PS- Wow! That's an amazing number of responses! Quite a community, here.

EDIT: @CameronWhite and @Yly have provided, perhaps the most elegant answers. And basis your votes, I'm going ahead to mark @Cameron's answer as accepted. No issues, I hope. Although, I did think @KeyboardWielder's answer was rather crafty. And @dcyfj... Well, that would have taken some time. Awesome answer. This really isn't an edit, is this?

EDIT: Perhaps I should have specified the level of out-of-the boxness allowed. Forgive my mistake. This was, my first post, and I was surprised by the number of answers. Thanks for taking the time.

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closed as too broad by Deusovi, Mike Earnest, Aza Jun 8 '16 at 23:46

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 5
    $\begingroup$ Mercy, downvoter. $\endgroup$ – Nikhil Itty Jun 6 '16 at 18:40
  • $\begingroup$ have you got a solution in mind? $\endgroup$ – JonMark Perry Jun 6 '16 at 19:05
  • $\begingroup$ I was originally thinking of it on the lines of @KeyboardWielder's answer, after defining a function went out the window. $\endgroup$ – Nikhil Itty Jun 6 '16 at 19:11
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    $\begingroup$ 3:3 - it's a football score*! $\endgroup$ – JonMark Perry Jun 7 '16 at 6:43
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    $\begingroup$ Answer: $3!=20_3$ (not enough reputation to post) $\endgroup$ – LGT Jun 8 '16 at 18:50

21 Answers 21

110
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If you are allowed to use decimals, then

$$\frac{3!}{.3} = 20$$

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113
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Does this count?

$\dbinom{3!}{3} = 20$

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    $\begingroup$ I like this one the best. $\endgroup$ – Joe Z. Jun 7 '16 at 14:21
  • $\begingroup$ So simple, I love it! $\endgroup$ – Martijn Jun 7 '16 at 17:46
  • $\begingroup$ It is just 2! How is this right? $\endgroup$ – Gaurav Agarwal Jun 8 '16 at 18:41
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    $\begingroup$ @GauravAgarwal: Read it as "6 choose 3", which is the binomial coefficient equating to the number of different ways to choose 3 elements from a set of 6. $\endgroup$ – Michael Myers Jun 8 '16 at 19:34
  • $\begingroup$ @MichaelMyers Thank you. In light of this knowledge, this is truly the most elegant answer. $\endgroup$ – Gaurav Agarwal Jun 8 '16 at 21:49
72
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Since the puzzle oddly and specifically mentions the symbol for the square root, I used this:

enter image description here

but rotated and reflected it giving:

enter image description here

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  • 1
    $\begingroup$ definitely "out-of-the-box" lol $\endgroup$ – dcfyj Jun 6 '16 at 18:57
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    $\begingroup$ Man, oh man... That, is rotating the box. But a very cool observation. $\endgroup$ – Nikhil Itty Jun 6 '16 at 19:01
  • $\begingroup$ you could do that with factorial if you just bend it a little bit and hide the dot $\endgroup$ – JonMark Perry Jun 6 '16 at 19:03
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    $\begingroup$ @RipTide You did encourage "out of the box, non-mathematical"! $\endgroup$ – Dan Russell Jun 6 '16 at 19:10
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    $\begingroup$ I wish I could +20 this. That note the symbol was driving me crazy. $\endgroup$ – Devsman Jun 7 '16 at 18:35
68
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Another answer could be

$\lfloor \sqrt{(3!)!}-3! \rfloor$ 
$= \lfloor \sqrt{6!}-3! \rfloor$ 
$= \lfloor \sqrt{720}-6 \rfloor$ 
$= \lfloor 26.8328...-6 \rfloor$ 
$= \lfloor 20.8328... \rfloor$ 
$= 20$

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  • $\begingroup$ That fits, as well. $\endgroup$ – Nikhil Itty Jun 6 '16 at 19:25
  • $\begingroup$ I would've put each equation on a separate line, but for the life of me I can't seem to figure it out. $\endgroup$ – dcfyj Jun 6 '16 at 19:26
  • $\begingroup$ You should add parentheses between the postfix symbols; see this Wiki article. (Tried to keep the comment spolier-free). $\endgroup$ – Angew Jun 7 '16 at 10:58
  • $\begingroup$ This uses approximation. Also, $\lfloor\sqrt{3!!}-3!\rfloor$ is not invalid. $\endgroup$ – EKons Jun 7 '16 at 16:17
  • $\begingroup$ @ΈρικΚωνσταντόπουλος: There is no mathematical approximation, it is just abbreviation in writing (filling up the page with an infinite or so number of digits would be a highly irrational thing to do). $\endgroup$ – KeyboardWielder Jun 7 '16 at 18:06
40
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The simplest one is:

three + three = 20 (in base 3)

