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Here's a fun (albeit difficult) one:

Make these equations true using arithmetic operations:

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

For example: 6 + 6 - 6 = 6 (I hope I did not spoil some of you :D)

Allowed operators are:

+, -, *, /, ! , ^, %

Setting parenthesis is also allowed.

The ^ operator is an exception as you are permitted to supply a second argument to it which may be any positive integer or the multiplicative inverse of it.

$x^{1/y}$ is always positive and real.

If you find an alternative solution using other operators you may post it but please also provide a solution using only these 7 operators.


For those of you who think this was easy, here is a bonus:

0 0 0 = 6
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Obviously not only -,+,*,/ are allowed. Please tell full list of allowed operations. –  klm123 Jul 26 at 22:12
    
"(x^0 + x^0 + x^0)!" - so you allowed to use additional numbers and ()? –  klm123 Jul 26 at 22:30
    
@klm123 Yes, you are allowed to use additional numbers, but only as second argument to the ^ operator –  ThreeFx Jul 26 at 22:31
1  
How about square roots? –  BitNinja Jul 26 at 23:28
1  
@Muqo For the sake of keeping everything nice and clean, we will only consider the positive real roots –  ThreeFx Jul 27 at 15:29

7 Answers 7

up vote 9 down vote accepted

1.

$(1+1+1)! = 6 $

2.

$2+2+2 = 6$

3.

$3*3-3 = 6$

4.

$(4-\frac 4 4)! = 6$

5.

$5+\frac 5 5 = 6$

6.

$6*\frac 6 6 = 6$

7.

$7-\frac 7 7 = 6$

8.

$(\sqrt{8+\frac 8 8})! = 6$

9.

$(\frac{\sqrt{9}\sqrt{9}}{\sqrt 9})! = 6$

Bonus:

No idea :)

share|improve this answer
    
Bonus: (0^0 + 0^0 + 0^0)! –  c0rp Jul 27 at 9:49
1  
@c0rp 0^0 is NaN. Also, you can only choose a positive exponent. –  ThreeFx Jul 27 at 11:38
5  
$0! = 1$, though. –  Ben Millwood Jul 28 at 0:09
    
@BenMillwood Sure but one has to know that in order to be able to use it –  ThreeFx Jul 30 at 16:27

I insist on using all the digits!

(1 + 1^1234567890 + 1)! = 6

(2 + (2^1234567890 % 2)!)! = 6

(3 + 3^1234567890 % 3)! = 6

(4 - (4^1234567890 % 4)!)! = 6

5 + (5^1234567890 % 5)! = 6

6 + 6^1234567890 % 6 = 6

7 - (7^1234567890 % 7)! = 6

(8^(1/3) + (8^1234567890 % 8)!)! = 6

(9^(1/2) + (9^1234567890 % 9))! = 6

(0! + (0^1234567890)! + 0!)! = 6

No, wait! How about if we take subtraction out and put subfactorial in? More exclamation points!!!!

((!1)! + (!1)! + (!1)!)! = 6

(!2 + !2 + !2)! = 6

!3 + !3 + !3 = 6

((!4)^(1/2) * 4 / 4)! = 6

!((!5 % 5)^(1/2)) + 5 = 6

!6 % 6 * 6 = 6

!7 % 7 % 7 = 6

(!8 % 8 + 8^(1/3))! = 6

(!9 % 9)^(1/3) * 9^(1/2) = 6

(!0 + !0 + !0)! = 6

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The bottom five (0 through 4) can all be solved using the same construction:

(0!+0!+0!)! = 6
(1 +1 +1 )! = 6
(2 +2 /2 )! = 6
(3 +3 %3 )! = 6
(4 -4 /4 )! = 6

For 6 and 7, there are slightly more funky solutions:

(6!)%(6!-6)=6
((7!)/7)%7=6

(I haven't found an interesting solution for 5, nor any square-root-free solutions for 8 or 9.)

share|improve this answer
    
Square roots are allowed. –  Emrakul Jul 27 at 1:54

I remember seeing this question in a few math curiosity books, and most recently on an episode of Scam School. However, they usually allow square roots. The answers I came up with are as follows:

$$(0! + 0! + 0!)! = 6$$

$$(1 + 1 + 1)! = 6$$

$$ 2 + 2 + 2 = 6$$

$$3 \times 3 - 3 = 6$$

$$4 + 4 - \sqrt{4} = 6$$

$$5 + 5/5 = 6$$

$$6 + 6 - 6 = 6$$

$$7 - 7/7 = 6$$

$$8 - \sqrt{\sqrt{8 + 8}} = 6$$

$$9 - 9 / \sqrt{9} = 6$$

As you can see, it's doable without any exponentiation or modular arithmetic, if square roots are allowed. You only need the four arithmetic operators, square roots, and factorials.

share|improve this answer
    
The square root of a number is equal to $x^{1/2}$ –  ThreeFx Jul 30 at 7:08
    
True. But it also has the benefit of not explicitly introducing any new numbers. –  Joe Z. Jul 30 at 15:09
    
@JoeZ. I like the idea of not introducing extra numbers (and operators) in order to help limit the number of valid answers. This question as it's currently stated seems far too broad. Would anyone know the original wording and parameters? –  Muqo Jul 30 at 17:29
    
There are probably infinitely many variations, but the basic rules always allows the four basic operators, factorials, and square roots. And as I've shown above, that's all you need. –  Joe Z. Jul 30 at 18:02
    
@JoeZ. You can improve on this particular variation more. Explicit multiplication is not required. –  Muqo Jul 30 at 18:25

1. (1+1+1)! = 6
2. 2+2+2 = 6
3. 3*3-3 = 6
4. (4^3)/(4^2)+(4^(1/2)) = 6
5. 5+(5/5) = 6
6. (6*6)/6 = 6
7. 7-(7/7) = 6
8. (8^3)/(8^2)-(8^(1/3)) = 6
9. (9+9)/(9^(1/2)) = 6

and the bonus

(0!+ 0! + 0!)! = 6

For more info on the bonus take a look here: http://en.wikipedia.org/wiki/Empty_product

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Having heard about this many times, I decided to give it a try. These are the answers that I came up with.

$$(1+1+1)!=6$$

$$2^2+2=6$$

$$3*3-3=6$$

$$4+(4/\sqrt4)=6$$

$$(5-5)!+5=6$$

$$6*6/6=6$$

$$7-(7-7)!=6$$

I couldn't figure out how to format long or double square roots for 8 8 8 = 6

$$(9+9)/(\sqrt9)=6$$

And finally,

$$(0!+0!+0!)!=6$$

share|improve this answer
    
Did you mean $\sqrt[3]{8}$? If so, it's $\sqrt[3]{8}$ –  Emrakul Aug 5 at 20:50
    
I mean double square roots as in fourth roots, like $\sqrt[4]{8}$, or two square roots. –  Vincent T Aug 6 at 18:34
    
Oh, you can actually do just $\sqrt{\sqrt{8}}$, or $\sqrt[4]{8}$ ($\sqrt{\sqrt{8}}$ or $\sqrt[4]{8}$). $\sqrt[n]{8}$ is $\sqrt[n]{8}. –  Emrakul Aug 6 at 18:58

For bonus one... ((0!)+(0!)+(0!))!

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