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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
share|improve this question
1  
Obviously not only -,+,*,/ are allowed. Please tell full list of allowed operations. – klm123 Jul 26 '14 at 22:12
    
"(x^0 + x^0 + x^0)!" - so you allowed to use additional numbers and ()? – klm123 Jul 26 '14 at 22:30
    
@klm123 Yes, you are allowed to use additional numbers, but only as second argument to the ^ operator – ThreeFx Jul 26 '14 at 22:31
1  
How about square roots? – BitNinja Jul 26 '14 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 '14 at 15:29

12 Answers 12

up vote 32 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:

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

share|improve this answer
2  
Bonus: (0^0 + 0^0 + 0^0)! – c0rp Jul 27 '14 at 9:49
2  
@c0rp 0^0 is NaN. Also, you can only choose a positive exponent. – ThreeFx Jul 27 '14 at 11:38
7  
$0! = 1$, though. – Ben Millwood Jul 28 '14 at 0:09
4  
@ThreeFx 0^0 is not always NaN depending on who you ask and what field you're in. It can also be set to 0^0=1 – Engineer Toast Apr 22 '15 at 15:18
1  
"one has to know that in order to be able to use it"? What on earth does that mean? – Lynn Nov 22 '15 at 1:05

I insist on using all the digits!

$(1 + 1^{1234567890} + 1)! = 6$

$(2 + (2^{1234567890}\ \text{mod}\ 2)!)! = 6$

$(3 + 3^{1234567890}\ \text{mod}\ 3)! = 6$

$(4 - (4^{1234567890}\ \text{mod}\ 4)!)! = 6$

$5 + (5^{1234567890}\ \text{mod}\ 5)! = 6$

$6 + 6^{1234567890}\ \text{mod}\ 6 = 6$

$7 - (7^{1234567890}\ \text{mod}\ 7)! = 6$

$(\sqrt[3]8 + (8^{1234567890}\ \text{mod}\ 8)!)! = 6$

$(\sqrt{9} + (9^{1234567890}\ \text{mod}\ 9))! = 6$

$(0! + (0^{1234567890})! + 0!)! = 6$

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

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

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

$!3 + !3 + !3 = 6$

$(\sqrt{!4} \times 4 \div 4)! = 6$

$!(\sqrt{!5\ \text{mod}\ 5}) + 5 = 6$

$!6\ \text{mod}\ 6 \times 6 = 6$

$!7\ \text{mod}\ 7\ \text{mod}\ 7 = 6$

$(!8\ \text{mod}\ 8 + \sqrt[3]8)! = 6$

$\sqrt[3]{!9\ \text{mod}\ 9} \times \sqrt9 = 6$

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

share|improve this answer
7  
????!!!!????!!!! – rand al'thor Dec 30 '14 at 11:20
    
@rand al'thor You look like you need some 's!! Hold on, is there a operator, too‽‽ This answer might need revision!! – Muqo Dec 30 '14 at 18:17

Here we go.

1:

$(1+1+1)! = 6$
This is the only possible one as far as I know.

2:

$2+2+2 = 6$

3:

$3*3-3 = 6$

4:

$4+(4/\sqrt{4}) = 6$

5:

$5+(5/5) = 6$

6:

$6*(6/6) = 6$

7:

$7-(7/7) = 6$

8:

8 - $\sqrt[4]{8 + 8} = 6$

9:

$(9+9)/\sqrt{9} = 6$

Bonus - 0:

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

share|improve this answer
    
Sorry, typo on the first one :3 – James Lynch Dec 30 '14 at 2:22
    
Ok. I got it fixed now :) – James Lynch Dec 30 '14 at 2:25
1  
Nice solutions, I particulary like the one to number 8, definitely worthy of an upvote. :D – ThreeFx Dec 30 '14 at 2:27

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 '14 at 1:54

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

share|improve this answer

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

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I am doing this for the eights only:

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

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

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

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

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

share|improve this answer
    
I deleted the invalid solutions. – Olive Stemforn Mar 14 at 4:52

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

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

$$2^2+2=6$$

$$3*3-3=6$$

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

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

$$6*6/6=6$$

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

$$\sqrt[3]{8}+\sqrt[3]{8}+\sqrt[3]{8}$$

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

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 '14 at 20:50
    
I mean double square roots as in fourth roots, like $\sqrt[4]{8}$, or two square roots. – Vincent Tang Aug 6 '14 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 '14 at 18:58

2+2+2=6

(3*3)-3=6

(4/sqrt4)+4=(4/2)+4=6

(5/5)+5=6

(6+6)-6=6

7-(7/7)=6

cubrt8+cubrt8+cubrt8=2+2+2=6

9-(9/sqrt9)=9-(9/3)=9-3=6

share|improve this answer
1  
Most of this is OK, but I think the cube root operator isn't allowed under the rules of the question. – rand al'thor Jul 8 '15 at 13:04
1  
@randal'thor: Actually, it is. The OP said that you can use ^ with any positive integer or multiplicative inverse. So you can do 8^(1/3). – mmking Jul 8 '15 at 13:55

$$2+2+2$$
$$3\times3-3$$
$$\sqrt{4}+\sqrt{4}+\sqrt{4}$$
$$\frac{5}{5}+5$$
$$6\times\frac{6}{6}$$
$$7-\frac{7}{7}$$
$$\sqrt[3]{8} +\sqrt[3]{8} +\sqrt[3]{8}$$
$$\sqrt{9}\times\sqrt{9}-\sqrt{9}$$

share|improve this answer
    
Hi, welcome to Puzzling.SE! I've cleaned up your answer a bit for you - hopefully you noticed that this question was answered a while ago and most of your answers are equivalent to the already accepted one. – Deusovi Oct 18 '15 at 13:54

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/(root of 9))=6

share|improve this answer
1  
2*2*2 is 8, not 6! – Bailey M Jul 9 '15 at 20:09
1  
Should be 2*2+2. – Victor Stafusa Jul 9 '15 at 20:46
1  
Or $2+2+2$. And your $4$s and $8$s are also wrong. – Kevin Jul 9 '15 at 20:58

(2+2+2)=6 (3+3+3)-3=6 (4+4+4)-6=6 (5+5+5)-9=6 (6+6+6)-12=6 (7+7+7)-15=6 (8+8+8)-18=6 (9+9+9)-21=6 Ans of above question.

share|improve this answer
3  
You are not allowed to use anything except the operators, so no extra numbers. – ThreeFx Oct 8 '15 at 18:38

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