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
  • 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

11 Answers 11

up vote 42 down vote accepted

1.

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

2.

$2+2+2 = 6$

3.

$3*3-3 = 6$

4.

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

5.

$5+\frac 5 5 = 6$

6.

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

7.

$7-\frac 7 7 = 6$

8.

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

9.

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

Bonus:

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

  • 4
    Bonus: (0^0 + 0^0 + 0^0)! – c0rp Jul 27 '14 at 9:49
  • 3
    @c0rp 0^0 is NaN. Also, you can only choose a positive exponent. – ThreeFx Jul 27 '14 at 11:38
  • 13
    $0! = 1$, though. – Ben Millwood Jul 28 '14 at 0:09
  • 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 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!!!!

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

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

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

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

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

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

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

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

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

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

  • 14
    ????!!!!????!!!! – 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

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.)

  • 1
    Square roots are allowed. – Zyerah Jul 27 '14 at 1:54
  • 2
    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 8:50

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:

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

  • 1
    Nice solutions, I particulary like the one to number 8, definitely worthy of an upvote. :D – ThreeFx Dec 30 '14 at 2:27
  • 1
    Well, is only if you allow roots, and the solution to #8 requires a "4" – HSuke Aug 26 '16 at 4:04
  • 1
    @HSuke Well, that's just taking square root twice – user58955 Sep 10 '16 at 4:39

I am doing this for the eights only:

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

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

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

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

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

  • I deleted the invalid solutions. – Olive Stemforn Mar 14 '16 at 4:52

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

  • @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 9 at 12:53
  • Sorry about that, I didn't see the time stamps, hahah; though you already had my upvote anyway :P – user477343 Sep 9 at 21:56

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,

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

  • Did you mean $\sqrt[3]{8}$? If so, it's $\sqrt[3]{8}$ – Zyerah 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}. – Zyerah Aug 6 '14 at 18:58

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

  • 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 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
  • @mmking despite this being old, you cant write any extra numbers based on the correct/original rules to this puzzle – mast3rd3mon Mar 15 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 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 at 8:56

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

  • 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

$2\times 2\times 2=6$

$3\times 3-3=6$

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

$5+(\frac55)=6$

$6+6-6=6$

$7-(\frac77)=6$

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

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

  • 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
  • $8*8/8=8$, not $8*8/8=6$. – EKons Apr 5 '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. – user477343 Sep 7 at 12:03

protected by Community Feb 14 '16 at 12:02

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.