# Maths! Maths! Maths! Roots and Squares.... Signs... What else??

Hello People!

Here's a problem where I can get maximum response from the community.

In the question, you can make the numbers undergo anything you wish and satisfy the conditions of the problem.

You can use any Fundamental Operations...

Eg : you can square, square root, cube root, cube the numbers...

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 are to use the operations in between the numbers.

You can make the numbers undergo any changes 4 years, 9 months ago

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$\dfrac{9+9}{\sqrt{9}} = 6$

$8^{\dfrac{1}{3}} + 8^{\dfrac{1}{3}} + 8^{\dfrac{1}{3}} =6$

$7 -\dfrac{7}{7} = 6$

$6 + 6 - 6 = 6$

$5 + \dfrac{5}{5} = 6$

$4 + 4 - \sqrt{4} = 6$

$\dfrac{{3^{2}} + {3^{2}}}{3} = 6$

$2 + 2 + 2 = 6$

Is this ok @Skanda Prasad ?

- 4 years, 9 months ago

Of course it is OK!!

- 4 years, 9 months ago

@Skanda Prasad can we use factorials,and powers of fraction etc.

- 4 years, 9 months ago

Are factorials fundamental operations??

- 4 years, 9 months ago

I do not really know.Are square and square roots a fundamentals?

- 4 years, 9 months ago

I think yes.

- 4 years, 9 months ago

Can you just google it and tell me?🐺

- 4 years, 9 months ago

oK!

- 4 years, 9 months ago

Cannot find any info :(

- 4 years, 9 months ago

- 4 years, 9 months ago

Great question for the brilliant community. . . By the way, it is up to you if you want me to post the answer @Skanda Prasad , 'cause it was us who solved in school . . .

- 4 years, 9 months ago

I don't recall giving this question to you in school... I found this out only yesterday when I was in Bangalore

- 4 years, 9 months ago

It was @Gokul Balaji who gave this! and we solved it in the art period . . . Recall recall . . .

- 4 years, 9 months ago

Yeah! I remember.... You know perfectly well that my memory power is not as brilliant as yours

- 4 years, 9 months ago

Who said that! You really have a very good and logical brain!

- 4 years, 9 months ago

Or its OK, You are most welcome to provide solutions to my questions and I am always ready to give an upvote..!!

- 4 years, 9 months ago

2*2+2=6

3!+3-3=6

4!/√4+√4=6

5!/[ $5^{2}$-5]

$(6*6*6)^{1/3}$

[$7^{2}$-7]/7

[8/$8^{1/3}$]!/$8^{2/3}$

√(9*9) - √9

I do not know if '!' etc are allowed but still I tried to be different.u Is it OK ?@Skanda Prasad

- 4 years, 9 months ago

Nice!

- 4 years, 9 months ago

Of course!! Nice one!!

- 4 years, 9 months ago

Nice one

- 4 years, 9 months ago

2+2+2=6 3×3-3=6 Sqrt (4)+sqrt(4)+sqrt (4)=6 5+5/5=6 6+6-6=6 7-7/7=6 Cuberoot (8)+cuberoot (8)+cuberoot (8)=6 9/sqrt (9)+sqrt (3)=6

- 4 years, 9 months ago

(2 x 2) + 2 = 6

(3 x 3) - 3 = 6

(4 + 4) - sqrt 4 = 6

(5 / 5) + 5 = 6

(6 x 6) / 6 = 6

7 - (7 / 7) = 6

cuberoot 8 + cuberoot 8 + cuberoot 8 = 6

(9 + 9) / sqrt9 = 6

- 4 years, 9 months ago