Do there exist operations that can be performed to make this equation true?

**
Note:
**
Any operations can be used, so get creative!

Yes
No

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(1+1)square+1/1= 5

Gnanasekaran Vijayam
- 5 years, 6 months ago

this is what i came up with......why cant u square it says to use ANY operations....?

John Turgeon
- 5 years, 6 months ago

I don't think this is allowed because of the square uses 2.

Frank Rodriguez
- 5 years, 6 months ago

Wow,you can really think broo

Mxolisi Sithole
- 5 years, 5 months ago

ah ha i understand why now!

John Turgeon
- 5 years, 6 months ago

@John Turgeon – I think you give up too easily. The 2 is an operator I'd say. For square root could we use ^0.5 instead of the sqrt or the square root "bracket". If I were programming the expression in many computer languages Sqr( ) would square an expression. I chose (1+1)!+1+1.

Robert Lucas
- 5 years, 5 months ago

@Robert Lucas – That makes 4

tom mcconnell
- 5 years, 4 months ago

@Tom Mcconnell – (1+1)! = 2! = 2+1 = 3 Then add 1+1 to that and it equals 5

Adam Buckley
- 5 years ago

@Adam Buckley – No 2! = 2*1 not 2+1

Paul Shaddick
- 5 years ago

you can't square, that's (1+1)^2 + 1/1 no 2s only 1s

Mr.Person 12345
- 5 years, 6 months ago

Not if you write it as "sq.", which many do. That's just as valid as using ! as factorial.

R. Chase Razabdouski
- 5 years, 6 months ago

@R. Chase Razabdouski – sq is a binary opn

Mohd Wasih
- 5 years, 6 months ago

@Mohd Wasih – multiplication is a binary operation, but if you define squaring as multiplying the same number with itself. you don't need two numbers to square

Zvonimir Baotić
- 5 years, 5 months ago

((1+1)^1)+1

Pinky Stiles
- 5 years, 6 months ago

@Pinky Stiles – ((1+1)^1)+1 = (2^1)+1 = 2+1 = 3 And 3 is not 5

Boyce Towell
- 5 years, 6 months ago

@Boyce Towell – I think "^" is a square she means :3

Luwi Fernandez
- 5 years, 5 months ago

@Luwi Fernandez – If so she would only have three "1"s, not the four in the problem.

Robert Lucas
- 5 years, 5 months ago

Squaring isn't an operation, but you can do (1+1)^(1+1)+1=5

A Former Brilliant Member
- 5 years, 4 months ago

That's using five 1s

Tom Franclemont
- 5 years, 2 months ago

i think we should use only signs, not numbers. square makes solution wrong.

Srujan Chary
- 5 years, 6 months ago

I think you can't use squares or powers because that's writing numbers

Moral Meshkot
- 5 years, 1 month ago

This is largely a matter of notation. Squaring a value could be considered as having a single operand whereas if it is considered more generically as raising to a power then the 2 is needed as an operand. The problem is that the notation to be used is not defined.

Paul Shaddick
- 5 years ago

1/.1/(1+1) is how i did it

Joseph O'Connor
- 5 years, 6 months ago

A decimal point isn't an operator...

Indelisio Prieto
- 5 years, 6 months ago

Decimal point is not an mathematical operator; you changed the number.

Bhaskar Basak
- 5 years, 5 months ago

Excellent!

Cathy Mac
- 5 years, 5 months ago

That's exactly how I did it

Ahmed Obaiedallah
- 5 years, 6 months ago

How about this??

(11-1)>>1 =5..Computer Science people will know!!

Aniruddha Bhattacharya
- 5 years, 6 months ago

[11/(1+1)]=5, where [] is the greatest integer function

Sachin Sharma
- 5 years, 6 months ago

@Sachin Sharma – Wouldn't that yield a 6?

Zoran Lalvani
- 5 years, 6 months ago

@Zoran Lalvani – How???

[11/(1+1)] = [5.5] = 5, since 5 is the greatest integer which is less than or equals to 5.5

Sachin Sharma
- 5 years, 6 months ago

@Sachin Sharma – ......... whaaattt

CarolAnn Tyson
- 5 years, 6 months ago

@Sachin Sharma – You mean the floor function.

