5
   

Can anyone anyone solve this math riddle

 
 
jnejne
 
Reply Mon 25 May, 2009 02:50 pm
Can anyone help me with this riddle?

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. I'm pretty sure you can only insert on the left side of the equal-symbol.

Here are the ones i have been able to solve so far:
2+2+2=6
3*3-3=6
√(4*4) +√4 = 6
6+6-6=6
sqrt(9)*sqrt(9)-sqrt(9)=6
7 - 7/7 = 6

And som half-solved:
(5 squared + 5)/5 = 6
cube root (8 * 8) + cube root 8

Hope you can help
  • Topic Stats
  • Top Replies
  • Link to this Topic
Type: Question • Score: 5 • Views: 97,361 • Replies: 33
No top replies

 
solipsister
 
  1  
Reply Mon 25 May, 2009 08:52 pm
@jnejne,
oo i love half solved
solipsister
 
  1  
Reply Mon 25 May, 2009 08:54 pm
@jnejne,
[ 0!+0!+0! ]! = 6
solipsister
 
  1  
Reply Mon 25 May, 2009 08:55 pm
@jnejne,
5 plus 5/5
0 Replies
 
jelt8549
 
  2  
Reply Sat 30 May, 2009 07:44 am
(0 cos + 0 cos + 0 cos )! = 6

(1 + 1 + 1 )!=6

5+ 5/5 =
5+ 1 = 6

8 - square root of (square root of [8+8]) = 6
8 - square root of (square root of 4) = 6
8 - square root of 4 = 6
8 - 2 = 6

anyone figured out 10
i think if u can put a decimal point between the one and zero then you can treat as a number one equation
so (1.0 + 1.0+ 1.0)!=6
raprap
 
  1  
Reply Sat 30 May, 2009 09:30 am
@jnejne,
(0!+0!+0!)!=6

(1+1+1)!=6

Rap
jelt8549
 
  1  
Reply Tue 2 Jun, 2009 08:47 am
@raprap,
(10squared)log + (10squared)log + (10squared)log =
100log+ 100log+ 100log=
2+ 2+2 = 6
0 Replies
 
dallasmath
 
  1  
Reply Sat 10 Apr, 2010 08:10 pm
(sqrt(9)*sqrt(9))-sqrt(9) = 6
0 Replies
 
dallasmath
 
  1  
Reply Sat 10 Apr, 2010 08:21 pm
((log 10)!+(log 10)!+(log 10)!)! = 6
where logs are to base 10
0 Replies
 
dallasmath
 
  1  
Reply Sat 10 Apr, 2010 08:24 pm
for the case of eleven, 11 11 11 = 6
1+1+1+1+1+1 = 6
dallasmath
 
  1  
Reply Sat 10 Apr, 2010 08:27 pm
@dallasmath,
for the case of twelve
1*2 + 1*2 + 1*2 = 6
0 Replies
 
dorothyp
 
  1  
Reply Mon 18 Jun, 2012 09:59 am
@jnejne,
maybe 4+4-1-4???

0 Replies
 
witswang
 
  1  
Reply Fri 29 Jun, 2012 03:28 am
@jnejne,
more directly:
sqrt(4)+sqrt(4)+sqrt(4)=6
0 Replies
 
bsingin64
 
  1  
Reply Wed 4 Dec, 2013 08:25 am
The original riddle doesnt allow the addition of any numbers. Squaring something requires that you write a 2.

others have correctly answered 0's and 1's and 8's, but there is a more elegant solution to 8's which make it act just like the rest of the questions

It has already been answered this way: 8-sqrt(sqrt(8+8) = 6
But after figuring out the 0's and 1's (learning to use factorial), another way to answer the 8's is:

SQRT( 8+(8/8) )! = 6
SQRT( 9 )! = 6
3! = 6
3 x 2 x 1 = 6
maJan
 
  1  
Reply Mon 22 Sep, 2014 09:46 pm
@bsingin64,
This one is a more preferable answer for me.
0 Replies
 
Crackles0100
 
  1  
Reply Thu 30 Oct, 2014 05:47 am
@jnejne,
Here is all of them solved to stop inconvenience:
(0!+0!+0!)!=6 (0!=1)
(1+1+1)!=6
2+2+2=6
3*3-3=6
Sqrt4+Sqrt4+Sqrt4=6
5/5+5=6
6+6-6=6
7+(7/7)=6
[Sqrt{8+(8/8)}]!=6
Sqrt9*Sqrt9-Sqrt9=6

Hope that helps.
troll
 
  1  
Reply Mon 10 Nov, 2014 08:35 am
@jnejne,
(0!+0!+0!)!=6
(1+1+1)!=6
2+2+2=6
(3+3-3)!=6
sqrt 4+ sqrt 4 + sqrt 4=6
5/5+5=6
6+6-6=6
7-(7/7)=6
(sqrt (8/8+8))!=6
9-(sqrt(sqrt 9+ sqrt 9))=6
ika3
 
  1  
Reply Sun 7 Dec, 2014 09:47 am
@solipsister,
for the math sake you can not use cube root. so when you have 8 8 8=6 the only whey to do it is sqr((8/8)+8)!)
0 Replies
 
StephenBotsford
 
  0  
Reply Wed 17 Dec, 2014 04:21 am
4+4-2=6
This is for 6.
0 Replies
 
medilea
 
  0  
Reply Wed 24 Dec, 2014 11:16 pm
@Crackles0100,
9+9/sqrt 9=6
 

Related Topics

Alternative Einstein's riddle answer - Discussion by cedor
Urgent !!! Puzzle / Riddle...Plz helpp - Question by zuzusheryl
Bottle - Question by Megha
"The World's Hardest Riddle" - Discussion by maxlovesmarie
Hard Riddle - Question by retsgned
Riddle Time - Question by Teddy Isaiah
riddle me this (easy) - Question by gree012
Riddle - Question by georgio7
Trick Question I think! - Question by sophocles
Answer my riddle - Question by DanDMan52
 
  1. Forums
  2. » Can anyone anyone solve this math riddle
Copyright © 2017 MadLab, LLC :: Terms of Service :: Privacy Policy :: Page generated in 0.03 seconds on 10/17/2017 at 06:07:27