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Multiples...

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Manager
Joined: 01 Jun 2006
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Multiples... [#permalink]  17 Dec 2006, 05:38
Q)
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Manager
Joined: 15 Nov 2006
Posts: 224
Location: Ohio
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If 450Y=X3
That means Y must be a factor of 450
Factors of 450=2X5X5X3X3
Thereofore, answer is (D). III only
Manager
Joined: 01 Jun 2006
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The ans is B.....anyone...can shed light on this...
Senior Manager
Joined: 08 Jun 2006
Posts: 341
Location: Washington DC
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450Y=X3
=>5^2 * 3^2 * 2 *Y = n^3
to make y a perfect cube, it has to be multiple of 5 * 3 * 2^2

so that n^3 = (5^2 * 3^2 * 2) * ( 5 * 3 * 2^2) K = (5 * 3 * 2)^3 * k
here k is another + ve integer

so Only B stands..
Manager
Joined: 01 Jun 2006
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great stuff...anindyat....
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Multiples...

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