Cameron Lunt
- 5 years, 6 months ago

@Sachin Sharma – We need operation like ! Not functions

Oussama Rhelimi
- 5 years, 6 months ago

@Oussama Rhelimi
–
**
An operator is an abstract thingy that takes another thingy as input and creates a third thingy as output!
**

a binary $\boxed {operation} ∗$ on a set $S$ is usually defined as $\boxed {function}$ $∗:S×S→S$

where for simplicity we write $a∗b$ instead of $∗(a,b)$ .

Some examples of operators:

1) Laplace, Fourier, and Hilbert transforms. These operators map functions into functions

2) derivatives and integrals. Ditto

3) differential and partial differential equations: the input function is mapped through the operator into an output function

4) Eigen-analysis is an operator mapping matrices into eigenvectors and eigenvalues

5) All matrix decompositions are operators, QR, tridiagonalization etc.

Note that things like ÷ is not an operator because the domain of ÷ is not $R^2$ .

Sachin Sharma
- 5 years, 6 months ago

@Sachin Sharma – What about accuracy?

A Former Brilliant Member
- 5 years, 5 months ago

@Zoran Lalvani – That would be 5.5

CarolAnn Tyson
- 5 years, 6 months ago

@Zoran Lalvani – No, it's greatest integer less than or equal to whatever is in the brackets.

anonymousgirl 3.14
- 5 years, 6 months ago

Or:

$11 + 1 + 1 = 5$

Where it is read in Base2 before the "=" and is read in Base10 after

Kyle Fairns
- 5 years, 6 months ago

@Kyle Fairns – I guess I would suggest just writing 11+1+1 = 101 then. I dunno, the problem used a 5 as a value so should make it at least a base 6 problem. ;)

Robert Lucas
- 5 years, 5 months ago

Indeed — my solution was a variant of your bit-shifting, namely ((1<<1)<<1)+1=4+1=5. Or ((1<<1)<<1)|1=4|1=5 if you truly want to geek out.

Bert Sierra
- 5 years, 6 months ago

@Bert Sierra – I had a similar solution ((1+1)<<1)+1

Robert McCain
- 5 years, 6 months ago

Well Done man

Gurinder Shergill
- 5 years, 6 months ago

Correct me if I'm wrong, but I don't think combining 1 and 1 to make 11 is a permitted step. Good idea though!!

Olivia Smaldone
- 5 years, 6 months ago

Nice!

>> is binary right shift in C style programming languages which is equivalent to integer division by 2.

Paul Shaddick
- 5 years ago

Yeah bro..!!

Rajesh Patil
- 5 years, 6 months ago

A little shifty, but nice.

Richard Fink
- 5 years, 5 months ago

plz explain how did u do it

Sarwar Akhtar
- 5 years, 6 months ago

11+1+1=101 (using binary operation)=5

Virupakshappa Nyamati
- 5 years, 5 months ago

Did that way too.

Roberto Passos
- 5 years, 6 months ago

Yes, that's exactly how i thought !

Mithun Santhikumar
- 5 years, 6 months ago

yep! that's it!

Kenneth Adair
- 5 years, 6 months ago

i think the use of arrays where the starting point is 0 can solve this..

Richard Makwabe
- 5 years, 6 months ago

Change the lhs to base-4:

11 in base-4 is 5...

Sqrt(11)*Sqrt(11) = 11 = 5 ( base 10)

Johnathon Sage
- 5 years, 6 months ago

I feel like its not okay to use sqrt since its translated into ^1/2. One other person couldn't use ^2 as an operation

Timothy Bennett
- 5 years, 4 months ago

(1+1)^2 + (1 x 1)

Vallab Nayak
- 5 years, 6 months ago

Very creative !

Andreas Liu
- 5 years, 6 months ago

(1/.1)/(1+1)=5

Brad Rodgers
- 5 years, 6 months ago

I cannot agree because 0.1 is not 1.

Moises Pereira
- 5 years, 6 months ago

(1+1)^2X1+1 = 5

Resat Uzmen
- 5 years, 6 months ago

i don't think using another digit is permitted

Vince Bander
- 5 years, 6 months ago

This isn't another digit it is square of (1+1)

Resat Uzmen
- 5 years, 6 months ago

@Resat Uzmen – you are using the digit '2' as an exponent which is not permitted; we can only use 4 digits & they are all ones

Vince Bander
- 5 years, 6 months ago

same answer... :)

Hicham Chalabi
- 5 years, 6 months ago

Are brckets included in operations?

