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# Can anyone anyone solve this math riddle

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
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Type: Question • Score: 8 • Views: 115,159 • Replies: 33
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solipsister

1
Mon 25 May, 2009 08:52 pm
@jnejne,
oo i love half solved
solipsister

1
Mon 25 May, 2009 08:54 pm
@jnejne,
[ 0!+0!+0! ]! = 6
solipsister

1
Mon 25 May, 2009 08:55 pm
@jnejne,
5 plus 5/5
0 Replies

jelt8549

2
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
Sat 30 May, 2009 09:30 am
@jnejne,
(0!+0!+0!)!=6

(1+1+1)!=6

Rap
jelt8549

1
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
Sat 10 Apr, 2010 08:10 pm
(sqrt(9)*sqrt(9))-sqrt(9) = 6
0 Replies

dallasmath

1
Sat 10 Apr, 2010 08:21 pm
((log 10)!+(log 10)!+(log 10)!)! = 6
where logs are to base 10
0 Replies

dallasmath

1
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
Sat 10 Apr, 2010 08:27 pm
@dallasmath,
for the case of twelve
1*2 + 1*2 + 1*2 = 6
0 Replies

dorothyp

1
Mon 18 Jun, 2012 09:59 am
@jnejne,
maybe 4+4-1-4???

0 Replies

witswang

1
Fri 29 Jun, 2012 03:28 am
@jnejne,
more directly:
sqrt(4)+sqrt(4)+sqrt(4)=6
0 Replies

bsingin64

0
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

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

0
Mon 22 Sep, 2014 09:46 pm
@bsingin64,
This one is a more preferable answer for me.
0 Replies

Crackles0100

0
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

0
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

0
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
Wed 17 Dec, 2014 04:21 am
4+4-2=6
This is for 6.
0 Replies

medilea

0
Wed 24 Dec, 2014 11:16 pm
@Crackles0100,
9+9/sqrt 9=6

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