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divya
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Post subject: if n and y are positive integers and 450y= n^3;  Posted: Thu Aug 14, 2008 4:53 pm |
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if n and y are positive integers and 450y= n^3; which of the following must be an integer:
I. y/(3 x 2^2 x5)
II. Y/( 3 ^ 2 X 2 X 5)
III. Y/( 3 X 2 X 5 ^ 3)
Answer I. only
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Raj
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Post subject: Re: if n and y are positive integers and 450y= n^3; Posted: Thu Aug 14, 2008 6:56 pm |
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Hi,
450 y = n^3
2*3*3*5*5 y = n^3
for this to be true, y has to have 2^2 & one 3 and one 5 so the cube is taken care of. So
y has to contain 2^2*3*5
With that said, only choice I has this combination which divides y perfectly.
Hope that helps,
-Raj.
divya wrote: if n and y are positive integers and 450y= n^3; which of the following must be an integer:
I. y/(3 x 2^2 x5)
II. Y/( 3 ^ 2 X 2 X 5)
III. Y/( 3 X 2 X 5 ^ 3)
Answer I. only
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divya
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Post subject: Posted: Thu Aug 14, 2008 7:29 pm |
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RonPurewal
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Post subject: Posted: Mon Aug 25, 2008 3:19 am |
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| ManhattanGMAT Staff |
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Posts: 7146
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good solution.
for any readers who don't understand Raj's derivation of what must be contained in 'y':
for the product 450y to be a perfect cube, its prime factors must come in threes. in other words, because the perfect cube is generated by multiplying together 3 identical copies of some number, you know that every prime factor occurs 3 times (or 6 times, or 9 times, or ...).
so because there's only one "2" in 450, there must be at least two more "2"s in y, in order to round out the perfect cube. (there could be more - there could be 5, or 8, or ..., but there must be at least 2.)
similar logic leads to the conclusion that there must be at least one "3" and at least one "5" in the factorization of y.
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Matt
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Post subject: Confused by this answer Posted: Tue Jan 20, 2009 5:23 pm |
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Hi, if I understand correctly, you are saying that as long as the prime factors in y contain "at least" 2^2 and 5^1 and 3^1, the product of 450y will be a cube. The "at least" part is what I don't understand, and I will show why:
450 x (2^2 x 3^1 x 5^1) = (2^3 x 3^3 x 5^3) ie. 27,000 ie. 30^3. A perfect cube, that much is clear.
Suppose I add another 2 into y:
450 x (2^3 x 3^1 x 5^1) = (2^4 x 3^3 x 5^3) ie. 54,000 ie. 37.77976315^3. This is obviously not a cube.
the same problem occurs when I add another 3 or 5 into y.
What is my error of comprehension? Is there some condition to your "at least" statement that I didn't understand?
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Matt
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Post subject: Posted: Tue Jan 20, 2009 5:39 pm |
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I just understood my error of comprehension: when you said "at least" two 2s, one 3, and one 5, you meant that the factors 2, 3, and 5 had to be added into y in such a way that the product of 450y would contain each of those factors in such a way that each one of those numbers had an exponent value that was a multiple of 3.
as in 450y could = (2^1 x 3^2 x 5^2) x (2^5 x 3^1 x 5^7) = (2^6 x 3^3 x 5^9) = 1500^3
It wasn't explicitly explained that way, and I'm not as good as you guys at math, so I didn't get that... my bad
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StaceyKoprince
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Post subject: Posted: Tue Jan 27, 2009 12:44 pm |
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| ManhattanGMAT Staff |
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Posts: 6077
Location: San Francisco
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Thanks, Matt, for explaining that a little bit further - and I'm glad you figured it out for yourself! Keep up the good work!
_________________
Stacey Koprince
Instructor
Director of Online Community
ManhattanGMAT
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