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# E and F are integers. Find them. (1) !=2 (2) E<0

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Joined: 03 Feb 2003
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E and F are integers. Find them. (1) !=2 (2) E<0 [#permalink]

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13 Aug 2003, 10:31
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E and F are integers. Find them.

(1) [(E-1)(F-1)]!=2
(2) E<0

Kudos [?]: 303 [0], given: 0

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 303 [0], given: 0

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14 Aug 2003, 04:40
Guys, do not get used to guess! Post your solutions!

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Intern
Joined: 30 Jul 2003
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14 Aug 2003, 06:15
It is C. E=-2/F=0 (and it can not be the other way around taking into account that E<0).

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SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 303 [0], given: 0

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14 Aug 2003, 06:25
hopeful wrote:
It is C. E=-2/F=0 (and it can not be the other way around taking into account that E<0).

-2/F=0 has no GMAT sense, F= +/-infitity

[E=0 and F=inf] is a wrong answer

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Intern
Joined: 30 Jul 2003
Posts: 6

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14 Aug 2003, 06:38
I mean E=2 and F=0 as the only solution taking into account 1 and 2 conditions. Therefore, the answer is C.

Stolyar, what is the right answer?

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Manager
Joined: 10 Jun 2003
Posts: 209

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Location: Maryland

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14 Aug 2003, 07:15
From (A):

You get the possibilities (E,F): (3,2), (2,3), (0,-1), (-1,0)

not suff

From (B):anything in the world
not suff

But taken together, if E is negative, then the pair must be (-1,0)

C

Kudos [?]: 7 [0], given: 0

SVP
Joined: 03 Feb 2003
Posts: 1603

Kudos [?]: 303 [0], given: 0

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14 Aug 2003, 08:10
mciatto wrote:
From (A):

You get the possibilities (E,F): (3,2), (2,3), (0,-1), (-1,0)

not suff

From (B):anything in the world
not suff

But taken together, if E is negative, then the pair must be (-1,0)

C

a great work! you all guys had to see that 2 is a prime, and since E and F are integers there are not so many combinations: 1 and 2 plus negatives.
C is correct.

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# E and F are integers. Find them. (1) !=2 (2) E<0

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