General Discussions

**clearvectra** · 02-08-2011, 02:55 AM

This is something interesting mathematical puzzle for you to be solved.

Below is an mathematical calculation puzzle for you. You can use any mathematical sign between the characters and you have to solve the puzzle.

Here it is:-

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

You can use any math sign between these three numerical digits and the answer must be 6.

Solve this puzzle and reply the answer, you can solve any part of this puzzle also.

**ChrisTurn** · 02-08-2011, 12:11 PM

2+2+2=6
3*3-3=6
5/5+5=6
6*6/6=6
7-7/7=6

**decadance** · 02-10-2011, 11:31 PM

^she's good

**Blake123** · 02-11-2011, 04:34 AM

I've solved this puzzle in school in the age of 9! Have you got some more harder? Share it....

**omd45** · 02-11-2011, 09:34 AM

(9+9)/square root of 9 = 6

boy this took me 10 minutes

**ChrisTurn** · 02-11-2011, 02:42 PM

Quote:

Originally Posted by decadance

^she's good

I'm a guy, Dosi

Just kidding with this puzzle

**deepaliw964** · 02-14-2011, 01:26 AM

Sorry but how to solve it?

**ChrisTurn** · 02-14-2011, 03:31 AM

You have to solve only with 4 and 8

**helltegal** · 02-14-2011, 04:33 AM

Math and Logic Puzzles

If you REALLY like exercising your brain, figuring things 'round and 'round till you explode, then this is the page for you !

**JeremyMiller** · 02-14-2011, 06:50 AM

Forgive the mathematician in me wanting to have some fun:

Define the operator ⊕ over the field of integers (ℤ) to be such that
x ⊕ x = 6 ∀ x ∈ ℤ

Then ∀ x ∈ ℤ, x ⊕ x ⊕ x = 6.

Code:

Proof:
x ⊕ x ⊕ x=(x ⊕ x) ⊕ x 
          =6 ⊕ x 
          =6 (since 6 ∈ ℤ)
or, since no implied order of operations has been assigned and ⊕ operates on only 2 operands (a left and a right),
x ⊕ x ⊕ x=x ⊕ (x ⊕ x) 
          =x ⊕ 6 
          =6 (since 6 ∈ ℤ)
□

Since 2, 3, ..., 9 are clearly ∈ ℤ, the ⊕ of any 3 of the same is trivially 6.

I was allowed to use any math sign, right?

**smoseley** · 10:19 AM

4*√4-√4 = 6 (where √ is "square root")

**smoseley** · 02-15-2011, 10:19 AM

√√8+√√8+√√8 = 6 (where √√ = "cubic root")

**NullPointer** · 02-15-2011, 09:17 PM

Quote:

Originally Posted by smoseley

√√8+√√8+√√8 = 6 (where √√ = "cubic root")

Is 'cubic root' an operator in and of itself? You could rewrite this as:
8^(1/3) + 8^(1/3) + 8^(1/3) = 6

Of course if you're allowed to define your own operators my point is moot

**decadance** · 02-15-2011, 11:49 PM

Quote:

Originally Posted by ChrisTurn

I'm a guy, Dosi

Just kidding with this puzzle

err..did I miss Oprah last week?

**smoseley** · 04:29 AM

Quote:

Originally Posted by NullPointer

Is 'cubic root' an operator in and of itself? You could rewrite this as:
8^(1/3) + 8^(1/3) + 8^(1/3) = 6

Of course if you're allowed to define your own operators my point is moot

Cubic Root is an operator in and of itself, at least as much as square root is... the following is the symbol for cubic root. The long-form version of square root is the same, supplanting the 3 for a 2 (as below), though the shorthand with no 2 is commonly used.

Technically, any root would be considered an operator, even if it contains a number in the symbol (even the 100th root, for example), as it's defining a process, not a value.

Cheers!

PS - interesting fact... the 100th root of Google is 10.

**smoseley** · 02-16-2011, 04:37 AM

So I guess here's the more appropriate way to show my answer:

³√8 + ³√8 + ³√8 = 6

**ChrisTurn** · 02-16-2011, 01:43 PM

I'm delighted, Mr. President *cool*

**Lemonie** · 02-17-2011, 07:25 AM

omg..no thanks i hate math

**NullPointer** · 08:02 AM

Quote:

Originally Posted by smoseley

Cubic Root is an operator in and of itself, at least as much as square root is

I would say square root isn't either.

Assuming they are, if ^(1/2) and ^(1/3) are distinct operators then why not any arbitrary ^(p)?. If this is the case then the problem becomes trivial. For any number N greater than 1, there is some p such that N^p = 6. In other words log_N(6) = p (log base n of six equals p). For example, 2^p = 6 when p ~ 2.585 (log_2(6) ~ 2.585).

Here is the answer for 8:

Code:

(8 + 8 - 8)^(log_8(6))

**ChrisTurn** · 02-17-2011, 09:53 AM

That's all, folks ! This is in Real, but in Complex ?