Can you make 1 1 1 1 = 5? [closed]

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.

Answer 1

How about

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

Answer 2

How about such a variant?

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

Answer 3

Here's another solution:

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

Where:

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

Answer 4

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

Answer 5

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).

Answer 6

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

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

Or, adding only a single symbol :

$1111 >= 5$

Answer 7

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.

Answer 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 $\ln$:

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

Answer 9

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...

Answer 10

$ 1+1+1-1 = ⌊\sqrt{5}⌋$

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