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Having had just checked into the Grand Hotel, Mr. Hilbert slumped into the hotel room armchair with relief. Finally he could have some peace and quiet and solve the quaint riddle his colleague gave him! He pulled a note out from his breastpocket, and on it was simply scribbled:

Fix this equation by adding three mathematical symbols:
$$2 \space\space\space 2 \space = \space9$$ NO letters, numbers, or tampering with the equal sign!

Unfortunately, an hour and thousands of incorrect symbols later, Mr. Hilbert remained hopelessly stuck. Can you help Mr. Hilbert?

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  • 23
    $\begingroup$ How can Mr. Hilbert spend an hour working on a problem in a hotel? Wouldn't he be requested to move to another room once in every few minutes? $\endgroup$
    – JiK
    Sep 2, 2015 at 9:47
  • 19
    $\begingroup$ @JiK The front desk was busy trying to complete Mr Gödel's check-in. $\endgroup$ Sep 2, 2015 at 20:11
  • $\begingroup$ While the Grand Hotel is large, the occupants move slowly =) Summer is also down season for them I suppose? $\endgroup$
    – Hackiisan
    Sep 3, 2015 at 0:34
  • $\begingroup$ Does 2 + 2 <= 9 qualify as "tampering with the equal sign"? It's still there, just has a < before it. $\endgroup$ Sep 3, 2015 at 15:40
  • $\begingroup$ You mean $\leq$? =) $\endgroup$
    – Hackiisan
    Sep 3, 2015 at 18:35

12 Answers 12

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I don't know how to do the formatting (thanks McMagister for the edit) but the answer is

$ 2\space\div\space .\overline{2} = 9 $

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  • 1
    $\begingroup$ Beautifully solved. $\endgroup$
    – alexmc
    Sep 2, 2015 at 3:09
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    $\begingroup$ That feel when grade school repeated decimal notation is more foreign to me than the gamma function... $\endgroup$
    – McMagister
    Sep 2, 2015 at 3:12
  • $\begingroup$ Well done, much faster than Mr. Hilbert! I had hidden a hint in the story text to be revealed later, but clearly it is unnecessary now =) $\endgroup$
    – Hackiisan
    Sep 2, 2015 at 3:15
  • $\begingroup$ The hint is not so much hidden as I was thinking about it just by reading the title of the question. There is not that many reasons why would one name the usually unnamed (or non-existent) protagonist after a mathematician, besides giving a hint :) $\endgroup$
    – zovits
    Sep 2, 2015 at 9:07
  • 2
    $\begingroup$ For the uninitiated what does the overbar notation mean? I thought that meant complex conjugate? $\endgroup$ Sep 2, 2015 at 12:21
24
+100
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$$ 2 \div 2 = .\overline{9} $$

Simply rearranging the symbols used in the intended solution.

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  • 5
    $\begingroup$ Nice! Mr. Hilbert agrees this could be a valid solution as well. $\endgroup$
    – Hackiisan
    Sep 2, 2015 at 7:19
17
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Probably fails the no letter criterion.

$-2+2={d \over dx} 9$

Or using Lagrange notation as a workaround (thanks to McMagister) we can also write

$-2+2= 9'$

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  • 3
    $\begingroup$ Use lagrange notation as a work-around? $\endgroup$
    – McMagister
    Sep 2, 2015 at 3:01
  • 4
    $\begingroup$ Yes. While technically you can use a "prime" symbol for differentiation, the intended answer is not a simple "cheat". Otherwise, most of similar type questions can be solved by the trivial $a' + b' = c' + d'$ $\endgroup$
    – Hackiisan
    Sep 2, 2015 at 3:01
14
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I came up with this:

$\lceil2\sqrt{2}\rceil = \sqrt{9}$

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2
  • 1
    $\begingroup$ implicit multiplcation - getting that fourth operation without a fourth math symbol - love it! $\endgroup$
    – corsiKa
    Sep 3, 2015 at 2:00
  • 3
    $\begingroup$ ...And then losing two math symbols on one math operation (ceil), ending up with one symbol too many... $\endgroup$
    – namey
    Sep 3, 2015 at 21:04
7
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My answer:

$$\Gamma(2) + 2 = \sqrt9$$

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  • 2
    $\begingroup$ Very good attempt, but unfortunately Mr. Hilbert would consider the use of the gamma function as "using letters". $\endgroup$
    – Hackiisan
    Sep 2, 2015 at 2:51
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    $\begingroup$ Do the brackets count as "mathematical symbols"? $\endgroup$
    – alexmc
    Sep 2, 2015 at 2:53
  • 1
    $\begingroup$ @alexmc Yes, they do. $\endgroup$
    – Hackiisan
    Sep 2, 2015 at 2:58
5
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Assuming that perfect formatting is not required:

$ \neg (2 \space 2 = 9) $

That is:

not (twenty-two equals nine)

