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

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## Comments

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TopNewest\[\dfrac{9+9}{\sqrt{9}} = 6\]

\[8^{\dfrac{1}{3}} + 8^{\dfrac{1}{3}} + 8^{\dfrac{1}{3}} =6 \]

\[ 7 -\dfrac{7}{7} = 6 \]

\[ 6 + 6 - 6 = 6 \]

\[ 5 + \dfrac{5}{5} = 6 \]

\[ 4 + 4 - \sqrt{4} = 6 \]

\[ \dfrac{{3^{2}} + {3^{2}}}{3} = 6\]

\[ 2 + 2 + 2 = 6\]

Is this ok @Skanda Prasad ?

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Of course it is OK!!

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

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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

uswho solved inschool. . .Please reply my friend!

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

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

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brain!logicalLog in to reply

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

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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

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Nice!

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Of course!! Nice one!!

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Nice one

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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

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(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

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