450y=n^3. Num Properties

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gmatruler: Junior | Next Rank: 30 Posts; Posts: 28; Joined: Mon Jul 05, 2010 6:24 am; Thanked: 1 times

450y=n^3. Num Properties

by gmatruler » Tue Jul 13, 2010 5:02 am

If n and y are positive integers and 450y=n^3, which of the following must be an integer?

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

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Source: Beat The GMAT — Problem Solving | Tue Jul 13, 2010 5:02 am

kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pm; Thanked: 215 times; Followed by:7 members

by kvcpk » Tue Jul 13, 2010 5:09 am

gmatruler wrote:If n and y are positive integers and 450y=n^3, which of the following must be an integer?

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

450y=n^3
5^2 * 3^2 * 2 * y = n^3
y can be 5*3*2^2
So only I is correct.

pick B

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Rahul@gurome: GMAT Instructor; Posts: 1179; Joined: Sun Apr 11, 2010 9:07 pm; Location: Milpitas, CA; Thanked: 447 times; Followed by:88 members

by Rahul@gurome » Tue Jul 13, 2010 5:10 am

450y=n^3 implies (3^2 * 5^2 * 2)y = n^3, which implies y should be 2^2 * 3 * 5 so that 450y=n^3
Then (I) will be an integer.
(II) and (III) will not be an integer always.

[spoiler]The correct answer is (B).[/spoiler]

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Patrick_GMATFix: GMAT Instructor; Posts: 1052; Joined: Fri May 21, 2010 1:30 am; Thanked: 335 times; Followed by:98 members

by Patrick_GMATFix » Tue Jul 13, 2010 5:16 am

The key here is to know how to make good use of 450y=n^3. When solving questions about divisibility, factors, multiples or primes, it is usually a good idea to break your numbers down into their prime factors.

450y=n^3 means that (2)(3^2)(5^2) y = (n)(n)(n) . Since the left side must equal the right, the left side must be 'breakable' into 3 equal factors. We already know that the prime numbers involved are 2, 3, and 5 (there may be more but we have no proof of this). Thus we can rewrite this equation as (2*3*5)(_*3*5)(_*_*_)=(n)(n)(n). The unknown variable y must fill in the missing primes for the equation to be valid. Thus y must contain 2*2*3*5 as a factor. Consequently, only statement II must be correct.

The correct answer is B. A more involved discussion, take-away lesson and video solution can be found at GMATPrep Question 1083. To practice similar questions, set topic='Number Properties' and difficulty='600-700 & 700+' in the Drill Generator

Good luck,
-Patrick

Check out my site: GMATFix.com

To prep my students I use this tool >> (screenshots, video)

Ask me about tutoring.

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fleckre: Junior | Next Rank: 30 Posts; Posts: 12; Joined: Wed Apr 07, 2010 8:32 am

by fleckre » Tue Jul 13, 2010 5:54 am

Thanks for the explanations. I'm still having a bit of trouble, though. I understand why we quickly break it down to prime factors, but I am not sure why

which implies y should be 2^2 * 3 * 5

Is it because you want to get the three numbers as cubes?

Thanks for your help.

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Patrick_GMATFix: GMAT Instructor; Posts: 1052; Joined: Fri May 21, 2010 1:30 am; Thanked: 335 times; Followed by:98 members

by Patrick_GMATFix » Tue Jul 13, 2010 6:01 am

Fleckre, any perfect cube can be broken down into a product of 3 identical factors. 27= (3)(3)(3). 216=(2*3)(2*3)(2*3)...

Since 450y is a perfect cube (it is n^3), it must be a product of 3 identical factors.

450 is (2*3*5) (2*5). For 450y to be a perfect cube, y must complete the trio, at a minimum to make (2*3*5)(2*3*5)(2*3*5). Thus y must have 2^2*3*5 as a factor.

Note that the trio could have a lot more numbers. y could also be 2^2*3*5*7^3, in which case the trio would be (2*3*5*7)(2*3*5*7)(2*3*5*7). There are an infinite number of possible values that y can take to make 450y a perfect cube, but what we know y must have is 2^2*3*5 at a minimum.

-Patrick