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Weird Integer Problem

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Manager
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Joined: 27 Oct 2009
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Weird Integer Problem [#permalink]

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New post 14 Nov 2009, 19:05
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Please explain your solution
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Re: Weird Integer Problem [#permalink]

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New post 14 Nov 2009, 19:30
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We know that 450 * y will equal a perfect cube, and we're told that n and y are postiive, so we do not have to worry about that.

450 * y = some cube. Break down 450 into primes

3 * 3 * 5 * 5 * 2

So 3 * 3 * 5 * 5 * 2 must equal some cube. We know that if we had three 3's, three 5's and three 2's, then that would be a perfect cube.

so if you let y equal what we are missing to give us three of each, then y must break down into the primes of 3 * 5 * 2 * 2. This is I. If we know that Y breaks doen to 2 * 5 * 2^2, and that is the denominator with y as the numerator, this will equal 1. for the same reason, we know that II and III will not result in integers.

ezinis wrote:
Please explain your solution
If n and y are positive integers and \(450y = n^3\), which of the following must be an integer?

I. \(\frac{y}{3 x 2^2 x 5}\)

II. \(\frac{y}{3^2 x 2 x 5}\)

III. \(\frac{y}{3 x 2 x 5^2}\)

a) None

b) I only

c) II only

d) III only

e I, II and III



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J Allen Morris
**I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

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Re: Weird Integer Problem [#permalink]

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New post 14 Nov 2009, 19:31
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ezinis wrote:
Please explain your solution


450 y = n^3

450 = 3*3*5*5*2

in order for 450 to have a cube root we need 3 of each number of 3*3*3*5*5*5*2*2*2

we already have two 3's and two 5's and one 2

therefore we need one 3, one 5, and 2 twos

or 3 x 5 x 2^2

that would give answer I
Manager
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Joined: 27 Oct 2009
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Re: Weird Integer Problem [#permalink]

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New post 15 Nov 2009, 05:36
Thanks jallenmorris +1 from me
Re: Weird Integer Problem   [#permalink] 15 Nov 2009, 05:36
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