# 6, the magic number

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 '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
How about square roots? – BitNinja Jul 26 '14 at 23:28
@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

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:

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

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Bonus: (0^0 + 0^0 + 0^0)! – c0rp Jul 27 '14 at 9:49
@c0rp 0^0 is NaN. Also, you can only choose a positive exponent. – ThreeFx Jul 27 '14 at 11:38
$0! = 1$, though. – Ben Millwood Jul 28 '14 at 0:09
@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
"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!!!!

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

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

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

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

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

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

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

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

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

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

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????!!!!????!!!! – 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:

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

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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
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.)

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Square roots are allowed. – Emrakul Jul 27 '14 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.

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The square root of a number is equal to $x^{1/2}$ – ThreeFx Jul 30 '14 at 7:08
True. But it also has the benefit of not explicitly introducing any new numbers. – Joe Z. Jul 30 '14 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 '14 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 '14 at 18:02
@JoeZ. You can improve on this particular variation more. Explicit multiplication is not required. – Muqo Jul 30 '14 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

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

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I am doing this for the eights only:

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

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

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

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

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

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I deleted the invalid solutions. – Olive Stemforn Mar 14 at 4:52

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

$$2^2+2=6$$

$$3*3-3=6$$

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

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

$$6*6/6=6$$

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

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

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

And finally,

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

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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 - 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 @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 $$2+2+2$$ $$3\times3-3$$ $$\sqrt{4}+\sqrt{4}+\sqrt{4}$$ $$\frac{5}{5}+5$$ $$6\times\frac{6}{6}$$ $$7-\frac{7}{7}$$ $$\sqrt[3]{8} +\sqrt[3]{8} +\sqrt[3]{8}$$ $$\sqrt{9}\times\sqrt{9}-\sqrt{9}$$ - 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 - 2*2*2 is 8, not 6! – Bailey M Jul 9 '15 at 20:09 Should be 2*2+2. – Victor Stafusa Jul 9 '15 at 20:46 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.

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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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