If n and y are positive integers and 450y = n^3, which of the following is true.
I) y/3*2^2*5
II) y/3^2*2*5
III) y/3*2*5^2
a) None
b) I only
c) II only
d) III only
e) I, II and III
Can someone please explian this problem. I will provide the OA soon.
Thanks in advance.
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The options -
are actually like -I) y/3*2^2*5
II) y/3^2*2*5
III) y/3*2*5^2
Is it clear to you gmatguy16, now?I) y/(3*2^2*5)
II) y/(3^2*2*5)
III) y/(3*2*5^2)
Correct me If I am wrong
Regards,
Amitava
Regards,
Amitava
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450y =n^3
so we need a value of y which will make the expr 450y a perfect cube
450 = 3^2 * 5^2 *2
i.e none of the peime factors have the power of 3
so y must provide this
i.e y should have a 3,5,& two 2's to have powers of 3 of all the prime factors in 450y
in general y should be of the form 3^n*5^n*2^n+1
where n should be a value such that n +2 is a multiple of 3
so we can say here that y will be divisible by 3*5*2^2
hence A I only
so we need a value of y which will make the expr 450y a perfect cube
450 = 3^2 * 5^2 *2
i.e none of the peime factors have the power of 3
so y must provide this
i.e y should have a 3,5,& two 2's to have powers of 3 of all the prime factors in 450y
in general y should be of the form 3^n*5^n*2^n+1
where n should be a value such that n +2 is a multiple of 3
so we can say here that y will be divisible by 3*5*2^2
hence A I only
Regards
Samir
Samir