Can You Make This Equation True?

Relevant wiki: Arithmetic Puzzles - Operator Search

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

(1+1)² + 1 x 1 = 5

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

a bit misleading perhaps. if any operation is allowed, I could simply define an operation, say &, such that 1&1 = 5, and then add and subtract the two other ones....

$(1+1)^2+1\times1=5$

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

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

11(base2)+1+1=5

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

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

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

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

1111 != 5 😂

$(1 / (1+1))^{-1}$

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

(1+1)^(1+1)+1

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

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

11-1-1=5 in base 6

1111 (mod 7) =5 ?

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

In base 2,

11+1+1=111

Wich represents 5 in base 10.

Best is to use factorial for sum of first three one and then subtract them (1+1+1)!-1.
(3)!-1
6-1=5. ... Hence,,, Proved

(1+1+1)!-1

(1÷.1)÷[1+1]=5

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

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

1x1x1x1 = 5^0

It's (1+1+1)! - 1 = 5

(1+1+1)! - 1

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

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

(1+1)2+1=5

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

(1+1)^2 + 1*1

((1<<1)<<1)+1

(1 x 1 x 1)! - 1

(1+1+1)!-1

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

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

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

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

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

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

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

You can use any one e.g. PEMDAS rules, FOIL. My friend use arc cosec of something to solve but it is too complicated. I use very simple math.

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

(1+1)^2+1/1

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

My solutions use only 3 ones.

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

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

(1×1)×1×1=5, 1×1=5, 1÷1=5÷1, 1=5,

(1+1)x(1x1)+1

((1+1)^2 + 1)×1=5

Simple. (1+1)^2+1×1

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

((1+1)^2 + 1) ÷1

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

[(1+1)^2 + 1]*1

Just use the 'Not Equal' sign :p

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

(1+1squared) +1÷1

(1+1)^2 +(1÷1)

gif(11/(1+1))=5

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

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

Floor function of 11/(1+1)

(1+1)^2+1÷1

(1<<(1+1)) +1

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

floor(11/(1+1))

(1+1)^3-(1 1)=5

(1*1)+(1+1)^2

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

((1+1)^2)*(1 * 1)

1+(1+1)^2 -1

(1+1)square + 1

(1+1)^2+(1x1)=5

Floor[11/(1+1)]=5

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

(1+1+1)! - 1= 5. I figured this out on my own, but saw the answer posted before 1 posted mine. Is there a way we can post our results before seeing anyone else's?

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

[(1+1)^2 X 1] + 1 = 5

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

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

Note: This is not a solution

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

(1+1)!+1+1

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

(1+1+1)!-1

In python

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

print(int(x))

5

log^(-1) (1) / (1+1)

(1+1)2 +1/1

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

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

1 EE 1 / (1 + 1)=5

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

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

((1+1)<<1)+1

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

[(1+1)^2]+[1x1]

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

1 = 1/2 +1/2

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

2+2+1=5

1111\neq 5

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

(1+1)^2 *1 +1

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

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

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

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

{1(1+1)}×1 where {} is fraction part.

(1+1)² + (1×1)

1+1+1+1++ Hell yeah

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 😊

(1+1) squared + (1x1) = 5

1(1+1)²+1 = 5

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

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

(1+1)×(1+1)+1=5

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

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

(1×1+1)^2+1=5

(1+1)²+1x1=5

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

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

(1+1)^2 +1/1

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

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

Of course there is. Let's define $\otimes$ such that $n\otimes m =n+m+1$ . Then $(1+1+1)\otimes 1=3+1+1=5$ .

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

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

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

(1+1)*2 + 1 = 5

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

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

[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.

(1+1)square + 1 x 1

[{(1+1)^2}+1]*1=5

(1/.1)/(1+1)

In excel if you have four cells with 1.25 in them,but you have the decimal point set to whole numbers each cell will show 1, but when you sum the answer is 5.

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

using this solution I only need three 1's

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

1 << 1 << 1 + 1

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

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.]

[ ( 1 + 1 ) squared + 1 ] x 1 = 5 Sorry I don't know how to type an exponent on this keyboard :) I believe squaring something is an operation. What are people's thoughts?

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

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

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

How about something like this?

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

Make up an operator that takes 4 1's and returns five. The question doesn't state that you must use standard arithmetic operators. Given this, the answer is obviously yes. Quite Easilly Done.

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.

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

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

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

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

(1+1)^2 + 1 = 5

Floor(11/(1+1))=5

(1+1)^2 + 1/1

(1+1+1)!-1

1+1+1=3 3! =6 6-1=5

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

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

x/x + 1 + 1 + 1 +1 = 5 :)

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

((1+1)^2) +1)x1=5

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

$\lfloor 1 + \overline{1\ 1}+ 1\rfloor = 5.$ 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.

Add first three ones and take the factorial of 3 which will equal to 6 and subtract remaining 1 from it which will give 5

(1+1)Squared+1*1=5

((1+1)^4)-11=5

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

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

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

(1+1)² x 1 + 1 = 5

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

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

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

1+1+1+1+1 1 1*1=5

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

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

[(1+1)^2×1]+1=5

(1+!+1)!-1

(1+1)*2 +1 ÷1

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

(((1+1)^2)+1)*1

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

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

The fifth root of 1= 1•1•1

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

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

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

(1+1) squared x 1 + 1 = 5

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

1+1+(11)bin=5

Floor(11/(1+1))

equal anyone side with other like 1+1+1+1= 5-1 bring negative one on the left side 1+1+1+1+1= 5 BE CREATIVE that's what the question says