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# 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

2 years, 1 month 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 ? · 2 years, 1 month ago

Of course it is OK!! · 2 years, 1 month ago

@Skanda Prasad can we use factorials,and powers of fraction etc. · 2 years, 1 month ago

Are factorials fundamental operations?? · 2 years, 1 month ago

I do not really know.Are square and square roots a fundamentals? · 2 years, 1 month ago

I think yes. · 2 years, 1 month ago

Can you just google it and tell me?🐺 · 2 years, 1 month ago

Cannot find any info :( · 2 years, 1 month ago

Just see my answers · 2 years, 1 month ago

oK! · 2 years, 1 month 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 · 2 years, 1 month ago

Of course!! Nice one!! · 2 years, 1 month ago

Nice! · 2 years, 1 month 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 . . .

Or its OK, You are most welcome to provide solutions to my questions and I am always ready to give an upvote..!! · 2 years, 1 month ago

I don't recall giving this question to you in school... I found this out only yesterday when I was in Bangalore · 2 years, 1 month ago

It was @Gokul Balaji who gave this! and we solved it in the art period . . . Recall recall . . . · 2 years, 1 month ago

Yeah! I remember.... You know perfectly well that my memory power is not as brilliant as yours · 2 years, 1 month ago

Who said that! You really have a very good and logical brain! · 2 years, 1 month 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 · 2 years, 1 month 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 · 2 years, 1 month ago