6, the magic number

Question

Here's a fun (albeit difficult) one:

Make these equations true using arithmetic operations:

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

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
Then "(x^0 + x^0 + x^0)!" satisfies. What does mean "be more creative"? You need to reformulate it using only objective criteria. — klm123, Jul 26 '14 at 22:33
I've seen this puzzle before, and I can (almost) guarantee it isn't possible without grouping, though it is possible without exponentiation. I think the operator to remove in this case is the exponentiation, not the parentheses, but that's just my opinion on the matter. Just be prepared to accept "this can't be done" as an answer otherwise. — , Jul 26 '14 at 22:45
@Emrakul I included exponentiation for more possibilities. Surely someone will come up with a solution not using it. — ThreeFx, Jul 26 '14 at 23:06
I have no idea how to do this for 5, 8, or 9. Hints, anyone? — Rhymoid, Jul 26 '14 at 23:54
If you're allowing multiplicative inverses of positive integers in exponents, that doesn't add unambiguous terms for potential solutions because it doesn't appear that you've provided a way to choose which root to use. — Muqo, Jul 27 '14 at 14:25
@Muqo Well you an choose between squareroot, cuberoot and so on. What is the problem with that? — ThreeFx, Jul 27 '14 at 14:31
@ThreeFx By "choose which root", I mean choosing which result given a particular root. Every positive integer has two different square roots, three different cube roots, and so on. For example, if I want to use 8^(1/3), how do I specify which of the one real and two complex results to use in my expression? Multiple solutions exist for 8 to get 6 using different cube roots. — Muqo, Jul 27 '14 at 15:08
@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
@Emrakul A bit late to reply, but I think that 8s and 9s aren't possible without exponentiation. — EKons, Apr 05 '17 at 12:07
@ThreeFx additional numbers are not allowed in the original/correct version of this puzzle — mast3rd3mon, Mar 15 '18 at 15:37

score 44 · Accepted Answer · edited Apr 30 '19 at 19:06

44

1.

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

2.

$2+2+2 = 6$

3.

$3*3-3 = 6$

4.

$\left(4-\frac 4 4\right)! = \sqrt 4+\sqrt 4+\sqrt 4=6$

5.

$5+\frac 5 5 = 6$

6.

$6*\frac 6 6 = 6 + 6 -6=6$

7.

$7-\frac 7 7 = 6$

8.

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

9.

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

Bonus:

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

edited Apr 30 '19 at 19:06

Brandon_J

9,307
1
16
67

answered Jul 26 '14 at 23:26

BitNinja

555
6
11

5

Bonus: (0^0 + 0^0 + 0^0)! – c0rp Jul 27 '14 at 09:49
3

@c0rp 0^0 is NaN. Also, you can only choose a positive exponent. – ThreeFx Jul 27 '14 at 11:38
14

$0! = 1$, though. – Ben Millwood Jul 28 '14 at 00:09
@BenMillwood Sure but one has to know that in order to be able to use it – ThreeFx Jul 30 '14 at 16:27
8

@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
2

"one has to know that in order to be able to use it"? What on earth does that mean? – Lynn Nov 22 '15 at 01:05
1

for 9, this should also be considered (9+9)/√9 – marmeladze Jan 24 '16 at 14:39
1

8: 8^1/3+8^1/3+8^1/3 – Jul 03 '16 at 12:41
1

@marmeladze or just to keep things simply written for above, $\sqrt{\small9\times 9\div 9}!$ – Mr Pie Sep 08 '18 at 07:10
@user477343 that's what already BitNinja did. √9 × √9 / √9 = √(9×9÷9) – marmeladze Sep 08 '18 at 09:27
1

@marmeladze my point exactly; it is simpler to write it with one radical as opposed to three :) – Mr Pie Sep 08 '18 at 09:30

score 32 · Answer 2 · edited Oct 07 '15 at 17:34

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!!!!

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

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

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

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

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

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

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

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

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

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

@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

Rhymoid · Answer 3 · 2016-11-13T08:51:17.237

8

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.)

edited Nov 13 '16 at 08:51

answered Jul 27 '14 at 00:07

Rhymoid

181
4

1

Square roots are allowed. – Jul 27 '14 at 01:54
3

I don't know who edited my answer or why, but I disagree with it. Why it was approved is a mystery to me. The added answer for 9 is incorrect. The answer for 8 uses the double factorial operator (not the same as the factorial of the factorial of its operand), which was not explicitly allowed by OP. To top it off, the markup was broken and would not properly hide the answers. – Rhymoid Nov 13 '16 at 08:50

score 7 · Answer 4 · edited Jun 17 '20 at 08:22

7

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:

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

edited Jun 17 '20 at 08:22

Community

1

answered Dec 30 '14 at 02:10

James Lynch

586
5
12

1

Nice solutions, I particulary like the one to number 8, definitely worthy of an upvote. :D – ThreeFx Dec 30 '14 at 02:27
1

Well, is only if you allow roots, and the solution to #8 requires a "4" – HSuke Aug 26 '16 at 04:04
2

