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# Weird Integer Problem

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Manager
Joined: 27 Oct 2009
Posts: 149
Location: Montreal
Schools: Harvard, Yale, HEC
Followers: 1

Kudos [?]: 70 [0], given: 18

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14 Nov 2009, 20:05
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SVP
Joined: 30 Apr 2008
Posts: 1888
Location: Oklahoma City
Schools: Hard Knocks
Followers: 40

Kudos [?]: 549 [1] , given: 32

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14 Nov 2009, 20:30
1
KUDOS
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:
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

_________________

------------------------------------
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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VP
Joined: 05 Mar 2008
Posts: 1473
Followers: 11

Kudos [?]: 256 [1] , given: 31

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14 Nov 2009, 20:31
1
KUDOS
ezinis wrote:

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

Manager
Joined: 27 Oct 2009
Posts: 149
Location: Montreal
Schools: Harvard, Yale, HEC
Followers: 1

Kudos [?]: 70 [0], given: 18

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15 Nov 2009, 06:36
Thanks jallenmorris +1 from me
Re: Weird Integer Problem   [#permalink] 15 Nov 2009, 06:36
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