[Guest]

What mathematical symbols can you insert to make the following equations true?

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

Not allowed:

1. Making an equal sign a "not equal".

2. Replacing or adding any numbers.

All other mathematical functions are allowed (see the hint for some ideas). Of course, some are easier than others. I found 1 the most challenging.

You can use roots, powers, and others - I don't consider the "squared" symbol as a number. Just don't want you to put 2 * (1+1+1) for the first one, for example.

[Guest]

2 + 2 + 2 = 6

3 * 3 - 3 = 6

[Guest]

4 + 4 - sqrt(4) = 6

6 + 6 - 6 = 6

Edited August 13, 2008 by Cherry Lane

[Guest]

7 - (7 / 7) = 6

cuberoot (8) + cuberoot (8) + cuberoot (8) = 6

Edited August 13, 2008 by Cherry Lane

[Guest]

the factorial symbol is your friend for the early ones. It makes it considerably easier that 6 = 3!

I'm not sure if you meant that we could use any power automatically, so 2 and 8 might have more elagant solutions, but:

(1+1+1)! = 6

2² + √2 × √2 = 6 (or 2³ - √2 × √2)

3! + 3 - 3 = 6

4! ÷ (√4 × √4) = 6

5 + (5÷5) = 6

6 + 6 - 6 = 6

7 - (7÷7) = 6

³√8 + ³√8 + ³√8 = 6

edit: mistake in 8!

Edited August 13, 2008 by foolonthehill

[Guest]

5 + (5 / 5) = 6

sqrt(9 * 9) - sqrt(9) = 6

[Guest]

2 + 2 + 2 = 6

[Guest]

I'm not sure if you meant that we could use any power automatically, so 2 and 8 might have more elagant solutions, but:

(1+1+1)! = 6

2² + √2 × √2 = 6 (or 2³ - √2 × √2)

3! + 3 - 3 = 6

4! ÷ (√4 × √4) = 6

5 + (5÷5) = 6

6 + 6 - 6 = 6

7 - (7÷7) = 6

³√8 + ³√8 + ³√8 = 6

edit: mistake in 8!

the factorial symbol is your friend for the early ones. It makes it considerably easier that 6 = 3!

Nice job with #1, FotH!

factorials!

[Guest]

1 1 1 = 6

2 + 2 + 2 = 6

(3 x 3) - 3 = 6

4 + 4 - (sqrt4) = 6

5 + (5/5) = 6

6 + 6 - 6 = 6

7 - (7/7) = 6

8 - (8^0)- (8^0) = 6

(9 + 9)/(sqrt 9) = 6

[Guest]

Great job, FotH! You got all - but some are more complex then need be. And Cherry Lane added the 9...

I thought I could confuse you all for longer than that!

Edited August 13, 2008 by IdoJava

[Guest]

Great job, FotH! You got all - but some are more complex then need be. And Cherry Lane added the 9...

I thought I could confuse you all for longer than that!

Oh dear, I didn't even spot that I had missed the 9! And I think I got a bit caught up with the !'s

Check out my problem, inspired by your problem

[Guest]

(1^0 + 1^0 + 1^0)! = 6

(2^0 + 2^0 + 2^0)! = 6

(3^0 + 3^0 + 3^0)! = 6

(4^0 + 4^0 + 4^0)! = 6

(5^0 + 5^0 + 5^0)! = 6

(6^0 + 6^0 + 6^0)! = 6

(7^0 + 7^0 + 7^0)! = 6

(8^0 + 8^0 + 8^0)! = 6

(9^0 + 9^0 + 9^0)! = 6

[TwoaDay]

wow

welcome to the den

Edited August 13, 2008 by TwoaDay

[Guest]

wow

welcome to the den

Thanks!

[Guest]

I also have a formula that solves all.

...that logarithms of any base are allowed, for any number n:

Log_{(n + n + n)}(n + n + n)⁶ = 6

Of course, it would work with just n but you must use it three times so just use the sum.

It's nowhere near as interesting as the factorial solution, but it works.

[Guest]

If the the base counts as adding a number and must be replaced by one of those given then this would work instead:

n⁰*log_nn⁶ = 6

[EventHorizon]

Repeat!

http://brainden.com/forum/index.php?showtopic=1502

Might want to look at my answers in that thread....they don't need extra 0's or the three's required for cube roots.