Can you make 1 1 1 1 = 5?

Question

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.

Why have these “find a way to make blah from blah” questions become so popular all of a sudden? — NL628, Jan 09 '18 at 07:05
Just consider $(1111)_n=5$
where

$$n=\frac{1}{3} \left(-2 \sqrt[3]{\frac{2}{115+3 \sqrt{1473}}}+\sqrt[3]{\frac{1}{2} \left(115+3 \sqrt{1473}\right)}-1\right)$$ — Display Name, Jan 09 '18 at 16:15
@phuzi too many 1s, you could have just said 1 +1 + 1+ 1+ 1 = 5 if you had that many to use ;-) — AidanO, Jan 10 '18 at 13:26
Man, I joined this exchange to answer this, but association rep doesn't count to answer this :(. From the perspective of someone who works with computers, I came up with these: 11/(1+1) = 5 as an int will implicitly drop remainders (seems a few people had variants of this answer). The other one is ((11b) << 1) - 1 = 5 as 11b is equal to 3, shift left by one bit gives us 110b, or 6, and then subtract 1 to get 5. — Christian Gibbons, Jan 10 '18 at 23:03

score 93 · Accepted Answer · edited May 06 '18 at 03:14

93

How about

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

edited May 06 '18 at 03:14

SteamCode

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answered Jan 09 '18 at 02:45

Reinis Mazeiks

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I'd say this is the better answer than mine, as mine concatenates two digits. – Phylyp Jan 09 '18 at 02:54
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@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. – Reinis Mazeiks Jan 09 '18 at 02:57
@R.M That is true. I never said anything about concatenation. The answer is perfectly valid. – SteamCode Jan 09 '18 at 03:52
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? – The_Sympathizer Jan 09 '18 at 08:34
@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! :) – Reinis Mazeiks Jan 10 '18 at 13:12

score 53 · Answer 2 · edited Jan 10 '18 at 00:33

53

How about such a variant?

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

edited Jan 10 '18 at 00:33

spacediver

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answered Jan 09 '18 at 03:01

KstuRoot

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Ladies and gents, we have a programmer amongst us! :-) – Phylyp Jan 09 '18 at 03:05
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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… – Ry- Jan 09 '18 at 03:47
Also most situations in which ++ means 'increment by 1', = will mean 'assign the item on the left to be equal to the item on the right' and == means 'is equal to' – Scott Jan 09 '18 at 04:47
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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. – Cem Kalyoncu Jan 09 '18 at 07:29
@CemKalyoncu What? This works almost every programming languages – I am the Most Stupid Person Jan 09 '18 at 10:29
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@IamtheMostStupidPerson At least it doesn't in C/C++: http://coliru.stacked-crooked.com/a/bde41bc9a5b1d16a – HolyBlackCat Jan 09 '18 at 10:54
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Works just fine in Pip! (The output of 1 signifies "true.") – DLosc Jan 09 '18 at 11:15
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Doesn't work in C# either (The operand of an increment or decrement operator must be a variable, property or indexer). – Chris Jan 09 '18 at 14:40
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It also doesn't work in c#, python, js, or java. With the addition of C/C++ this covers quite a few important languages. – bob0the0mighty Jan 09 '18 at 14:42

score 42 · Answer 3 · edited May 06 '18 at 04:38

42

Here's another solution:

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

Where:

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

edited May 06 '18 at 04:38

SteamCode

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answered Jan 09 '18 at 06:22

Ergwun

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score 36 · Answer 4 · edited Jan 29 '19 at 00:27

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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

edited Jan 29 '19 at 00:27

SteamCode

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answered Jan 09 '18 at 06:09

Jonah Haney

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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. – Lolgast Jan 09 '18 at 06:25
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@Lolgast. Nope. I just really derped on that one – Jonah Haney Jan 09 '18 at 06:44
@Lolgast. Found a way to fix it :) – Jonah Haney Jan 09 '18 at 07:38
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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 – Lolgast Jan 09 '18 at 07:45
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you may need parens in some programming languages such as Python: (1<<1<<1)+1 – jfs Jan 09 '18 at 13:03
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@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. – Jeppe Stig Nielsen Jan 09 '18 at 14:58

score 35 · Answer 5 · edited May 06 '18 at 03:48

35

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).

edited May 06 '18 at 03:48

SteamCode

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answered Jan 09 '18 at 02:53

Phylyp

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in Python: 11 // (1+1) – jfs Jan 09 '18 at 13:08
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@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. – Arthur Jan 09 '18 at 13:24
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. – Cloud Jan 09 '18 at 23:47
@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. – chepner Jan 10 '18 at 01:21

score 27 · Answer 6 · answered Jan 09 '18 at 08:51

27

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.

answered Jan 09 '18 at 08:51

nickgard

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Upvoted for the second suggestion. Very nice. – J.R. Jan 10 '18 at 01:25

score 26 · Answer 7 · edited Aug 27 '18 at 02:34

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Slightly stretching the definition of 'adding a symbol' here:

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

Or, adding only a single symbol :

$1111 >= 5$

edited Aug 27 '18 at 02:34

SteamCode

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answered Jan 09 '18 at 04:51

Scott

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How about $1111 \ne 5$? – John Gowers Jan 10 '18 at 15:09

score 8 · Answer 8 · edited Jan 09 '18 at 15:57

8

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 $\ln$:

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

edited Jan 09 '18 at 15:57

jfs

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answered Jan 09 '18 at 04:10

Yly

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Using a different $\phi$, the golden ratio, $\left((1+1)\phi-1\right)^{1+1}$ is five -- but that's five 1s. – David Hammen Jan 10 '18 at 00:48

P. Siehr · Answer 9 · 2018-01-09T09:30:05.957

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...

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 — Mr Pie, Aug 27 '18 at 11:40

score 2 · Answer 10 · answered Jan 10 '18 at 14:29

2

$ 1+1+1-1 = ⌊\sqrt{5}⌋$

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

answered Jan 10 '18 at 14:29

Klaws

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Can you make 1 1 1 1 = 5?

10 Answers10