@jelt8549,
(sqrt(10-(10/10)))!
and for the 8
((sqrt(8+8))!/8)!
xD that's my version, and for the 0, (0!+0!+0!)!
@troll,
Please can you explain full equation 999
@jnejne,
Please can you explain this equation full 999
@solipsister,
That doesn't work
0! is 1
so you were saying 1+1+1=6
@jnejne,
You said that you can insert everything but numbers so surely you can't add a squared because the symbol for squared is a 2. here I am referring to the half solved one for 5 5 5 = 6
@lottymouse,
Quote:Re: solipsister (Post 3659728)
That doesn't work
0! is 1
so you were saying 1+1+1=6
You factorial fact wit , you woke we with that ?
What I said was :
[ 0!+0!+0! ]! = 6
See your opthalmological surgeon immediately with a view to a singular plucking.
@jelt8549,
(Square root of ( 10-(10/10))!=6
@solipsister,
Damn girl, you are smart. I thought the zeros couldn't be done.
I cry "foul" on the case for 11. 11 is an integral entity, we only write it as two 1's as an accident of our system for expressing numbers. I think, however, the following may work: 11 (mod 9) + 11 (mod 9) + 11 (mod 9) = 6.
I realize the notation is fucked up, but am pretty sure everybody gets the idea.
@jnejne,
!0! + 0! + 0!)!
(1+1+1)!
2+2+2
sqrt(3+3+3)
sqrt(4*4) + sqrt(4)
5/5 + 5
6 + 6 - 6
7 + )7/7)
(sqrt(8 + (8/8)))!
(sqrt(sqrt(9) + sqrt(9) + sqrt(9)))!
@lottymouse,
Where the factorial [0] is 0, the value becomes 1. So consequently, it's asking 1 1 1=6. This can be read as 3=6. And therefore turned into 3!=6.
