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.
$\begingroup$
$\endgroup$
11
-
13$\begingroup$ Why have these “find a way to make blah from blah” questions become so popular all of a sudden? $\endgroup$– NL628Jan 9, 2018 at 7:05
-
41$\begingroup$ 1111 != 5 , you mad :) ? $\endgroup$– nl-xJan 9, 2018 at 15:01
-
25$\begingroup$ Just consider $(1111)_n=5$ where $$n=\frac{1}{3} \left(-2 \sqrt[3]{\frac{2}{115+3 \sqrt{1473}}}+\sqrt[3]{\frac{1}{2} \left(115+3 \sqrt{1473}\right)}-1\right)$$ $\endgroup$– Second Person ShooterJan 9, 2018 at 16:15
-
2$\begingroup$ Kinda similar: 1(1-1)1 = 101 = 5 (in base 2). $\endgroup$– userNaNJan 10, 2018 at 3:10
-
2$\begingroup$ (1+1)(1+1)+1 = 5 $\endgroup$– phuziJan 10, 2018 at 9:46
|
Show 6 more comments
10 Answers
$\begingroup$
$\endgroup$
5
How about
$(1+1+1)!-1$ = $6 -1$ = $5$
answered Jan 9, 2018 at 2:45
-
$\begingroup$ I'd say this is the better answer than mine, as mine concatenates two digits. $\endgroup$– PhylypJan 9, 2018 at 2:54
-
3$\begingroup$ @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. $\endgroup$ Jan 9, 2018 at 2:57
-
$\begingroup$ @R.M That is true. I never said anything about concatenation. The answer is perfectly valid. $\endgroup$ Jan 9, 2018 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? $\endgroup$ Jan 9, 2018 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! :) $\endgroup$ Jan 10, 2018 at 13:12
$\begingroup$
$\endgroup$
9
How about such a variant?
++(1+1+1+1)=5
-
61$\begingroup$ Ladies and gents, we have a programmer amongst us! :-) $\endgroup$– PhylypJan 9, 2018 at 3:05
-
27$\begingroup$ 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, 2018 at 3:47 -
13$\begingroup$ 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. $\endgroup$ Jan 9, 2018 at 7:29
-
1$\begingroup$ @IamtheMostStupidPerson At least it doesn't in C/C++: coliru.stacked-crooked.com/a/bde41bc9a5b1d16a $\endgroup$ Jan 9, 2018 at 10:54
-
2$\begingroup$ It also doesn't work in c#, python, js, or java. With the addition of C/C++ this covers quite a few important languages. $\endgroup$ Jan 9, 2018 at 14:42
$\begingroup$
$\endgroup$
Here's another solution:
$1$E$1$ $/ (1 + 1)$ = $5$
Where:
E is the exponential notation symbol. i.e. 1E1 = $1 \times 10^1$
$\begingroup$
$\endgroup$
6
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
-
$\begingroup$ 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. $\endgroup$– LolgastJan 9, 2018 at 6:25
-
1$\begingroup$ @Lolgast. Nope. I just really derped on that one $\endgroup$ Jan 9, 2018 at 6:44
-
1$\begingroup$ 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 $\endgroup$– LolgastJan 9, 2018 at 7:45
-
1$\begingroup$ you may need parens in some programming languages such as Python:
(1<<1<<1)+1$\endgroup$– jfsJan 9, 2018 at 13:03 -
1$\begingroup$ @jfs Agree! Otherwise we may get
(1 << 1) << (1 + 1)or1 << (1 << (1 + 1)), for example, depending on the language. That is8or16, respectively. $\endgroup$ Jan 9, 2018 at 14:58
$\begingroup$
$\endgroup$
4
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).
-
2
-
1$\begingroup$ @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. $\endgroup$– ArthurJan 9, 2018 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. $\endgroup$– CloudJan 9, 2018 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. $\endgroup$– chepnerJan 10, 2018 at 1:21
$\begingroup$
$\endgroup$
1
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.
-
1
$\begingroup$
$\endgroup$
1
Slightly stretching the definition of 'adding a symbol' here:
$1+1+1+1 <= 5$
Or, adding only a single symbol :
$1111 >= 5$
-
1
$\begingroup$
$\endgroup$
1
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 $\ln$:
$\lceil \ln(11)\times (1+1)\rceil = 5$
-
$\begingroup$ Using a different $\phi$, the golden ratio, $\left((1+1)\phi-1\right)^{1+1}$ is five -- but that's five 1s. $\endgroup$ Jan 10, 2018 at 0:48
$\begingroup$
$\endgroup$
1
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...
-
$\begingroup$ 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 PieAug 27, 2018 at 11:40
$\begingroup$
$\endgroup$
$ 1+1+1-1 = ⌊\sqrt{5}⌋$
Of course, with enough nested square roots and rounding you can turn any positive number to 1.
