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Can you find a way to make 1 1 1 1 = 5 by adding any operations or symbols? You can use symbols such as these: +, -, *, !, ^, (). It is not limited to this list.

Good Luck!

P.S. You can not add any other numbers to the equation.

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    $\begingroup$ Why have these “find a way to make blah from blah” questions become so popular all of a sudden? $\endgroup$ – NL628 Jan 9 '18 at 7:05
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    $\begingroup$ 1111 != 5 , you mad :) ? $\endgroup$ – nl-x Jan 9 '18 at 15:01
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    $\begingroup$ Just consider $(1111)_n=5$ where $$n=\frac{1}{3} \left(-2 \sqrt[3]{\frac{2}{115+3 \sqrt{1473}}}+\sqrt[3]{\frac{1}{2} \left(115+3 \sqrt{1473}\right)}-1\right)$$ $\endgroup$ – user69652 Jan 9 '18 at 16:15
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    $\begingroup$ Kinda similar: 1(1-1)1 = 101 = 5 (in base 2). $\endgroup$ – userNaN Jan 10 '18 at 3:10
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    $\begingroup$ (1+1)(1+1)+1 = 5 $\endgroup$ – phuzi Jan 10 '18 at 9:46

10 Answers 10

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How about

$(1+1+1)!-1$ = $6 -1$ = $5$

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  • $\begingroup$ I'd say this is the better answer than mine, as mine concatenates two digits. $\endgroup$ – Phylyp Jan 9 '18 at 2:54
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    $\begingroup$ @Phylyp, that doesn't necessarily make your answer worse; the OP said nothing about concatenation, so we can just assume that we can leave a space blank if we want to :). Your answer is perfectly valid, I think. $\endgroup$ – Reinis Mazeiks Jan 9 '18 at 2:57
  • $\begingroup$ @R.M That is true. I never said anything about concatenation. The answer is perfectly valid. $\endgroup$ – kraby15 Jan 9 '18 at 3:52
  • $\begingroup$ The interesting question is if you can do any better with the given symbols, or at least alternatively - and if not, how can you prove this? $\endgroup$ – The_Sympathizer Jan 9 '18 at 8:34
  • $\begingroup$ @StrangePhoton, I saw your edit attempt – but my answer uses only 4 ones: (1+1+1)!-1. You might be confused about the 1 that appeared when I was solving the question (the expression is equivalent to 6-1). Only the numbers in the first expression are my answer. (And welcome to Puzzling SE! :) $\endgroup$ – Reinis Mazeiks Jan 10 '18 at 13:12
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How about such a variant?

++(1+1+1+1)=5

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    $\begingroup$ Ladies and gents, we have a programmer amongst us! :-) $\endgroup$ – Phylyp Jan 9 '18 at 3:05
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    $\begingroup$ This doesn’t typically work in programming languages, though. And if it were valid in this type of challenge, you could just add the appropriate number of ++ to get anything… $\endgroup$ – Ry- Jan 9 '18 at 3:47
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    $\begingroup$ This won't work in most (all of I know) programming languages as increment operator requires a l-value whereas 4 is an r-value. $\endgroup$ – Cem Kalyoncu Jan 9 '18 at 7:29
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    $\begingroup$ @IamtheMostStupidPerson At least it doesn't in C/C++: coliru.stacked-crooked.com/a/bde41bc9a5b1d16a $\endgroup$ – HolyBlackCat Jan 9 '18 at 10:54
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    $\begingroup$ It also doesn't work in c#, python, js, or java. With the addition of C/C++ this covers quite a few important languages. $\endgroup$ – bob0the0mighty Jan 9 '18 at 14:42
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Here's another solution:

$1$E$1$ $/ (1 + 1)$ = $5$

Where:

E is the exponential notation symbol. i.e. 1E1 = $1 \times 10^1$

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1 << 1 << $1 + 1 = 5$
where << is a bitwise-shift

1 << 1 = (binary) $10 = 2$
2 << 1 = (binary) $100 = 4$, then finally add 1 to get 5

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  • $\begingroup$ Welcome to Puzzling.SE! Note, however, that the question asks for making the result of the operation to be 5, not 6. If you were aware of this and wanted to post it anyway for some reason, please address that in your answer. $\endgroup$ – Lolgast Jan 9 '18 at 6:25
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    $\begingroup$ @Lolgast. Nope. I just really derped on that one $\endgroup$ – Jonah Haney Jan 9 '18 at 6:44
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    $\begingroup$ Ah, we all derp sometimes :) Hopefully the downvoter will be kind enough to remove his downvote since this is now a valid answer (assuming he downvoted for not being valid). Eitherway I've upvoted $\endgroup$ – Lolgast Jan 9 '18 at 7:45
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    $\begingroup$ you may need parens in some programming languages such as Python: (1<<1<<1)+1 $\endgroup$ – jfs Jan 9 '18 at 13:03
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    $\begingroup$ @jfs Agree! Otherwise we may get (1 << 1) << (1 + 1) or 1 << (1 << (1 + 1)), for example, depending on the language. That is 8 or 16, respectively. $\endgroup$ – Jeppe Stig Nielsen Jan 9 '18 at 14:58
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A little contrived:

