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# a, b, c are integers. Is (a*b*c)! even? (1) a^2-5a=-6 (2)

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SVP
Joined: 03 Feb 2003
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a, b, c are integers. Is (a*b*c)! even? (1) a^2-5a=-6 (2) [#permalink]  13 Aug 2003, 09:23
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a, b, c are integers. Is (a*b*c)! even?

(1) a^2-5a=-6
(2) bc>0
Manager
Joined: 14 Aug 2003
Posts: 88
Location: barcelona
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Kudos [?]: 0 [0], given: 0

(a*b*c)! can only be odd if a*b*c=1...
A) tells us that a=2 or a=3... b, c integers... a*b*c<>1... so A suf...
B) bc>0... not suf...
SVP
Joined: 03 Feb 2003
Posts: 1608
Followers: 6

Kudos [?]: 76 [0], given: 0

javropu wrote:
(a*b*c)! can only be odd if a*b*c=1...
A) tells us that a=2 or a=3... b, c integers... a*b*c<>1... so A suf...
B) bc>0... not suf...

(a*b*c)! can also be odd if a*b*c=0 since 0!=1
Manager
Joined: 11 Jun 2003
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Location: Moscow, Russia
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stolyar wrote:
Ok, it is getting hot...
Why E?

Upsss
sorry did not notice "!"
Manager
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0!=1??? i didnt know that... thx pal... in that case, id go for C...
(1) tells us a=2 or a=3... not suf
(2) bc>0 not suf
but (1) & (2) suf, coz we know a=2 or a=3, b, c<>0=> a*b*c<>0 and a*b*c<>1... suf
SVP
Joined: 03 Feb 2003
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Dmitry wrote:
stolyar wrote:
Ok, it is getting hot...
Why E?

Than C!!!

C is correct, but give a clear rationale.
the foregoing is pretty vague.
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
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stolyar wrote:
Dmitry wrote:
stolyar wrote:
Ok, it is getting hot...
Why E?

Than C!!!

C is correct, but give a clear rationale.
the foregoing is pretty vague.

n! is always even except when n=1 or 0

Now from (1), we know that a is a non zero number and that its not 1

From (2) we know that b and c are non zero

Therefore, from (1) and (2) together we can conclude that (a*b*c)! is even

Hence C
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