Number properties Roman Numeral from GMATPREP

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 105
Joined: Tue Oct 16, 2007 5:57 am
Thanked: 3 times

Number properties Roman Numeral from GMATPREP

by abhi75 » Sun Dec 16, 2007 1:05 pm
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.

Master | Next Rank: 500 Posts
Posts: 195
Joined: Sun Oct 21, 2007 4:33 am
Thanked: 10 times

Master | Next Rank: 500 Posts
Posts: 124
Joined: Thu Aug 23, 2007 5:11 am
Thanked: 2 times

by gmatguy16 » Mon Dec 17, 2007 3:34 pm
I am not able to understand the answer options..can someone please put appropriate brackets so that the options can be interpreted.

Legendary Member
Posts: 645
Joined: Wed Sep 05, 2007 4:37 am
Location: India
Thanked: 34 times
Followed by:5 members

by camitava » Mon Dec 17, 2007 8:35 pm
The options -
I) y/3*2^2*5
II) y/3^2*2*5
III) y/3*2*5^2
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?
Correct me If I am wrong


Regards,

Amitava

Master | Next Rank: 500 Posts
Posts: 460
Joined: Sun Mar 25, 2007 7:42 am
Thanked: 27 times

by samirpandeyit62 » Tue Dec 18, 2007 3:02 am
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
Regards
Samir