$⌊ 11 ÷ (1 + 1) ⌋ $

Explanation:

$⌊ 11 ÷ (1 + 1) ⌋$ = $⌊ 11 ÷ 2 ⌋$ = $⌊ 5.5 ⌋ = 5$
Where ⌊ ⌋ is the round down operator (a.k.a. floor operator).

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    $\begingroup$ in Python: 11 // (1+1) $\endgroup$ – jfs Jan 9 '18 at 13:08
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    $\begingroup$ @jfs That would be Python 3. In Python 2 you can get away with 11 / (1+1) That being said, these comments spoil a bit, though. $\endgroup$ – Arthur Jan 9 '18 at 13:24
  • $\begingroup$ Not contrived: you could just concatenate the first two 1s if lexical operations are permitted. OP just suggests symbols and does specify if operations must be logical vs lexical. $\endgroup$ – Cloud Jan 9 '18 at 23:47
  • $\begingroup$ @Arthur // has been in Python since 2.2. It just wasn't necessary until / was redefined to always mean floating-point division in 3.0 (or via the __future__ mechanism. $\endgroup$ – chepner Jan 10 '18 at 1:21
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Slightly stretching the definition of 'adding a symbol' here:

$1+1+1+1 <= 5$

Or, adding only a single symbol :

$1111 >= 5$

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    $\begingroup$ How about $1111 \ne 5$? $\endgroup$ – John Gowers Jan 10 '18 at 15:09
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Here's another way:

$\frac{1}{\sqrt{.\bar1}}+1+1=5$

Inside the square root is $.\bar1$, which is a recurring decimal equal to $1/9$.
The square root of $1/9$ is $1/3$, so the first fraction equals $3$.

Or, using just three 1s:

$\frac{1}{.1+.1}=5$

Lots of ways to add a fourth $1$ to that, of course.

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    $\begingroup$ Upvoted for the second suggestion. Very nice. $\endgroup$ – J.R. Jan 10 '18 at 1:25
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Using the Euler totient $\varphi$:

$\varphi(11)/(1+1) = 5$

Using $\lim_{x\rightarrow\infty}$ (which contains no numbers :) there are all sorts of solutions, e.g.

$\lim_{x\rightarrow\infty} (1+1+1+1)/x = \lim_{x\rightarrow\infty} 5/x$

Using the symbol "/" gives a subtle solution:

$1111 \neq 5$

Edit: Using $\ln$:

$\lceil \ln(11)\times (1+1)\rceil = 5$

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  • $\begingroup$ Using a different $\phi$, the golden ratio, $\left((1+1)\phi-1\right)^{1+1}$ is five -- but that's five 1s. $\endgroup$ – David Hammen Jan 10 '18 at 0:48
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Since the question allows to add symbols, we could do:

$1111 := 5$
Which defines 1111 to be 5. That can be seen as overloading a number to be a variable. In the same way as you use other symbols ($x,a,α,…$) as names for variables.

Another solution is:

$\bar{1} = \bar{5}$
which uses the fact that these residue classes are equal in $ℤ/2ℤ$. I couldn't find a good English wikipedia page for explanation. On this wiki page the second example explains what $ℤ/2ℤ$ is. If you can speak German this wikipedia article is pretty good.
Edit: Maybe you are more familiar with the notation $[1]=[5]$, I just favor the bar as you rarely can confuse it with other bar-notation such as complex numbers.

A third solution idea, which is not working so far, is:

We know that $\sum_{i=0}^{∞}\frac{1}{2^n}= 2$, so $\sum_{i=-1}^∞\frac{1}{(1+1)^n}=3$. The problem is - I already used all of my four $1$s for this...

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  • $\begingroup$ Yeah, in my notation $[n]=\{1,2,\ldots , n\}$. Some people might use $[n]$ to mean "to round off to the closest integer to $n$" but I use $\lfloor n\rceil$ for that :P $\endgroup$ – Mr Pie Aug 27 '18 at 11:40
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$ 1+1+1-1 = ⌊\sqrt{5}⌋$

Of course, with enough nested square roots and rounding you can turn any positive number to 1.

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