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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Sign up to join this communityCan 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.
How about
$(1+1+1)!-1$ = $6 -1$ = $5$
How about such a variant?
++(1+1+1+1)=5
++
to get anything…
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– Ry-
Jan 9 '18 at 3:47
Here's another solution:
$1$E$1$ $/ (1 + 1)$ = $5$
Where:
E is the exponential notation symbol. i.e. 1E1 = $1 \times 10^1$
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
(1<<1<<1)+1
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– jfs
Jan 9 '18 at 13:03
(1 << 1) << (1 + 1)
or 1 << (1 << (1 + 1))
, for example, depending on the language. That is 8
or 16
, respectively.
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– Jeppe Stig Nielsen
Jan 9 '18 at 14:58
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).
11 / (1+1)
That being said, these comments spoil a bit, though.
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– Arthur
Jan 9 '18 at 13:24
1
s if lexical operations are permitted. OP just suggests symbols and does specify if operations must be logical vs lexical.
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– Cloud
Jan 9 '18 at 23:47
//
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.
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– chepner
Jan 10 '18 at 1:21
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.
Slightly stretching the definition of 'adding a symbol' here:
$1+1+1+1 <= 5$
Or, adding only a single symbol :
$1111 >= 5$
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 $\ln$:
$\lceil \ln(11)\times (1+1)\rceil = 5$
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...
$ 1+1+1-1 = ⌊\sqrt{5}⌋$
Of course, with enough nested square roots and rounding you can turn any positive number to 1.