perfect squre

no of positive integers of x such that 2x and 3x are to be a perfcct square

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Live question?

Aditya Parson - 8 years, 1 month ago

last week algebra level 3

Tan Li Xuan - 8 years, 1 month ago

Their product, $6x^2$ is a perfect square. If $x$ is a non-zero integer, then $6$ is a perfect square, which is a contradiction.

Qi Huan Tan - 8 years, 1 month ago

YEAH!! im thinking that also!!

Soham Chanda - 8 years, 1 month ago

I think 0. If a = 2x and b = 3x then a = 2/3 * b So it would mean that root(2/3 * b) needs to be an integer. root(b) is an integer (cause it's a perfect square) but there is no integer that multiplied by an irrational number is also an integer. I'm not sure tho.

Sam Segers - 8 years, 1 month ago

no,1 is the answer.the only integer that can solve this is 0.

Tan Li Xuan - 8 years, 1 month ago

Correct

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – With other questions I have found that they don't count 0 as a positive integer.

Sam Segers - 8 years, 1 month ago

@Sam Segers – Read his comment, he said integer not positive integer, and 0 is an integer.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – But the Question wasn't.

Sam Segers - 8 years, 1 month ago

@Sam Segers – He answered it in a different context. Since last week the same question was asked on brilliant, and he assumed it asked integers and not positive. Just let it go.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – Then I'm not the one making it difficult here. You could have just told that instead. Indeed just leave it.

Sam Segers - 8 years, 1 month ago

@Sam Segers – the answer for the problem was 1

Tan Li Xuan - 8 years, 1 month ago

Copy paste this brute force code to jsbin.com function looker(){

var count=0;

for(i=0;i<100000000;i++){

if(perfect(2i)===true && perfect(3i)===true){

count++

}

alert(count)

}

function perfect(N){

if(Math.floor(Math.sqrt(N))==Math.sqrt(N)){

return true

}

else{return false

}

looker()

i checked up to 100000000 but only counts one

Mharfe Micaroz - 8 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$