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# DS 7

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26 May 2004, 20:21
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Senior Manager
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26 May 2004, 21:57
xyz = 1 (2)
or x = y = 1 (2)

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

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27 May 2004, 05:34
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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27 May 2004, 09:16
Ndidi,

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

Thanks.
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27 May 2004, 10:48
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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27 May 2004, 12:06
Hi ndidi,

Thanks   [#permalink] 27 May 2004, 12:06
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