Rishabh Mehta
- 5 years, 6 months ago

I did not like this question. Why is a factorial more valid than a ^2? If anything ^2 is more valid, because I am not actually using the number two. I am only denoting that I am multiplying the number by itself once. If I go ^3, then I am multiplying the number by itself twice, and so on.

A factorial is just a way of bringing other numbers into the equation while pretending that you are not bringing other numbers into the equation. If you use the factorial function, you have actually used other numbers than the numbers provided. It is implicit in the factorial function that you are using other numbers than the number itself.

When you square something you are still only using the numbers give, you just use a 2 or 3 or 4 to indicate how many times the self multiplication function is applied.

George Dobbs
- 5 years, 6 months ago

You're assuming that ^2 is a basic operation which is not. It's the most common number for that operation.

^2 is just the most common exponential there is I guess, but that does not mean every exponential out there ends with a 2. You can do ^3, ^4, ^x...

Point is that's just like multiplying something by 2 and saying your not using the number 2 when you actually are...

Kevin Luigi Reyes
- 5 years, 5 months ago

I was in the process of explaining why I disagree with being allowed to use exponents, however after giving it more thought, I think it is allowed. Raising something to a power is simply multiplying it by itself several times, and exponents are operation, I mean Pemdas, so no rule is broken, right?

Aiko Alvarez-Gibson
- 4 years, 11 months ago

that's exactly i came up with. Similar to this can 3 0's ever form a 6? (an easy self devised question)

Srivats Anshumaali
- 5 years, 5 months ago

Am I wrong ?
1. (.1/1 + .1)^-1 = 5
2. |
*
sqrt(11)
*
| + 1 + 1 = 5
3. (smallest integer function(sqrt(11)) + 1)/1 = 5

Devesh Ratna Singh
- 5 years, 5 months ago

(((1+1)^2)+1)/1

Alex Mazin
- 5 years, 1 month ago

Why not just derive them?

Michael Gaudio
- 5 years ago

(1+1)^2×1+1

Nat Thampipop
- 5 years ago

I think this should work... because it doesn't have any extra #s

Avery Sliwak
- 4 years, 12 months ago

What does this ! mark mean???

Mansi Singhal
- 5 years, 6 months ago

it's called factorial. which means 3! = 1x2x3 =6, 5! = 1x2x3x4x5 like wise n! = 1x2x3x....(n-1)xn

atchu atchu
- 5 years, 6 months ago

thank you very much ^_^

Mansi Singhal
- 5 years, 6 months ago

https://en.wikipedia.org/wiki/Factorial

Andrey Andreev
- 5 years, 6 months ago

thanx a lot

Mansi Singhal
- 5 years, 6 months ago

! stands for factorial

mimi omano
- 5 years, 6 months ago

Factorial (1+1+1)!-1=5

Oussama Rhelimi
- 5 years, 6 months ago

conguralations

Aydin Dinçer
- 5 years, 6 months ago

1+1=2 2^2=4 4/1=4 4+1=5

Sam Nouri-Zad
- 5 years, 6 months ago

I disagree with this solution. The factorial operation is shortened as a ! but in reality is several numbers, none of which were listed in the numbers being used. In addition, the single spacing between each number 1 implies their is ONE operation between every number, which is obviously not the case in the solution. This is misleading and does not fall in line with operator search rules.

T.j. Pankiewicz
- 5 years, 6 months ago

If you think the use of factorial is cheating, then you may as well disqualify the use of exponentiation, and multiplication, since exponentiation is a series of multiplications, and multiplication is a series of additions.

Oh wait a minute, that doesn't sound like a very good idea...

Colin Xu
- 5 years, 6 months ago

Finally someone who actually understands what is going on, so many uneducated people in here thinking there are mathematical geniuses

harry williams
- 5 years, 5 months ago

first get this - an Operator is a FUNCTION from a predefined set to another set.

Soham Halder
- 5 years, 6 months ago

So first off, where in the question does it say that you can't use more than one operation in between spaces, oh wait never mind, I can clearly see you have just made it up your head, how ignorantly stupid of you

harry williams
- 5 years, 5 months ago

Wow, there's really no need to be mean. People are here to learn and if you feel the need to insult people that know less about mathematics than you do then you are truly the ignorant one.

