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gmat prep algebra

Expert replies
Source: — Problem Solving |

by pepeprepa » Thu Aug 28, 2008 3:13 pm
450y=n^3
3²*5²*2*y=n^3

y must at least contain 3,5, and 2² to suit the form n^3
Imagine y=1, 3²*5²*2=n^3 does not exist at least for n as an integer, we need to state that y contains integers to suit n^3
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by niraj_a » Fri Aug 29, 2008 4:05 pm
i still don't understand this
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450y = n cubed problem

by cramya » Fri Aug 29, 2008 4:59 pm
Given 450y= n cubed

break down 450 = 2 * 3 squared * 5 squared

To get to the cubed form for n, y must be atleast equal to 2 squared * 3 * 5

450 y
so 450y becomes (2 * 3 squared * 5 squared) * (2 squared * 3 * 5)



If we assume take y= 2 squared * 3 * 5
we can only have (I ) y/ 2 * 3 squared * 5 squared as an integer

If y = 2 squared * 3 * 5 then (II) would be 2/3 which is not an integer

If y = 2 squared * 3 * 5 then (III) would be 2/5 which is not an integer
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by lunarpower » Sat Aug 30, 2008 11:01 pm
first, rephrase the problem. if y divided by a bunch of factors is an integer, that's just a roundabout way of saying that all those factors go into y.
so, this question is really asking: which of the following sets of factors must divide into y?
(i) two 2's, one 3, one 5; (ii) two 3's, one 2, one 5; (iii) two 5's, one 2, one 3.

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 ...).
if you don't see why this is true, try factoring out some perfect cubes for yourself. for instance, 8 = 2^3 = 2x2x2 (three 2's). 1728 = 12^3 = 2x2x2x2x2x2x3x3x3 (six 2's, three 3's). etc.

note that 450 = 10x45 = 2x5x5x3x3, but 450y = 2x5x5x3x3xY is a perfect cube. this means that y must contain ALL the missing factors to round out the sets of 3 to make a perfect cube.
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.
so it's (i) only.

--

<random comment>
it's a bit unusual of the gmat writers to construct the denominators with the 2, 3, and 5 out of numerical order. normally, their question writing is very systematic, and i'd expect the 2, 3, and 5 to appear in that order.
</random comment>
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