Possible or impossible

Logic Level 1

3 + 4 = 5 3 + 4 = 5

Above given a wrong equation. Is it possible for you to place mathematical operators like 2 , 0 , etc ^2, \sqrt{\phantom0}, \text{etc} into the equation above so that it becomes true?

No Yes

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6 solutions

Munem Shahriar
Aug 2, 2017

3 2 = 9 3^2 = 9

4 2 = 16 4^2 = 16

9 + 16 = 25 9 + 16 = 25

25 = 5 \sqrt{25} = 5

3 + √4 =5 the simplest method.

Md Mehedi Hasan
Nov 30, 2017

We can input 2 squares and a root to correct the equation.

3 2 + 4 2 = 5 \LARGE {\color{#D61F06}\sqrt{\color{#333333}3^{\color{#D61F06}2}+4^{\color{#D61F06}2}}}=5

This is not correct.

9 + 16 5 9 + 16 \ne 5

Munem Shahriar - 3 years, 6 months ago

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I put 5 2 5^2

Md Mehedi Hasan - 3 years, 6 months ago

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I was trying say that we need to make 5 not 25

Munem Shahriar - 3 years, 6 months ago

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@Munem Shahriar ok I have edited. you can clear it in question that we can only change in left side and have to be equal 5.

Md Mehedi Hasan - 3 years, 6 months ago
Saksham Jain
Nov 11, 2017

3 +square root of 4=5

3 +square root of 2=5

This is wrong. 3 + 2 4.414 3 + \sqrt2 \approx 4.414 .

Munem Shahriar - 3 years, 7 months ago

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thanks it was a typing mistake

Saksham Jain - 3 years, 7 months ago
Genis Dude
Aug 14, 2017

Just put the '√' in 4

Therefore,

The equation changes to

3+√4=5 ,which is correct

Sándor Daróczi
Aug 2, 2017

The equation is true ( m o d 2 ) \pmod{2} , since

3 + 4 7 1 ( m o d 2 ) 3+4 \equiv 7 \equiv 1 \pmod{2}

and

5 1 ( m o d 2 ) 5 \equiv 1 \pmod{2} .

That's a congruence, however, not an equation.

Zach Abueg - 3 years, 10 months ago

Yes, you are right. Although, in some point of view, it can be interpreted as an equation. At least I believe.

Sándor Daróczi - 3 years, 10 months ago

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