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DS 7

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SVP
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DS 7 [#permalink] New post 26 May 2004, 19:21
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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Senior Manager
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 [#permalink] New post 26 May 2004, 20:57
xyz = 1 (2)
or x = y = 1 (2)

x(y^2+z^3) = 0 (2), or even number

B is the answer.
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 [#permalink] New post 27 May 2004, 04:34
B is sufficient.

If xyz is odd, then x, y and z are odd and x(y^2 + z^3) is even.

Statement 1 is insufficent; we don't know anything about y and z.
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answer [#permalink] New post 27 May 2004, 08:16
Ndidi,

Coud you please elaborate your sentence,
"If xyz is odd, then x, y and z are odd and x(y^2 + z^3) is even. "


Thanks.
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 [#permalink] New post 27 May 2004, 09:48
Hi Abacus,

If xyz is odd, then x, y and z are odd and x(y^2 + z^3) is even.


If xyz is odd, it means that x is odd, y is odd and z is odd. If any one of the three is even, then xyz would be even. Remember that 2 is a factor of every even number so x, y or z cannot be even.

If x, y and z are odd, then:

odd(odd^2 + odd^3) = even

odd^2 + odd^3 is even since the sum of 2 odd numbers is even.

odd * even is also even.
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Thanks [#permalink] New post 27 May 2004, 11:06
Hi ndidi,

Thank you for your explanation.
Thanks   [#permalink] 27 May 2004, 11:06
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