8
   

Can anyone anyone solve this math riddle

 
 
buaja
 
  0  
Thu 12 Feb, 2015 08:16 pm
@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!)!
0 Replies
 
vicky07391
 
  -1  
Thu 15 Oct, 2015 02:14 pm
@medilea,
(1*2+1*2+1*2)/1
(2*2+2*2+2*2)/2
(3*2+3*2+3*2)/3
..so on....
0 Replies
 
Ojus mishra
 
  -1  
Thu 4 Feb, 2016 07:46 pm
@troll,
Please can you explain full equation 999
0 Replies
 
Ojus mishra
 
  -1  
Thu 4 Feb, 2016 07:48 pm
@jnejne,
Please can you explain this equation full 999
0 Replies
 
selectmytutor
 
  -2  
Thu 17 Mar, 2016 03:49 am
@jnejne,
your answer is 6.
0 Replies
 
kloraston
 
  -3  
Thu 24 Mar, 2016 03:41 pm
Try to solve this riddle
0 Replies
 
lottymouse
 
  -2  
Wed 14 Dec, 2016 02:59 pm
@solipsister,
That doesn't work
0! is 1
so you were saying 1+1+1=6
lottymouse
 
  -1  
Wed 14 Dec, 2016 03:01 pm
@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
0 Replies
 
solipsister
 
  1  
Wed 4 Jan, 2017 11:02 pm
@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.
0 Replies
 
MrAgeless
 
  -1  
Fri 6 Jan, 2017 07:42 am
@jelt8549,
(Square root of ( 10-(10/10))!=6
0 Replies
 
hibbitus
 
  1  
Sat 9 Sep, 2017 03:16 pm
@solipsister,
Damn girl, you are smart. I thought the zeros couldn't be done.
0 Replies
 
hibbitus
 
  0  
Sat 9 Sep, 2017 04:00 pm
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.
0 Replies
 
radsqrt
 
  -1  
Sat 23 Sep, 2017 02:03 am
@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)))!
0 Replies
 
Christian0912
 
  -1  
Fri 5 Jan, 2018 11:05 pm
@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.
0 Replies
 
 

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