Victor Blancard
- 5 years ago

How dear.....give me detail

Atul Gupta
- 5 years, 6 months ago

Factorial is not an operator. It is a representation of product of natural numbers less than and equal to itself.

Aravind Pradhyumnan
- 5 years, 5 months ago

I still don't understand...😅 isn't that 2 ???

Femke Dewulf
- 4 years, 11 months ago

Or (1+1)!+1+1

Mark Zwicker
- 5 years, 6 months ago

That is equal to four.

Colin Xu
- 5 years, 6 months ago

(1+1)^2+(1+1)^0 or better than this => (1+1)^(1+1)+0^0

Rafael Vale de Sousa
- 5 years, 6 months ago

0^0 is nothing. It is an indeterminate form.

Vardaan Aggarwal
- 5 years, 6 months ago

not by convention

Rafael Vale de Sousa
- 5 years, 6 months ago

You are adding a digit "2" and "0" though? Should only be using the 4 provided "1's"

Seth Brumpton
- 5 years, 6 months ago

So we cant use ! cuz it adds other numbers to the problem. Factorial isnt an operator, its just a simplified form to write.

Rafael Vale de Sousa
- 5 years, 6 months ago

@Rafael Vale de Sousa – It is a function. Functions are a name for binary operators. Factorial is a valid form here, but the people trying to use '^2' are using an invalid form. Squaring is not an operation. Exponentiation is the operation. But it's a vaguely phrased question, regardless.

Nolan Steinhart
- 5 years, 5 months ago

@Rafael Vale de Sousa – Let U, V be two vector spaces. Any mapping from U to V is called an operator. Therefore, using factorial maps a vector value to another vector value by carrying out a series of multiplications does it not? (That is rhetorical as obviously that is the case) Therefore, it is quite blatantly true that factorial is an operator. You could use your poor argument to say multiplication isn't an operation as A multiplied by B, is just a simplified version of adding B numbers of A together, so you could say multiplication is a simplified way of addition, does that stop being an operator, obviously not, get educated

harry williams
- 5 years, 5 months ago

@Harry Williams – Despite your useless explanation with vectors, that could trick people without knowledge, for math ! continues to be a function, not operator, im not your teacher nor you are mine, its my opinion that as yours wont make any change to the world but arrogance and intolerance will ;)

Rafael Vale de Sousa
- 5 years, 5 months ago

my solution was very easy 1+(1+1)2 / 1 = 5

Jhean Nheaj
- 5 years, 6 months ago

I don't think we can put the power two....coz its not an operation

Kuduvan Vaishnav
- 5 years, 6 months ago

thats square

Jhean Nheaj
- 5 years, 6 months ago

Where did that 2 come from?

Jim Rickman
- 5 years, 6 months ago

Mr. Jim that means square

Jhean Nheaj
- 5 years, 6 months ago

@Jhean Nheaj – You can use exponent function but not 2 as an exponent unless it's obtained by operations on the four 1s. Otherwise what's the problem in multiplying with 2 or any number obtained from your head?

Punit Thanki
- 5 years, 6 months ago

Why not (1+1)*2+1?

Sam Ellington
- 5 years, 6 months ago

Wow, why didn't I think about changing the problem?

Tom Brauer
- 5 years, 6 months ago

I don't think u can use 2

Anwesha Maharana
- 5 years, 6 months ago

That won't work because 2 is not part of the problem

Jim Rickman
- 5 years, 6 months ago

cuz you are using three 1's !

Ahsan Memon
- 5 years, 6 months ago

Oh... I think it was like 3 AM when I wrote that... Just now I saw instantly what the issue was haha

Sam Ellington
- 5 years, 6 months ago

(1+1)² + 1 x 1 = 5

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0.1 is not the same as 1

Oscar Janda
- 4 years, 11 months ago

how can 1+1+1+1=5? it is = to 4

Veer Singh
- 4 years, 11 months ago

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Then you should be able to define the operation on all real numbers successfully without having to define it for each number explicitly. Get the point?

