Question

Simple Arithmetic Problem:

Using any Maths Symbols or Signs prove it...

0 0 0 = 6
1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6
8 8 8 = 6
9 9 9 = 6

You can Insert Everything but Numbers,
And you can't change everything.

Answer 1

\(\ne\)

Answer 2

can i use that sign

Answer 3

no. just math operations

Answer 4

6-6+6 = 6 :D

Answer 5

that's one

Answer 6

2+2+2=6. :D

Answer 7

that's two

Answer 8

3*3-3=6. :D
I got 3 points so far!

Answer 9

( 0! + 0! + 0!)! = 6
( 1 + 1 +1) ! =6

Answer 10

good good keep it coming

Answer 11

signum function and successor function will take us anywhere! :D
but we have to use only operators..hmm

Answer 12

4 down

Answer 13

Square root of 4 + Square root of 4+ Square root of 4 = 6 :D

Answer 14

(3!- 3+ 3) = 6

Answer 15

5 down

Answer 16

that seems legit

Answer 17

6

Answer 18

6*6/6 = 6

Answer 19

7- (7/7)

Answer 20

5+(5/5) = 6 :D

Answer 21

i think 8 and 9 are the last

Answer 22

Where in math will this help?

Answer 23

@skullpatrol more than you know. many top users disappeared because of the disappearance of these things. and the disappearance of the top users became the disappearance of the good helpers here. but i wouldn't want to delve on those subjective matters

Answer 24

\(\log_2 4 + \log_2 4 + \log_2 4\) = 6

Answer 25

lol wow. logs got in

Answer 26

but i don't think that's legit

Answer 27

you put in numbers

Answer 28

also in sqrt4 + sqrt4 + sqrt4 , wont you consider "2" being used ?

Answer 29

in 4^(1/2)

Answer 30

sqrt is a math symbol...but if you want to reject that then that leaves 4 8 and 9

Answer 31

Cube root of 8 + Cube root of 8+ Cube Root of 8 = 6 :D
I'm on the role!

Answer 32

well yes cube root can also be taken into account, and then any root should be taken into account ?

Answer 33

shubhamsburg disqualified roots though

Answer 34

that was an extra bu

Answer 35

Aww man.

Answer 36

oh roots we cant abuse

Answer 37

sometime you can write binary log (ie base 2) \( \log_2(x) =\text{lb}(x)\)

Answer 38

lol..

Answer 39

i'd prefer ( 4- (4/4))!

Answer 40

8 and 9 then

Answer 41

I disqualify this question as having no use in math.

Answer 42

It's like saying that algebra has no use in math. :|

Answer 43

8 and 9 seem difficult without roots

Answer 44

we can use operators hmm
how about ++ and -- ? :P

Answer 45

isn't that a logical operator?

Answer 46

not a genuine operator we use in theoretical maths.hmm
we use that in computer programs

Answer 47

nvm

Answer 48

i dont see anything for 8 or 9

Answer 49

because you disqualified the roots :p lol

Answer 50

Use the power of square root! :|

Answer 51

since you answered 8 with roots how would you answer 9 @GoldPhenoix

Answer 52

here is my rather late and lame contribution
\[\sqrt{9}!+9-9\]

Answer 53

this has to be the best question i have seen all week

Answer 54

Not bad, satellite.

Answer 55

it's nice to have some math riddles sometimes

Answer 56

Yes, totally agree.

Answer 57

Ok, I'll agree it is "recreational" arithmetic :)

Answer 58

10+10+10=6 (binary)

Answer 59

right! :D

Answer 60

11 + 11 + 11 = 6 (unary)

Answer 61

nice

Answer 62

reminds me of a quote,
"there are only 10 types of people in the world, one who knows binary, other who doesn't! " :P

Answer 63

lol

Answer 64

Thanks for the question :D