@HSuke Well, that's just taking square root twice – user58955 Sep 10 '16 at 04:39

score 3 · Answer 5 · edited May 06 '18 at 03:34

3

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

edited May 06 '18 at 03:34

SteamCode

3,181
3
15
32

answered Aug 04 '14 at 13:11

KBusc

237
1
7

@user477343 Uhhh probably? This was 4 years ago and looking at the time stamps only 4 comments were before my answer and none of those comments affected my answer, thanks for your concern though. – KBusc Sep 09 '18 at 12:53
Sorry about that, I didn't see the time stamps, hahah; though you already had my upvote anyway :P – Mr Pie Sep 09 '18 at 21:56

Olive Stemforn · Answer 6 · 2016-03-14T04:47:43.767

3

I am doing this for the eights only:

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

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

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

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

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

edited Mar 14 '16 at 04:47

answered Mar 14 '16 at 03:40

Olive Stemforn

1,788
12
26

I deleted the invalid solutions. – Olive Stemforn Mar 14 '16 at 04:52
Another solution: 8!! / 8 / 8 – Keavon Apr 14 '20 at 06:37

Vincent Tang · Answer 7 · 2016-04-27T03:12:23.880

2

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,

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

edited Apr 27 '16 at 03:12

answered Aug 05 '14 at 17:32

Vincent Tang

417
6
13

Did you mean $\sqrt[3]{8}$? If so, it's $\sqrt[3]{8}$ – Aug 05 '14 at 20:50
I mean double square roots as in fourth roots, like $\sqrt[4]{8}$, or two square roots. – Vincent Tang Aug 06 '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}. – Aug 06 '14 at 18:58

score 1 · Answer 8 · answered Aug 20 '14 at 19:49

1

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

answered Aug 20 '14 at 19:49

nilesAnand

21
1

score 0 · Answer 9 · answered Jul 08 '15 at 12:42

0

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

answered Jul 08 '15 at 12:42

yeshwanth

1

2

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 08 '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 08 '15 at 13:55
@mmking despite this being old, you cant write any extra numbers based on the correct/original rules to this puzzle – mast3rd3mon Mar 15 '18 at 15:42
@mast3rd3mon Not to split hairs but: 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.. 1/3 is the multiplicative inverse of 3, which is an integer. – mmking Mar 15 '18 at 19:45
@mmking not true, you have to supply an extra number which isnt allowed, which is why you can only square root a number, not cube route it – mast3rd3mon Mar 16 '18 at 08:56
@mast3rd3mon 3 is not an integer? – mmking Mar 16 '18 at 13:56
@mmking yes, it is, which is why i said you can $\sqrt{x}$ but not $\sqrt[3]{x}$ – mast3rd3mon Mar 16 '18 at 13:58
@mast3rd3mon I don't follow. 3 is an integer. You are allowed to use an integer or its inverse with the power operator. Therefore ^ (1/3) power is allowed. – mmking Mar 16 '18 at 13:59
3 is an integer so it is not allowed. you may not write any extra numbers beyond the original – mast3rd3mon Mar 16 '18 at 14:00
@Randal'Thor you're the math guy, you explain it :) – mmking Mar 20 '18 at 00:26

score -1 · Answer 10 · edited Jan 09 '16 at 18:16

-1

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

edited Jan 09 '16 at 18:16

wythagoras

4,133
17
54

answered Oct 18 '15 at 13:20

satish

1

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
The ones with the cube roots use 3s. There can only be 8s used in that line for digits. – Olive Stemforn Apr 15 '21 at 00:04

score -3 · Answer 11 · edited May 06 '18 at 03:03

-3

$2\times 2\times 2=6$

$3\times 3-3=6$

$\frac{(4\times 4)}4=6$

$5+(\frac55)=6$

$6+6-6=6$

$7-(\frac77)=6$

$\frac{(8\times 8)}8=6$

$9-(\frac9{\sqrt{9}})=6$

edited May 06 '18 at 03:03

JMP

35,612
7
78
151

answered Jul 09 '15 at 20:00

Zoe

1

1

222 is 8, not 6! – Bailey M Jul 09 '15 at 20:09
1

Should be 2*2+2. – Victor Stafusa - BozoNaCadeia Jul 09 '15 at 20:46
1

Or $2+2+2$. And your $4$s and $8$s are also wrong. – Kevin Jul 09 '15 at 20:58
$88/8=8$, not $88/8=6$. – EKons Apr 05 '17 at 12:05
Yikes! I will not downvote now... but I may later on if this does not get fixed soon. Please fix your errors (e.g. $2\times 2\times 2 = 8\neq 6$ as @BaileyM mentioned before and $(4\times 4)\div 4=4\neq 6$ and $(8\times 8)\div 8=8\neq 6$ too. This is because of very basic (not necessarily simple) mathematical rules (including basic products like $4\times 4 = 16\neq 24$ and $8\times 8 = 64\neq 48$). So once again, please fix this errors; otherwise, this is not an answer, even though it attempts to answer the puzzle. I apologise for having said this... but sadly, it is true. – Mr Pie Sep 07 '18 at 12:03

6, the magic number

11 Answers11

1:

2:

3:

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5:

6:

7:

8:

9:

Bonus - 0:

Linked

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