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

Oldest

Best Reply

Author

Message

TAGS:

SVP

Joined: 03 Feb 2003

Posts: 1683

Followers: 4

Kudos [?]: 16 [0], given: 0

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

13 Aug 2003, 10:31

00:00

Question Stats:

0% (00:00) correct

0% (00:00) wrong

based on 0 sessions

E and F are integers. Find them.

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

SVP

Joined: 03 Feb 2003

Posts: 1683

Followers: 4

Kudos [?]: 16 [0], given: 0

[#permalink]

14 Aug 2003, 04:40

Guys, do not get used to guess! Post your solutions!

Intern

Joined: 30 Jul 2003

Posts: 6

Followers: 0

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

[#permalink]

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).

SVP

Joined: 03 Feb 2003

Posts: 1683

Followers: 4

Kudos [?]: 16 [0], given: 0

[#permalink]

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

Intern

Joined: 30 Jul 2003

Posts: 6

Followers: 0

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

[#permalink]

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?

Manager

Joined: 10 Jun 2003

Posts: 216

Location: Maryland

Followers: 2

Kudos [?]: 1 [0], given: 0

Answer [#permalink]

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

SVP

Joined: 03 Feb 2003

Posts: 1683

Followers: 4

Kudos [?]: 16 [0], given: 0

Re: Answer [#permalink]

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.

Re: Answer [#permalink] 14 Aug 2003, 08:10

		Similar topics	Author	Replies	Last post
Similar Topics:		Find integers E and F. (1) E^F=4 (2) F^E=1	stolyar	3	31 Aug 2003, 12:17
		Is (a/d)(b/e)(c/f) > (b/d)*(c/e) ? (1) bc > de (2) a	forumsmba	4	02 Dec 2004, 16:48
	3	P= dX^2+eX+f; d,e,f are real number, d<0 what are d, e,	vcbabu	11	22 Jul 2009, 12:52
		Find Integer E (1) E = 1 ^ E (2) E ^ 0 = E The OA is D My	rxs0005	1	22 Jul 2010, 08:54
	2	Is cdef < 0? 1) cd2e3f4< 0 2) c2d3e4f5 > 0 (A)	MitDavidDv	8	04 May 2011, 10:37

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

Main navigation

GMAT Resources

Partners

MBA Resources

GMAT ® is a registered trademark of the Graduate Management Admission Council ® (GMAC ®). GMAT Club's website has not been reviewed or endorsed by GMAC.

GMAT Courses

Admissions Consulting

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

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

Main navigation

GMAT Resources

Partners

MBA Resources