49
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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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20
  • 1
    $\begingroup$ Obviously not only -,+,*,/ are allowed. Please tell full list of allowed operations. $\endgroup$
    – klm123
    Commented Jul 26, 2014 at 22:12
  • $\begingroup$ "(x^0 + x^0 + x^0)!" - so you allowed to use additional numbers and ()? $\endgroup$
    – klm123
    Commented Jul 26, 2014 at 22:30
  • $\begingroup$ @klm123 Yes, you are allowed to use additional numbers, but only as second argument to the ^ operator $\endgroup$
    – ThreeFx
    Commented Jul 26, 2014 at 22:31
  • 1
    $\begingroup$ How about square roots? $\endgroup$
    – BitNinja
    Commented Jul 26, 2014 at 23:28
  • 1
    $\begingroup$ @Muqo For the sake of keeping everything nice and clean, we will only consider the positive real roots $\endgroup$
    – ThreeFx
    Commented Jul 27, 2014 at 15:29

11 Answers 11

44
$\begingroup$

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:

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

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11
  • 5
    $\begingroup$ Bonus: (0^0 + 0^0 + 0^0)! $\endgroup$
    – c0rp
    Commented Jul 27, 2014 at 9:49
  • 3
    $\begingroup$ @c0rp 0^0 is NaN. Also, you can only choose a positive exponent. $\endgroup$
    – ThreeFx
    Commented Jul 27, 2014 at 11:38
  • 14
    $\begingroup$ $0! = 1$, though. $\endgroup$ Commented Jul 28, 2014 at 0:09
  • 8
    $\begingroup$ @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 $\endgroup$ Commented Apr 22, 2015 at 15:18
  • 2
    $\begingroup$ "one has to know that in order to be able to use it"? What on earth does that mean? $\endgroup$
    – lynn
    Commented Nov 22, 2015 at 1:05
32
$\begingroup$

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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2
  • 20
    $\begingroup$ ????!!!!????!!!! $\endgroup$ Commented Dec 30, 2014 at 11:20
  • 2
    $\begingroup$ @rand al'thor You look like you need some 's!! Hold on, is there a operator, too‽‽ This answer might need revision!! $\endgroup$
    – Muqo
    Commented Dec 30, 2014 at 18:17
8
$\begingroup$

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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2
  • 1
    $\begingroup$ Square roots are allowed. $\endgroup$
    – user20
    Commented Jul 27, 2014 at 1:54
  • 3
    $\begingroup$ 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. $\endgroup$
    – Rhymoid
    Commented Nov 13, 2016 at 8:50
7
$\begingroup$

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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3
  • 1
    $\begingroup$ Nice solutions, I particulary like the one to number 8, definitely worthy of an upvote. :D $\endgroup$
    – ThreeFx
    Commented Dec 30, 2014 at 2:27
  • 1
    $\begingroup$ Well, is only if you allow roots, and the solution to #8 requires a "4" $\endgroup$
    – HSuke
    Commented Aug 26, 2016 at 4:04
  • 2
    $\begingroup$ @HSuke Well, that's just taking square root twice $\endgroup$
    – user58955
    Commented Sep 10, 2016 at 4:39
3
$\begingroup$

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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2
  • $\begingroup$ I deleted the invalid solutions. $\endgroup$ Commented Mar 14, 2016 at 4:52
  • $\begingroup$ Another solution: 8!! / 8 / 8 $\endgroup$
    – Keavon
    Commented Apr 14, 2020 at 6:37
3
$\begingroup$

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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2
  • $\begingroup$ @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. $\endgroup$
    – KBusc
    Commented Sep 9, 2018 at 12:53
  • $\begingroup$ Sorry about that, I didn't see the time stamps, hahah; though you already had my upvote anyway :P $\endgroup$
    – Mr Pie
    Commented Sep 9, 2018 at 21:56
2
$\begingroup$

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,

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

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3
  • $\begingroup$ Did you mean $\sqrt[3]{8}$? If so, it's $\sqrt[3]{8}$ $\endgroup$
    – user20
    Commented Aug 5, 2014 at 20:50
  • $\begingroup$ I mean double square roots as in fourth roots, like $\sqrt[4]{8}$, or two square roots. $\endgroup$ Commented Aug 6, 2014 at 18:34
  • $\begingroup$ 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}. $\endgroup$
    – user20
    Commented Aug 6, 2014 at 18:58
1
$\begingroup$

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

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0
$\begingroup$

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

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10
  • 2
    $\begingroup$ Most of this is OK, but I think the cube root operator isn't allowed under the rules of the question. $\endgroup$ Commented Jul 8, 2015 at 13:04
  • 1
    $\begingroup$ @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). $\endgroup$
    – mmking
    Commented Jul 8, 2015 at 13:55
  • $\begingroup$ @mmking despite this being old, you cant write any extra numbers based on the correct/original rules to this puzzle $\endgroup$ Commented Mar 15, 2018 at 15:42
  • $\begingroup$ @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. $\endgroup$
    – mmking
    Commented Mar 15, 2018 at 19:45
  • $\begingroup$ @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 $\endgroup$ Commented Mar 16, 2018 at 8:56
-1
$\begingroup$

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

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2
  • $\begingroup$ 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. $\endgroup$
    – Deusovi
    Commented Oct 18, 2015 at 13:54
  • $\begingroup$ The ones with the cube roots use 3s. There can only be 8s used in that line for digits. $\endgroup$ Commented Apr 15, 2021 at 0:04
-3
$\begingroup$

$2\times 2\times 2=6$

$3\times 3-3=6$

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

$5+(\frac55)=6$

$6+6-6=6$

$7-(\frac77)=6$

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

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

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5
  • 1
    $\begingroup$ 2*2*2 is 8, not 6! $\endgroup$
    – Bailey M
    Commented Jul 9, 2015 at 20:09
  • 1
    $\begingroup$ Should be 2*2+2. $\endgroup$ Commented Jul 9, 2015 at 20:46
  • 1
    $\begingroup$ Or $2+2+2$. And your $4$s and $8$s are also wrong. $\endgroup$
    – Kevin
    Commented Jul 9, 2015 at 20:58
  • $\begingroup$ $8*8/8=8$, not $8*8/8=6$. $\endgroup$
    – EKons
    Commented Apr 5, 2017 at 12:05
  • $\begingroup$ 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. $\endgroup$
    – Mr Pie
    Commented Sep 7, 2018 at 12:03

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