48
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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
    Jul 26 '14 at 22:12
  • $\begingroup$ "(x^0 + x^0 + x^0)!" - so you allowed to use additional numbers and ()? $\endgroup$
    – klm123
    Jul 26 '14 at 22:30
  • $\begingroup$ @klm123 Yes, you are allowed to use additional numbers, but only as second argument to the ^ operator $\endgroup$
    – ThreeFx
    Jul 26 '14 at 22:31
  • 1
    $\begingroup$ How about square roots? $\endgroup$
    – BitNinja
    Jul 26 '14 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
    Jul 27 '14 at 15:29

11 Answers 11

45
$\begingroup$

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
    Jul 27 '14 at 9:49
  • 3
    $\begingroup$ @c0rp 0^0 is NaN. Also, you can only choose a positive exponent. $\endgroup$
    – ThreeFx
    Jul 27 '14 at 11:38
  • 14
    $\begingroup$ $0! = 1$, though. $\endgroup$ Jul 28 '14 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 $\endgroup$ Apr 22 '15 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
    Nov 22 '15 at 1:05
33
$\begingroup$

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
  • 19
    $\begingroup$ ????!!!!????!!!! $\endgroup$ Dec 30 '14 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
    Dec 30 '14 at 18:17
9
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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.)

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2
  • 1
    $\begingroup$ Square roots are allowed. $\endgroup$
    – user20
    Jul 27 '14 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
    Nov 13 '16 at 8:50
7
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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
    Dec 30 '14 at 2:27
  • 1
    $\begingroup$ Well, is only if you allow roots, and the solution to #8 requires a "4" $\endgroup$
    – HSuke
    Aug 26 '16 at 4:04
  • 2
    $\begingroup$ @HSuke Well, that's just taking square root twice $\endgroup$
    – user58955
    Sep 10 '16 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. $\endgroup$ Mar 14 '16 at 4:52
  • $\begingroup$ Another solution: 8!! / 8 / 8 $\endgroup$
    – Keavon
    Apr 14 '20 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
    Sep 9 '18 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
    Sep 9 '18 at 21:56
2
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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.

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

$$2^2+2=6$$

$$3*3-3=6$$

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

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

$$6*6/6=6$$

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

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

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

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
    Aug 5 '14 at 20:50
  • $\begingroup$ I mean double square roots as in fourth roots, like $\sqrt[4]{8}$, or two square roots. $\endgroup$ Aug 6 '14 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
    Aug 6 '14 at 18:58
1
$\begingroup$

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

$\endgroup$
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. $\endgroup$ Jul 8 '15 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
    Jul 8 '15 at 13:55
  • $\begingroup$ @mmking despite this being old, you cant write any extra numbers based on the correct/original rules to this puzzle $\endgroup$ Mar 15 '18 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
    Mar 15 '18 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 $\endgroup$ Mar 16 '18 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
    Oct 18 '15 at 13:54
  • $\begingroup$ The ones with the cube roots use 3s. There can only be 8s used in that line for digits. $\endgroup$ Apr 15 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
    Jul 9 '15 at 20:09
  • 1
    $\begingroup$ Should be 2*2+2. $\endgroup$ Jul 9 '15 at 20:46
  • 1
    $\begingroup$ Or $2+2+2$. And your $4$s and $8$s are also wrong. $\endgroup$
    – Kevin
    Jul 9 '15 at 20:58
  • $\begingroup$ $8*8/8=8$, not $8*8/8=6$. $\endgroup$
    – EKons
    Apr 5 '17 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
    Sep 7 '18 at 12:03

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