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  • 16
    $\begingroup$ This would work for all bases greater than 3, too. $x+x=20_x$. $\endgroup$ – Engineer Toast Jun 6 '16 at 19:13
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    $\begingroup$ The question title specifically says "twenty", which is a specific value, not a representation of a value in any base. three + three will always equal six, not twenty. The question does, however, says "20" $\endgroup$ – Suppen Jun 7 '16 at 6:55
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    $\begingroup$ three + three = 20 but 20 is not twenty in base 3. $\endgroup$ – immibis Jun 7 '16 at 7:07
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    $\begingroup$ @EngineerToast That's why I've started to increase the base of my age instead of the years. I'm 21 now. Base 17. $\endgroup$ – Hendrik Wiese Jun 7 '16 at 8:27
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    $\begingroup$ @Suppen Number 20 is pronounced as "twenty" regardless of the base. So three + three = twenty in base three. However, the problem of this solution is that in order to be correct, it requires third three on the right side, which makes it incorrect regarding the puzzle limitations. $\endgroup$ – fbxmg Jun 7 '16 at 14:32
35
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How would you get 20, using two threes?

Any out of the box, non-mathematical solutions are accepted.

  How about an “in the box” solution?
  You'd be lucky to not get 20, following the explicit formula published in U.S.[A.] Code, Title 18:

§471. Obligations or securities of United States.   Whoever, with intent to defraud, falsely makes, forges, counterfeits, or alters any obligation or other security of the United States, shall be fined under this title or imprisoned not more than 20 years, or both.

  Naturally, the threes are ...

... USA three-dollar bills ...

  This formula can be generalized for any positive number of threes, of course, but may not translate exactly to all international units of measure.

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  • $\begingroup$ Ooops, to be really sure to get 20, you'd have to first reassemble pieces of the threes to more closely resemble a denomination actually in circulation $\endgroup$ – humn Jun 8 '16 at 3:24
27
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Here's another one:

enter image description here

Sorry for MS Paint skills

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  • $\begingroup$ I was thinking of using a couple of threes drawn with four matchsticks each (e.g. two stacked ">" each) but turning the first left 90 degrees and the second right 90 degrees, and then stacking them to yield "XX", but leaving the threes "intact". $\endgroup$ – supercat Jun 6 '16 at 20:12
  • $\begingroup$ Opening this spoiler crashes my iPhone app :O $\endgroup$ – Marco Bonelli Jun 7 '16 at 6:55
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    $\begingroup$ @MarcoBonelli: That's very strange, it's just a png image. $\endgroup$ – Business Cat Jun 7 '16 at 12:41
  • $\begingroup$ You can also arrange your halves to form $\rm XX$ if you allow curved $\rm X$'s. $\endgroup$ – yo' Jun 7 '16 at 22:06
22
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If we're allowing chopping up the numbers, how about some Roman Numerals enter image description here

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21
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As you specifically said to think outside the box, I drew a box, and two three's (one Roman numeral and one Arabic numeral) which extend "outside the box". As my answer note that they divide the box into 20 (yellow regions).

diagram

Therefore: Box / two three's = 20

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  • 1
    $\begingroup$ Isn't this an "inside the box" solution :) $\endgroup$ – user21939 Jun 8 '16 at 13:25
  • $\begingroup$ I <3 this one... haha, made my day. $\endgroup$ – tfrascaroli Jun 8 '16 at 13:46
20
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Two threes:

enter image description here

How would you make 20, using two threes?

Solution:

enter image description here

With apologies.

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14
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How would you make 20, using two threes?

Here's the two: 2
and here are the threes: 3 - 3 = 0
giving 2 0.

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13
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A little bit silly, but it gets 20 without using any digits other than the two '3's

$\lfloor3+3+\pi+\pi+\pi+e+e\rfloor$

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  • $\begingroup$ Let me understand, are you claiming that irrational numbers are rational? ah, got it XD $\endgroup$ – GameDeveloper Jun 7 '16 at 9:12
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    $\begingroup$ @DarioOO He's just using the floor function (wolfram). $\endgroup$ – Marco Bonelli Jun 7 '16 at 12:55
  • $\begingroup$ @MarcoBonelli He meant $\pi$ and $e$. $\endgroup$ – EKons Jun 7 '16 at 16:25
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Since the OP says "any operation is allowed", I find it convenient to use this one: 3 ʭ 3, where the operator ʭ is defined as follows: ∀(x∈Z) ∀(y∈Z) x ʭ y = 20

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  • 5
    $\begingroup$ Ah, the old GiveTwenty operator. $\endgroup$ – Phil H Jun 8 '16 at 13:43
10
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output(++(++(++3))*(++3)); Consider this as a C expression

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  • 6
    $\begingroup$ error: lvalue required as increment operand $\endgroup$ – immibis Jun 7 '16 at 7:10
  • $\begingroup$ for safety, use pre-increment $\endgroup$ – JonMark Perry Jun 7 '16 at 7:21
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    $\begingroup$ For correctness, use pre-increment. As written, the result, if the code were valid at all, would be 9. $\endgroup$ – Christopher Creutzig Jun 7 '16 at 8:37
  • $\begingroup$ I tried it with pre-increments and it still gives me an error, at least with clang: 3++.c:4:21: error: expression is not assignable $\endgroup$ – Joe Z. Jun 7 '16 at 14:27
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    $\begingroup$ On top of the compilation errors that others mentioned, this would produce 24, not 20. $\endgroup$ – Kevin Jun 8 '16 at 12:28
7
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I would sell the first $3$ to person A for $10$, and the second to Person B, again for $10$, giving me $20$

