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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

Note by Skanda Prasad
2 years, 3 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 ? Azhaghu Roopesh M · 2 years, 3 months ago

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@Azhaghu Roopesh M Of course it is OK!! Skanda Prasad · 2 years, 3 months ago

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@Skanda Prasad @Skanda Prasad can we use factorials,and powers of fraction etc. Kalash Verma · 2 years, 3 months ago

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@Kalash Verma Are factorials fundamental operations?? Harsh Shrivastava · 2 years, 3 months ago

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@Harsh Shrivastava I do not really know.Are square and square roots a fundamentals? Kalash Verma · 2 years, 3 months ago

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@Kalash Verma I think yes. Harsh Shrivastava · 2 years, 3 months ago

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@Harsh Shrivastava Can you just google it and tell me?🐺 Kalash Verma · 2 years, 3 months ago

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@Kalash Verma Cannot find any info :( Harsh Shrivastava · 2 years, 3 months ago

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@Harsh Shrivastava Just see my answers Kalash Verma · 2 years, 3 months ago

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@Kalash Verma oK! Harsh Shrivastava · 2 years, 3 months ago

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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 Kalash Verma · 2 years, 3 months ago

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@Kalash Verma Of course!! Nice one!! Skanda Prasad · 2 years, 3 months ago

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@Kalash Verma Nice! Harsh Shrivastava · 2 years, 3 months ago

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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 us who solved in school . . .

Please reply my friend! Sravanth Chebrolu · 2 years, 3 months ago

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@Sravanth Chebrolu Or its OK, You are most welcome to provide solutions to my questions and I am always ready to give an upvote..!! Skanda Prasad · 2 years, 3 months ago

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@Sravanth Chebrolu I don't recall giving this question to you in school... I found this out only yesterday when I was in Bangalore Skanda Prasad · 2 years, 3 months ago

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@Skanda Prasad It was @Gokul Balaji who gave this! and we solved it in the art period . . . Recall recall . . . Sravanth Chebrolu · 2 years, 3 months ago

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@Sravanth Chebrolu Yeah! I remember.... You know perfectly well that my memory power is not as brilliant as yours Skanda Prasad · 2 years, 3 months ago

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@Skanda Prasad Who said that! You really have a very good and logical brain! Sravanth Chebrolu · 2 years, 3 months ago

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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 Angela Fajardo · 2 years, 3 months ago

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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 Rei Vilo · 2 years, 3 months ago

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Nice one Suprem S.Nalkund · 2 years, 3 months ago

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