If we count a pair of parentheses as a single unit, then:

$ \neg (2 + 2 = 9) $

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  • 2
    $\begingroup$ A solution in the similar vein: $2+2=9\to\bot$ $\endgroup$
    – Taemyr
    Sep 2, 2015 at 19:29
5
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While the accepted answer was also the first one I thought of, there's also a nice solution with subfactorials:

$$2\;!2 = !\sqrt{9}$$

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  • $\begingroup$ What does a factorial sign in front of a number mean in math? $\endgroup$ Jan 10, 2016 at 22:26
  • $\begingroup$ @user1717828, see the linked Mathworld page. $\endgroup$ Jan 10, 2016 at 22:47
  • $\begingroup$ Didn't see that on mobile. My bad. $\endgroup$ Jan 10, 2016 at 23:40
  • 2
    $\begingroup$ Fascinating. Also $2+!2 = \sqrt{9}$. $\endgroup$
    – GOTO 0
    Jan 12, 2016 at 13:29
4
$\begingroup$

What about :

$2\div2=\#\{9\}$

In case I got the symbols wrong, what I am trying to say is:

two divided by two equals the cardinality of the set of numbers that just contains the number nine

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8
  • $\begingroup$ Nice! While I can't say this is wrong (similar to Anachor's answer), it goes against the spirit of this puzzle by ignoring the value of the number. Otherwise, many other trivial answers exist, such as $(\#(2)+2)! = 9$ or $\#(2) + 2 = \sqrt{9}$. I would have edited the question to reflect this, but the correct solution has already appeared =P $\endgroup$
    – Hackiisan
    Sep 2, 2015 at 7:15
  • $\begingroup$ This solution also uses 4 symbols - not the allowed 3. $\endgroup$
    – Eborbob
    Sep 2, 2015 at 12:50
  • $\begingroup$ @Eborbob Parentheses and brackets are always used a unit and I would consider them to be one symbol. Based on his comments on other answers, Hackiisan seems to agree with me. $\endgroup$
    – Kevin
    Sep 2, 2015 at 13:00
  • $\begingroup$ @Kevin Fair enough, I'll remember that for future use! $\endgroup$
    – Eborbob
    Sep 2, 2015 at 13:04
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    $\begingroup$ @Hackiisan $\#(2)+2=\sqrt{9}$ isn't trivial, and doesn't ignore the value of any number but the first. One could argue the 'accepted' answer ignores the value of one of the numbers, because any two equal numbers will result in an acceptable result. $\endgroup$
    – corsiKa
    Sep 2, 2015 at 16:42
4
$\begingroup$

How about this?

$\lfloor\sqrt{2}\rfloor + 2 = \sqrt{9}$

Oops, that is 4 symbols. Thanks @corsiKa.

This one uses three symbols:

$2 - 2 = \lfloor9\%\rfloor$

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1
  • 2
    $\begingroup$ That's four - floor, sqrt, plus, sqrt $\endgroup$
    – corsiKa
    Sep 2, 2015 at 16:37
2
$\begingroup$

How about

$2'\cdot 2' = 9$, where $(\,\cdot\,)'$ denotes the successor function.

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5
  • $\begingroup$ By that logic, I can always define (⋅)′ to always return 3, irrespective of what is inside. (⋅)′ is not standard notation for the successor function, which means you will have to add letters after these 3 operators to explain what you mean. $\endgroup$
    – CodeNewbie
    Sep 3, 2015 at 13:08
  • $\begingroup$ I'm pretty sure that $(\,\cdot\,)'$ is a standard notation for the successor function. At least this is the notation I learned when I was in school. And it is not far fetched to recognise it as such. $\endgroup$
    – Claudius
    Sep 3, 2015 at 13:11
  • $\begingroup$ Can you provide any links to text where such a notation is used? (Hopefully one that isn't too obscure) $\endgroup$
    – CodeNewbie
    Sep 3, 2015 at 13:16
  • $\begingroup$ see this question on math.stackexchange: math.stackexchange.com/questions/643374/… There is a reference to Kleene’s Mathematical Logic that uses this notation. $\endgroup$
    – Claudius
    Sep 3, 2015 at 13:28
  • $\begingroup$ I stand corrected. Although this is not as common as you expect, indicated by the accepted answer in the above question pointing out that S(n) is a more common notation. $\endgroup$
    – CodeNewbie
    Sep 3, 2015 at 13:31
2
$\begingroup$

If we can assume these digits are measured we get:

2 + √2 = √9 (true to one significant figure)

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2
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With 2 mathematical operations:

$ 2-2=\{9\} $

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1
  • 3
    $\begingroup$ {} is the fractional part of a number. Clarifying because a set is used in another answer $\endgroup$
    – Artemmm
    Apr 8, 2019 at 2:01

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