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

#NumberTheory #Numbers #Logic #Easy #Skanda

Note by Skanda Prasad
6 years, 2 months ago

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1 vote

  Easy Math Editor

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Comments

9+99=6\dfrac{9+9}{\sqrt{9}} = 6

813+813+813=68^{\dfrac{1}{3}} + 8^{\dfrac{1}{3}} + 8^{\dfrac{1}{3}} =6

777=6 7 -\dfrac{7}{7} = 6

6+66=6 6 + 6 - 6 = 6

5+55=6 5 + \dfrac{5}{5} = 6

4+44=6 4 + 4 - \sqrt{4} = 6

32+323=6 \dfrac{{3^{2}} + {3^{2}}}{3} = 6

2+2+2=6 2 + 2 + 2 = 6

Is this ok @Skanda Prasad ?

A Former Brilliant Member - 6 years, 2 months ago

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Of course it is OK!!

Skanda Prasad - 6 years, 2 months ago

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

A Former Brilliant Member - 6 years, 2 months ago

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@A Former Brilliant Member Are factorials fundamental operations??

Harsh Shrivastava - 6 years, 2 months ago

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@Harsh Shrivastava I do not really know.Are square and square roots a fundamentals?

A Former Brilliant Member - 6 years, 2 months ago

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@A Former Brilliant Member I think yes.

Harsh Shrivastava - 6 years, 2 months ago

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@Harsh Shrivastava Can you just google it and tell me?🐺

A Former Brilliant Member - 6 years, 2 months ago

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@A Former Brilliant Member oK!

Harsh Shrivastava - 6 years, 2 months ago

@A Former Brilliant Member Cannot find any info :(

Harsh Shrivastava - 6 years, 2 months ago

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@Harsh Shrivastava Just see my answers

A Former Brilliant Member - 6 years, 2 months 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 . . .

Please reply my friend!

Sravanth C. - 6 years, 2 months ago

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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

Skanda Prasad - 6 years, 2 months ago

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

Sravanth C. - 6 years, 2 months ago

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@Sravanth C. Yeah! I remember.... You know perfectly well that my memory power is not as brilliant as yours

Skanda Prasad - 6 years, 2 months ago

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@Skanda Prasad Who said that! You really have a very good and logical brain!

Sravanth C. - 6 years, 2 months ago

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

Skanda Prasad - 6 years, 2 months ago

2*2+2=6

3!+3-3=6

4!/√4+√4=6

5!/[ 525^{2}-5]

(666)1/3(6*6*6)^{1/3}

[727^{2}-7]/7

[8/81/38^{1/3}]!/82/38^{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

A Former Brilliant Member - 6 years, 2 months ago

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Nice!

Harsh Shrivastava - 6 years, 2 months ago

Of course!! Nice one!!

Skanda Prasad - 6 years, 2 months ago

Nice one

Suprem s.nalkund - 6 years, 2 months 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

Rei Vilo - 6 years, 2 months 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

Angela Fajardo - 6 years, 2 months ago
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