Aman Deep Singh
- 5 years, 6 months ago

Easy: A&B&C&D = 5 if A = B = C = D; else, A&B&C&D = 0

Rajat Mehndiratta
- 5 years, 6 months ago

+1 Never thought of that :-D Sorry

Aman Deep Singh
- 5 years, 6 months ago

No, because what you have done there is called, 'creating a function' and that is basically the one very clearly defined thing that is not an operator, functions are completely different and are usually made up of a lot of operators, good try though with your arrogant conclusion, shame you are ignorantly stupid and uneducated

harry williams
- 5 years, 5 months ago

You cannot use extra digits.. :/

Rvy Pandey
- 5 years, 6 months ago

squaring is an operation, not a digit.

K.k. Sarma
- 5 years, 6 months ago

Yes it is. Raising it to a power an operation (^) but adding a digit to decide which power it's raised to, is!

Rachel Collander-Brown
- 5 years, 6 months ago

Squaring is not an operation...its transformation...u're transforming the number to its square

Kuduvan Vaishnav
- 5 years, 6 months ago

@Kuduvan Vaishnav – Let U, V be two vector spaces. Any mapping from U to V is called an operator. Therefore, using factorial maps a vector value to another vector, or 'translates' it, still making it an operator clearly. The only problem is that the operator is the '^' and the number placed afterwards is not included as the operator, so you can't just decide you want to add a two and call it squared, and that making it and operator, that is like using a function that inverts a value, then calling it the 'inverse' and deciding as you gave it a nice name it now becomes an operation, it clearly doesn't

harry williams
- 5 years, 5 months ago

How about: ${\sqrt{}}^{-1}(1+1)+1$ ,

where ${\sqrt{}}^{-1}$ stands for the inverse of the square root function, i.e. the square :)

Arjen Vreugdenhil
- 5 years, 6 months ago

This is incorrect, you can't just use the square root operator without giving it a power, the power here for the square root is 2 which you haven't stated but just merely assumed, you can't just include a 2 and keep it discrete in order to solve the equation, you must state all numbers used, which you haven't, and if you did you would see that the root operator is of power 2 meaning not all '1's have been used

harry williams
- 5 years, 5 months ago

@Harry Williams – The square root $\sqrt x$ is the same as the $x^{1/2}$ but that does not make my approach invalid... Moreover, it is customary to use $f^{-1}$ for the inverse of a function $f$ , e.g. $\sin^{-1} x$ is the inverse of the sine function. So I don't really see what the problem is. And you will notice that I used the numeral "1" four times.

Maybe you don't like create solutions and I am willing to admit that this one is contrived, but isn't that the fun of it? The problem said "so get creative"-- and I did!

Arjen Vreugdenhil
- 4 years, 11 months ago

That was not a rule in the given problem. Plus it said to be creative.

Cauchy Sequence
- 5 years, 6 months ago

My bad. I did not read the note that says "
**
Any
**
operations can be used". And yes,
squaring is an operation
.

Rvy Pandey
- 5 years, 6 months ago

Really, then i can double any number and say doubling or multiplying by two is an operation just like raising to the power 2.

Kuduvan Vaishnav
- 5 years, 6 months ago

@Kuduvan Vaishnav – Someone whom understands maths, this is a breathe of fresh air😂

harry williams
- 5 years, 5 months ago

Someone whom doesn't know how to use 'whom'

Monty Wain
- 4 years, 11 months ago

Well done gautam. Keep it up

Sadullah Baig
- 5 years, 6 months ago

(1+1+1)! - 1 = 3! - 1 = 1x2x3 - 1 =6 - 1 =5

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[11/(1+1)] Where [x] is integral part of x.

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(1<<(1+1))+1 Bitwise left shift is beautiful operation.

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1111≠5 [ps. dont complicate things.. read the question carefully]

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The question says you are supposed to make the EQUAtion true. An equation must have an equal sign. You should read the question more carefully. '-'

Breno Lemos
- 5 years, 6 months ago

its not so much complicated. its very simple. just add one diagonal tally mark

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Haha.....everyone doesn't like the diagonal line through the equal sign....so make it a diagonal tally.

Tom Adams
- 5 years, 5 months ago

That yields 2, though.

Kieran Kaempen
- 5 years, 6 months ago

I also thought of this 👍🏻

Esteban Valdez
- 5 years, 6 months ago

you should use prefix operator instead of post fix (ie ++1 in place of 1++)

Chaitnya Shrivastava
- 5 years, 6 months ago

Let *: Int -> Int be defined as the constant operation: a * b = 5. for all integers.

