### 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`

"(x^0 + x^0 + x^0)!" - so you allowed to use additional numbers and ()?

@klm123 Yes, you are allowed to use additional numbers, but only as second argument to the ``^`` operator

Then "(x^0 + x^0 + x^0)!" satisfies. What does mean "be more creative"? You need to reformulate it using only objective criteria.

@klm123 Eliminated that possibility

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.

Parenthesis? Allowed?

@kaine Yes, parenthesis are allowed

@Emrakul I included exponentiation for more possibilities. Surely someone will come up with a solution not using it.

How about square roots?

@BitNinja Allowed, I'll add it in a sec

I have no idea how to do this for 5, 8, or 9. Hints, anyone?

@Rhymoid 5 is only 1 away from 6...

@ThreeFx Right. Overlooked that one pretty badly.

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 Well you an choose between squareroot, cuberoot and so on. What is the problem with that?

@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 For the sake of keeping everything nice and clean, we will only consider the positive real roots

@Emrakul A bit late to reply, but I think that 8s and 9s aren't possible without exponentiation.

@ThreeFx additional numbers are not allowed in the original/correct version of this puzzle

BitNinja Correct answer

6 years ago1.

$(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:

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

Bonus: (0^0 + 0^0 + 0^0)!

@c0rp ``0^0`` is NaN. Also, you can only choose a **positive exponent**.

$0! = 1$, though.

@BenMillwood Sure but one has to know that in order to be able to use it

@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`

"one has to know that in order to be able to use it"? What on earth does that mean?

for 9, this should also be considered (9+9)/√9

8: 8^1/3+8^1/3+8^1/3

@marmeladze or just to keep things simply written for above, $\sqrt{\small9\times 9\div 9}!$

@user477343 that's what already BitNinja did. √9 × √9 / √9 = √(9×9÷9)

@marmeladze my point exactly; it is simpler to write it with one radical as opposed to three :)

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Content dated before 6/26/2020 9:53 AM

klm123 6 years ago

Obviously not only -,+,*,/ are allowed. Please tell full list of allowed operations.