Brain Teasers
Sixes, Sixes, Everywhere!
The following equations are currently incorrect. Using these rules, correct them.
You can't write any other numbers.
You cannot use any form of an "equals" sign, a "greater than" sign, or a "less than" sign. Good luck.
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
0 0 0 =6
There can be multiple answers for one equation. I shall be putting only one of each in the answer. Post more answers in the comments.
You can't write any other numbers.
You cannot use any form of an "equals" sign, a "greater than" sign, or a "less than" sign. Good luck.
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
0 0 0 =6
There can be multiple answers for one equation. I shall be putting only one of each in the answer. Post more answers in the comments.
Hint
You can use factorials (!) and square roots.A factorial of a number is defined as:
N*(N-1)*(N-2)*...3*2*1
The factorial of three is:
3!=3*2*1=6
Answer
(1+1+1)!2+2+2
3*3-3
4+4- SqRt (4) SqRt = Square Root
5+5/5
6-6+6
7-7/7
8- SqRt (SqRt (8+8))
((SqRt (9)*SqRt(9))-SqRt (9)
(0!+0!+0!)!
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Comments
Nice puzzle!
3 3 3 => (3+3-3)! =3! =6
4 4 4 => (4-4/4)! =3! =6
8 8 8 => SqRt((8/8+)! =3! =6
9 9 9 => (SqRt(9)+SqRt(9)-SqRt(9))! =3! =6
3 3 3 => (3+3-3)! =3! =6
4 4 4 => (4-4/4)! =3! =6
8 8 8 => SqRt((8/8+)! =3! =6
9 9 9 => (SqRt(9)+SqRt(9)-SqRt(9))! =3! =6
The emoticon with sunglasses in my preceding comment should be an 8 followed by a right parenthesis.
Very nice!!
Some additional possibilities:
SqRt(4) + SqRt (4) + SqRt (4)
5 + (5 - 5)!
5 + (5mod5)!
6*6/6
7 - (7 - 7)!
7 - (7mod7)!
SqRt(4) + SqRt (4) + SqRt (4)
5 + (5 - 5)!
5 + (5mod5)!
6*6/6
7 - (7 - 7)!
7 - (7mod7)!
Either I can or I can't use other numbers. Factorials are other numbers. And 1 has no factorials to start with.
1 1 1 =6
(1+1+1) ! does not fall into the parameters set forth.
You simplify (1+1+1) ! to 3! which becomes 1x2x3. There are no longer 3 one's being used. You're using one 3.
2 2 2 =6 Obviously 2+2+2
3 3 3 =6 3x3-3
4 4 4 =6 4+4-(SqRt 4)
5 5 5 =6 5/5+5
6 6 6 =6 6+6-6
7 7 7 =6 7-(7/7)
8 8 8 =6 8-(SqRt (SqRt (8+))
9 9 9 =6 (9+9)/(SqRt 9)
0 0 0 =6 No way to get to one let alone 6. No matter whether you add 0+0 subtract 0-0 multiply 0x0 or divide 0/0 Zero is always Zero.
And there are no factorials of 0 either.
1 1 1 =6
(1+1+1) ! does not fall into the parameters set forth.
You simplify (1+1+1) ! to 3! which becomes 1x2x3. There are no longer 3 one's being used. You're using one 3.
2 2 2 =6 Obviously 2+2+2
3 3 3 =6 3x3-3
4 4 4 =6 4+4-(SqRt 4)
5 5 5 =6 5/5+5
6 6 6 =6 6+6-6
7 7 7 =6 7-(7/7)
8 8 8 =6 8-(SqRt (SqRt (8+))
9 9 9 =6 (9+9)/(SqRt 9)
0 0 0 =6 No way to get to one let alone 6. No matter whether you add 0+0 subtract 0-0 multiply 0x0 or divide 0/0 Zero is always Zero.
And there are no factorials of 0 either.
i disagree with stan. a mathematical Symbol is no number at all. and of course you get during the calculation process new and other numbers ...
for short, i like it
for short, i like it
Interesting!
(0!+0!+0!)! = 3! = 6
Since 0!=1 by definition
Since 0!=1 by definition
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