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

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

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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
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14 Aug 2003, 01:38
(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...
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14 Aug 2003, 01:55
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
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14 Aug 2003, 02:03
stolyar wrote:
Ok, it is getting hot...
Why E?

Upsss
sorry did not notice "!"
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14 Aug 2003, 02:04
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
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14 Aug 2003, 03:34
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.
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15 Aug 2003, 00:52
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
15 Aug 2003, 00:52
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