Question

Here's a fun puzzle to try :)

Answer 1

where it at tho >.>

Answer 2

yay xD

Answer 3

where they at doe

Answer 4

?

Answer 5

i like puzzles c:

Answer 6

Make each of the following equations true. You may use any mathematical symbols, but cannot use numbers (no cube roots, exponents, etc).

\(\Large 0 ~~0~~0=6 \)
\(\Large 1~~1~~1=6 \)
\(\Large 2~~2~~2=6 \)
\(\Large 3~~3~~3=6 \)
\(\Large 4~~4~~4=6 \)
\(\Large 5~~5~~5=6 \)
\(\Large 6~~6~~6=6 \)
\(\Large 7~~7~~7=6 \)
\(\Large 8~~8~~8=6 \)
\(\Large 9~~9~~9=6 \)

I'll start you off:

\(\Large 0 ~~0~~0=6 \)
\(\Large 1~~1~~1=6 \)
\(\Large 2~~2~~2=6 \)
\(\Large 3~~3~~3=6 \)
\(\Large 4~~4~~4=6 \)
\(\Large 5~~5~~5=6 \)
\(\Large 6\color{red}{+}6\color{red}{-}6=6 \)
\(\Large 7~~7~~7=6 \)
\(\Large 8~~8~~8=6 \)
\(\Large 9~~9~~9=6 \)

Sorry, OS keeps lagging me off :/

Answer 7

5/5+5 ;-;

Answer 8

Oh God, what a good Scenario.

Answer 9

2*2 + 2
3*3 - 3
7 - 7/7
5 + 5/5

Answer 10

0/0/0 = 6
since its infinity and it contains 6?

Answer 11

1/(1-1) = 6
infinity concept

Answer 12

lol is this allowed?

Answer 13

"Make each of the following equations true. You may use \(\color{red}{any}\) mathematical symbols, but cannot use numbers (no cube roots, exponents, etc)."
;))

Answer 14

The equal cannot change xD

Only in the blank spaces

Answer 15

2+2+2 = 6
(3*3)-3 = 6
5/5 + 5 =6
7-7/7 = 6

Answer 16

lol okay e.e

Answer 17

im not good wit dis puzzl stuff. cx

Answer 18

\(\Large 0 ~~0~~0<=6 \)
\(\Large 1~~1~~1<=6 \)
\(\Large 2~~2~~2=6 \)
\(\Large 3~~3~~3>=6 \)
\(\Large 4~~4~~4>=6 \)
\(\Large 5~~5~~5>=6 \)
\(\Large 6~~6~~6>=6 \)
\(\Large 7~~7~~7>=6 \)
\(\Large 8~~8~~8>=6 \)
\(\Large 9~~9~~9>=6 \)

Answer 19

is it OK if I just use this for all?

\[n+n-n=n\]
since \(n-n=0\)

Answer 20

\[5+5-5=6\]hmm

Answer 21

lolol whoops I totally didn't read the last column

Answer 22

\[3! + 3! -3!=6\]

Answer 23

What about this

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

Answer 24

This was hard:

\[(0!+0!+0! )!= 6\]

Answer 25

NOOOOOOOOOOOOOOOOOOOO

Answer 26

Lol

Answer 27

\[(4!\div (4+4))!\]

Answer 28

@Kainui \(\div\)

Answer 29

@ParthKohli \(/\)*

Answer 30

we're left with 8 and 9

Answer 31

I feel like we should set up a scoring system though like I bet there's an answer for 4 that requires fewer symbols.

Answer 32

$(\sqrt{9})!\times(9/9)$

Answer 33

hmm but he said that there are no exponents allowed
not sure though

Answer 34

im on my phone doing this..... Not easy ;)

Answer 35

Square roots are allowed :)

Answer 36

\[(\frac{d}{dx}(9)!+\frac{d}{dx}(9)!+\frac{d}{dx}(9)!)!\]

Answer 37

i'm not sure why square roots are allowed but cube roots are not

Answer 38

\[((8' + 8') \div 8)!\] https://en.wikipedia.org/wiki/Arithmetic_derivative

Answer 39

\[9' + 9' - 9'\]

Answer 40

Alternate solution \[((4+4)' \div 4)!\]

Answer 41

\[(3+3+3)'\]

Answer 42

Not sure about 1 and 0...
2*2+2=6
3*3-3=6
5/5+5=6
6+6-6=6

That's as far as I got ;-;