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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`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

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Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## 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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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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Of course!! Nice one!!

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Nice!

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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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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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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

(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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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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Nice one

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