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Bunuel
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Bunuel
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Is integer zz odd?
(1) z3 is odd

(2) 3z is odd

Second Statement says 3z is odd. According to statement 2, z could (1/3) then 3z=1, which is still odd.

Is integer z odd?

The stem defines z as integer, so z cannot be 1/3.
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For those interested, another way of solving this is to use the following definitions:
any odd number can be written in the form 2n+1 with n integer
any even number can be written 2n with n integer

(1) tells us that z/3 is of the form 2k+1 with k integer
This means that z = 3 * (2k+1) = 6k+3 = 2(3k+1) + 1. This is of the format 2n+1 with n integer therefore z is odd. Sufficient

(2) 3z is odd
z is either odd or even.
If z is even, then it can be written z=2k with k integer. In this case 3z = 3 x 2k = 2(3k) with 3k an integer, meaning that 3z would be even, which contradicts statement (2).
Therefore z must be odd! Sufficient
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But if z=0?
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mariql
But if z=0?

z = 0 does not satisfy any of the statements: z/3 = 0/3 = 0 = even and 3z = 3*0 = 0 = even.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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