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Ceiling(Sqrt(11 x (1 + 1))) or Replace Ceiling with Round.

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Or Floor(Sqrt(Sqrt(1111))

Anh Vu
- 5 years, 6 months ago

Or this. Ceiling(Arctan(Sqrt(Cot(Sin(Cos(11!))) + Cos(11 $^\circ$ )))). The first trig argument is in radian, the second is in degree. I propose some change in this problem. Let's find the maximum amount of operations without using programming LOL.

Anh Vu
- 5 years, 6 months ago

I took it as tally marks, and closed the gate (put in a slash to make it 5)

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In base 2,

11+1+1=111

Wich represents 5 in base 10.

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(3)!-1

6-1=5. ... Hence,,, Proved

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1-1+1+1=5 upside down of reads 2=1+1+1-1

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Ackermann(1,1+1)+1 OR 1×Ackermann(1+1,1)

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This is the only ideal solutionsolution

Bhaskar Mehta
- 5 years, 6 months ago

Define five(x)=5 and zero(x)=0. five(1)+zero(1+1+1)=5

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\frac {d(1+1+1+1)}{dx} = \frac {d(5)}{dx}

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Yes, that evaluates to 5. However, you have also used a $2$ , which is not allowed!

Pranshu Gaba
- 5 years, 1 month ago

.1^(- 1)/(1+1)

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

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(1+1+1+1)^0 = 5^0

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(1+1)^2+1×1=5 I guess, looking at others comments I wasn't allowed to square

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(1-1)/(1-1) Could be basically anything, including 5

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This is the most genius of all the solutions! :-) although sadly, it doesn't work. But screw the rules - I still think is the most genius in a diabolically creative way. well done!

Rev. Huber
- 5 years, 6 months ago

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All the top poi TX on the "ones" are equal to all the points of the five.

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$\large \ln 1 + 1 - 1 + \displaystyle \sum_{i=1}^5 1 = 5$

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My solutions use only 3 ones.

$\lceil \sqrt \frac {1}{. 1} \rceil + 1 = 5,$

$\lfloor \sqrt \frac {1}{.\overline{1}}! \rfloor - 1 = 5$

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(1×1)×1×1=5, 1×1=5, 1÷1=5÷1, 1=5,

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0111- 0010 = 0101 in binary 7 -2 =5 in Base 10

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Just use the 'Not Equal' sign :p

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\left \lfloor{11÷(1+1)}\right \rfloor = 5

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f(1)+ 1 + 1 + 1 =5 where f(x)= x+1

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11/11=5 5*11 is 55. 55/11 is 5

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((1+1)^-1) exp 1 = 5. Isn't that right?

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This expression doesn't even result on 5. 1 + 1 = 2, 2² = 4, 4 * 1 * 1=4. And you shouldn't use a number 2

Pietro Cavassin
- 5 years, 5 months ago

I just said 1+1+1+1 does not equal 5 you know. The equal with forward slash through it.

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Could ((1+1)^(1+1))+1 be 5. I think it could

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It took no time to get to the solution as $(1+1+1)!-1=5$

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$\huge{1+1+1+1\neq 5}$

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In python

x=sqrt(11)+(1+1)

print(int(x))

5

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(1-1)/(1-1)=0/0=all real numbers.

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Why it won't work (1/.1)/(1+1)

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Abusing the C/C++/Java programming language:

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

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floor[11/(1+1)] = floor(11/2) = floor(5.5) = 5

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1 (1-1) 1 = 101 = 5 in binary

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1 = 1/2 +1/2

(1/1/2)+(1/1/2)+1

2+2+1=5

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(1<<(1<<1)) | 1 = 0b00001001= 0x05

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1+1+(11base2)=5 ;whereby 11base2 is 3 XD

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Yes, it's not that hard: (1+1+1)!-1=5

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{1(1+1)}×1 where {} is fraction part.

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1 + 1 + 1 +1 +1 - 1 = 5 1 + 1 + 1 × (1 + 1) - 1 = 5 3 × 2 - 1 = 5 6 - 1 = 5 5 = 5 hence proved 😊

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(1+1)×(1+1)+1=5

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

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(1+1+1)! - 1 ==> 3! - 1 ==> 1
*
2
*
3 - 1 ==> 6-1=5 ==> 5=5

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The best solution is 1/.1/(1+1)

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How about (1+1)^2 + (1+1)^0 = 5?

