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#2 (permalink) |
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Eager!
Join Date: Jun 2006
Posts: 39
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Question seems insufficient somehow, but based on the information you provided, I would answer no, because:
the product of all the numbers in S is a negative number => there are either 1, 3 or 5 negative numbers in the bunch. Obviously then S can contain either more negative numbers, or fewer, or equal numbers |
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#4 (permalink) |
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Eager!
Join Date: Feb 2005
Posts: 65
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I would say E
from 1: Product of all number is negetive -> there are odd (1,3,5) number of negetive number(s). So there could be 1 ngetive and 5 positive number or there could be 5 negetive and 1 positive number; both series would have same result (a negetive value) From 2 There are 6 numbers does not add much to it. So, E |
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#5 (permalink) |
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Within my grasp!
![]() ![]() Join Date: May 2004
Posts: 323
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Agreed E.
Obviously A,B and D are not right. Now let's try and see if it is C or E. Examples of 6 number set satisfying both conditions that make it go either way: {1200, -1, 1, 1 ,1 ,1} more pos. {1200, -1, -1, -1 -,1 ,-1} more neg. So E.
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