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

## 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 ? – Azhaghu Roopesh M · 2 years, 5 months ago

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– Skanda Prasad · 2 years, 5 months ago

Of course it is OK!!Log in to reply

@Skanda Prasad can we use factorials,and powers of fraction etc. – Kalash Verma · 2 years, 5 months ago

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– Harsh Shrivastava · 2 years, 5 months ago

Are factorials fundamental operations??Log in to reply

– Kalash Verma · 2 years, 5 months ago

I do not really know.Are square and square roots a fundamentals?Log in to reply

– Harsh Shrivastava · 2 years, 5 months ago

I think yes.Log in to reply

– Kalash Verma · 2 years, 5 months ago

Can you just google it and tell me?🐺Log in to reply

– Harsh Shrivastava · 2 years, 5 months ago

Cannot find any info :(Log in to reply

– Kalash Verma · 2 years, 5 months ago

Just see my answersLog in to reply

– Harsh Shrivastava · 2 years, 5 months ago

oK!Log in to reply

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, 5 months ago

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– Skanda Prasad · 2 years, 5 months ago

Of course!! Nice one!!Log in to reply

– Harsh Shrivastava · 2 years, 5 months ago

Nice!Log in to reply

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! – Sravanth Chebrolu · 2 years, 5 months ago

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– Skanda Prasad · 2 years, 5 months ago

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

– Skanda Prasad · 2 years, 5 months ago

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

@Gokul Balaji who gave this! and we solved it in the art period . . . Recall recall . . . – Sravanth Chebrolu · 2 years, 5 months ago

It wasLog in to reply

– Skanda Prasad · 2 years, 5 months ago

Yeah! I remember.... You know perfectly well that my memory power is not as brilliant as yoursLog in to reply

brain! – Sravanth Chebrolu · 2 years, 5 months agologicalLog 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 – Angela Fajardo · 2 years, 5 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, 5 months ago

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Nice one – Suprem S.Nalkund · 2 years, 5 months ago

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