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(1+1)x2+1=5 (1+1)=2x2=4+1=5

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[1+1+1]!-1=5 the only solution left.

Take this new challenge

1*1= 1

2*2=2

3*3=3

4*4=4

5*5=5

6*6=6

7*7=7

8*8=8

9*9=9

Any operation can be uses except cube root. If finally not possible then use cube root.

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antilog(1)*1/(1+1) = 5

using this solution I only need three 1's

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It strikes me that the (1+1+1)! – 1 = 5 solution is the most elegant and mathematical. [from Abdur Rehman Zahid]

I started out as a C programmer, so to me my bag of operators would include bitwise operators in addition to the arithmetic operators, in which case there are a couple of ways to make the equation true:

((1<<1)<<1)+1 = 4+1 = 5 ((1<<1)<<1)|1 = 4 | 1 = 5

…but then if you allow for the expressiveness of C, there is also the trivial case below:

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

which is most certainly true and would also lead to a large number of trivial variants in which you flip false to true at the end.

You run into a bit of a semantic problem because it’s not clear if the ‘operations’ should be applied to the 1s only, or could be applied to the equation as a whole. Also, to be a bit more picky, the word ‘operation’ isn’t quite ‘operator’, which might be a language issue — so presumably if a function could be considered an operation you could solve it trivially this way:

Given: f(x) = 5

Then f(1+1+1+1) = 5. QED.

[short answer — YES… there are several combinations of ‘operations’ which will work. The question is somewhat “Internetty” and so by exploiting vagueness in the question you can most certainly find a solution.]

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write it on binaries (11)=3+1+1=5

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How about something like this?

$(1+1)^{2} + (1+1)^{0} = 5$

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That's exactly what I got

tj weiler
- 5 years, 6 months ago

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I have no idea if this is right or cheating or not but me and my classmates thought of this way 1 1 1 1 = 5 You take the first 1 witch is equal to 0,5 and 0,5 then you divide the second 1 with 0.5 ( 1/0.5) = 2 then you do the same with the second 1 (1/0.5) = 2 and add the remaining 1 So (1/0.5) + (1/0.5) + 1 = 5

Or the easier sulotion would be (1+1)^2 x 1 + 1 = 5 But we thought that was alittle too easy.

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(1+1)^2 + (1*1) =2^2 + 1 =4+1 =5

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1x10^1 (ie 1e10)=10 and so 1e1/(1+1)

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Not completely valid because of the exponents...But I still had fun coming up with it...

(1+1)^(2)+(((1+1)^(2))^(0))=5

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That's pretty much what I got, (1+1)^2 + (1+1)^0, not sure if those operations are allowed though

tj weiler
- 5 years, 6 months ago

(1+1)2 (2 being squared) +1x1

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Ha-ha I thought it was asking for something like 1111 - 1106= 5

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$\frac{1\cdot1}{.1+.1}$

$\lfloor 1 + \overline{1\ 1}+ 1\rfloor = 5.$ Here, $\overline{1\ 1}$ is of course the letter $\pi$ . $\left\lfloor\sqrt{\sqrt{1111}}\right\rfloor = 5.$

$|1+i+1|^{1+1} = 5.$

$1^1+\phi(\phi(11)) = 5.$ Here, $\phi(n)$ is the totient function.

$[\pi^{\sqrt{1+1}}] = 5.$ The $[ ]$ may be viewed as variations on the digit 1.

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Lets assume these numbers are in base 2 not base 10 hence (1+1)+1+1=5

(1+1)base2 + (1)base2 + (1)base2

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(1+ 1) squared, + 1, / 1 = 5

Daniel Rothas
- 5 years, 6 months ago

(1exp1)/(1+1) = (10/2) = 5

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(Log(1+1+1+1))/ Log(x) = 5, does it works?

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11 (base ten) - 11 (base five) = 5 (base ten)

Joseph Novick
- 5 years, 6 months ago

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Relevant wiki: Arithmetic Puzzles - Operator SearchIn this operator search puzzle, the big hint is the exclamation mark at the end of the note! It's suggesting that we use factorial! $\large{(1+1+1)!-1=5}$