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
·
4 years, 3 months ago

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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
·
4 years, 3 months ago

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@Sam Segers
–
no,1 is the answer.the only integer that can solve this is 0.
–
Tan Li Xuan
·
4 years, 3 months 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
·
4 years, 3 months ago

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@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
·
4 years, 3 months ago

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YEAH!! im thinking that also!!
–
Soham Chanda
·
4 years, 3 months ago

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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
·
4 years, 3 months ago

## Comments

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TopNewestLive question? – Aditya Parson · 4 years, 3 months ago

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– Tan Li Xuan · 4 years, 3 months ago

last week algebra level 3Log in to reply

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 · 4 years, 3 months ago

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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 · 4 years, 3 months ago

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– Tan Li Xuan · 4 years, 3 months ago

no,1 is the answer.the only integer that can solve this is 0.Log in to reply

– Aditya Parson · 4 years, 3 months ago

CorrectLog in to reply

– Sam Segers · 4 years, 3 months ago

With other questions I have found that they don't count 0 as a positive integer.Log in to reply

– Aditya Parson · 4 years, 3 months ago

Read his comment, he said integer not positive integer, and 0 is an integer.Log in to reply

– Sam Segers · 4 years, 3 months ago

But the Question wasn't.Log in to reply

– Tan Li Xuan · 4 years, 3 months ago

the answer for the problem was 1Log in to reply

– Aditya Parson · 4 years, 3 months ago

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.Log in to reply

– Sam Segers · 4 years, 3 months ago

Then I'm not the one making it difficult here. You could have just told that instead. Indeed just leave it.Log in to reply

YEAH!! im thinking that also!! – Soham Chanda · 4 years, 3 months ago

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

var count=0;

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

if(perfect(2

i)===true && perfect(3i)===true){}

}

alert(count)

}

function perfect(N){

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

}

else{return false

}

looker()

i checked up to 100000000 but only counts one – Mharfe Micaroz · 4 years, 3 months ago

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