y1 , y2 , y3 .... are all distinct integers and represent : PS Archive
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# y1 , y2 , y3 .... are all distinct integers and represent

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VP
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y1 , y2 , y3 .... are all distinct integers and represent [#permalink]

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03 Oct 2008, 20:59
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y1 , y2 , y3 .... are all distinct integers and represent the entire set of integers. P is a non negative number such that
P = (x - y1 ) (x - y2 ) (x - y3 ) .... up to infinity, where x is an integer.

Which of the following value can x take so that p has the least value?
I. 0
II. 1
III. -1

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) any of I, II or III
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03 Oct 2008, 21:09
I don't think that infinite sets are in the GMAT scope

but the answer should be (E).

Georg Cantor would have loved this problem !!

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03 Oct 2008, 21:14
why do i have this strange feeling that it should definitely be A
SVP
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03 Oct 2008, 21:42
amitdgr wrote:
y1 , y2 , y3 .... are all distinct integers and represent the entire set of integers. P is a non negative number such that P = (x - y1 ) (x - y2 ) (x - y3) .... up to infinity, where x is an integer. Which of the following value can x take so that p has the least value?

I. 0
II. 1
III. -1

(A) I only
(B) II only
(C) I and II only
(D) I and III only
(E) any of I, II or III

not sure i got it properly.

what is the possible least value of p if it is not -ve? 0.
when is p = 0? if any of (x - y1 ) or (x - y2 ) or (x - y3) is 0.
so if one of either y1 or y2 or y3 is 1, then x has to be -1 so that p becoms 0.
similarly, if y1 or y2 or y3 is -1, then x has to be 1 to make p = 0.
if any of y1 or y2 or y3 is 0, then x has to be 0, then only p becomes 0.

so for me E.
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VP
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03 Oct 2008, 21:46
OA is E. Thanks for your explanation peeps
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Re: PS: sequence   [#permalink] 03 Oct 2008, 21:46
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