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# Find A and B. (1) (A*B)! is odd (2) A*B is odd

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SVP
Joined: 03 Feb 2003
Posts: 1607
Followers: 7

Kudos [?]: 87 [0], given: 0

Find A and B. (1) (A*B)! is odd (2) A*B is odd [#permalink]  31 Aug 2003, 07:31
Find A and B.

(1) (A*B)! is odd
(2) A*B is odd
SVP
Joined: 03 Feb 2003
Posts: 1607
Followers: 7

Kudos [?]: 87 [0], given: 0

Intern
Joined: 17 Aug 2003
Posts: 38
Location: USA
Followers: 0

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

A=B=1 can be deduced from (A*B)! is odd. but not from A*B is odd
Manager
Joined: 25 Jun 2003
Posts: 95
Followers: 1

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

I go with C

From (1) , A = 0 and B = 0
OR A = 1 and B = 0
OR A = 0 and B = 1
OR A = 1 and B = 1 , thus insufficient

From (2) A = 1 and B =1 is one of the many possible solutions

To satisfy both (1) and (2) , A = 1 and B = 1 are the only possible values
_________________

Brainless

Manager
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
Followers: 1

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

ups, i thought that was a PS, not a DS, sorry
i go for C as well
Intern
Joined: 19 Aug 2003
Posts: 8
Location: Austin, TX
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Kudos [?]: 0 [0], given: 0

Since there is no stipulation on whether A or B are integers, then A=0.25 and B=4 (or vice versa) can satisfy both conditions. The answer would have to be E.
Intern
Joined: 17 Aug 2003
Posts: 38
Location: USA
Followers: 0

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

Brainless wrote:
I go with C

From (1) , A = 0 and B = 0
OR A = 1 and B = 0
OR A = 0 and B = 1
OR A = 1 and B = 1 , thus insufficient

From (2) A = 1 and B =1 is one of the many possible solutions

To satisfy both (1) and (2) , A = 1 and B = 1 are the only possible values

I think you are right, since 0!=1
SVP
Joined: 03 Feb 2003
Posts: 1607
Followers: 7

Kudos [?]: 87 [0], given: 0

EssQ wrote:
Since there is no stipulation on whether A or B are integers, then A=0.25 and B=4 (or vice versa) can satisfy both conditions. The answer would have to be E.

RIGHT ON THE MONEY!!!
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