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  • 7
    $\begingroup$ Wouldn't you then have 14 since you had to give 6 away? :p $\endgroup$ – Inazuma Jun 7 '16 at 0:39
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    $\begingroup$ you would have 20 and sold 6. you could sell at 13 i suppose, but you would then have 26. depends how you look at it. $\endgroup$ – JonMark Perry Jun 7 '16 at 3:42
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    $\begingroup$ i had 6, i made 14 profit, now i have 20 $\endgroup$ – JonMark Perry Jun 8 '16 at 6:10
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In C++:

#include <cassert>
int ConvertThreesToTwenty(const int first_three, const int second_three)
{
    // verify preconditions
    assert(first_three == second_three && first_three == 3);
    return 20;
}
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5
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Just buy two \$3 things from ebay. That will be \$20 easy.

Not answering within the spirit of this, sorry, but I'm waiting with baited breath as I can't figure out a correct answer.

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  • 1
    $\begingroup$ You can display the \$ sign by writing \\\$ $\endgroup$ – Anton Jun 7 '16 at 14:15
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    $\begingroup$ baited breath -- ರ_ರ $\endgroup$ – user1717828 Jun 7 '16 at 18:38
  • $\begingroup$ @user1717828 I second that ರ_ರ $\endgroup$ – Kevin Jun 8 '16 at 12:30
5
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enter image description here

Very non-mathematical and taking a leaf out of the Car Talk match-stick puzzlers, we can shift around some lines and get a 2 and a 0 from a 3 and a 3.

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  • $\begingroup$ I would call that lateral thinking, for sure! $\endgroup$ – Engineer Toast Jun 7 '16 at 21:17
5
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Step 1: locate a person with a 20 dollar bill

Step 2: offer $33 for the 20 dollar bill

Step 3: acquire the 20 dollar bill in exchange for $33

Step 4: profit?? (probably not, unless it was some kind of vintage 20)

Congratulations you have converted \$33 into \$20!

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  • 4
    $\begingroup$ "Step 4: profit???" just looks weird when we know that successful business models have "step 2: ???; step 3: profit" $\endgroup$ – Hagen von Eitzen Jun 8 '16 at 12:45
4
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If we consider the function :$$f(x) = (x-1)\sum_{i=1}^{x+1}i$$ Then we see that $f(3) = 20$. So in this case we get away with using just one 3, but maybe $-1$ and $+1$ inside the function would be considered cheating. If we must use two threes then function: $$g(x,y) = (x-1)\sum_{i=1}^{y+1}i$$ then $$g(3,3) = 20$$

Aren't the functions simple enough to be considered elementary operations?

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  • 2
    $\begingroup$ Then f(x,y) = x+y+14 is also simple enough. :) $\endgroup$ – KeyboardWielder Jun 7 '16 at 18:12
  • $\begingroup$ Yes a first degree polynomial is simpler than my second/third degree. But 14 is not as "basic" as +1 or -1. :) $\endgroup$ – mathreadler Jun 7 '16 at 18:32
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    $\begingroup$ ^vote for using both variables where one could simply have been ignored. Could further substitute $\frac x y$ for each $1$ . . . $\endgroup$ – humn Jun 7 '16 at 19:30
4
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Using buttons found on most any calculator, hit the keys:

$3$ $!$ $!$ $\div$ $3$ $!$ $^2$ $=$

The result will be $20$.

I know that's a stretch on the definition of basic operations, but it works.

More in keeping with the rules, the following also works (from comment by humm):

$3$ $!$ $!$ $\div$ $3$ $!$ $=$ $=$

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  • 1
    $\begingroup$ OP specifically stated that you can't directly raise to a digit other than 3. But perhaps you could abuse the features which repeat the previous operation/operand or use the M+/MR keys. ;) $\endgroup$ – KeyboardWielder Jun 7 '16 at 18:16
  • 2
    $\begingroup$ Your workaround works great, @KeyboardWielder, and compelled me ^vote this answer after trying: 3 n! n! / 3 n! = = (could be edited into the answer) $\endgroup$ – humn Jun 7 '16 at 19:19
  • $\begingroup$ You can always write just floor of e instead of 2. I had to do that in my solution. $\endgroup$ – BoltKey Jun 7 '16 at 21:50
  • $\begingroup$ I agree with @BoltKey. Floor e. $\endgroup$ – Nikhil Itty Jun 9 '16 at 3:25
  • $\begingroup$ Just a note on notation. $n!!\ne (n!)!$ - the former is called semifactorial and is a notation for product of all positive integers up to $n$ which have the same parity as $n$. $\endgroup$ – elias Feb 26 '17 at 22:53

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