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GRE 1: Q170 V170
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(A) for me :)

From 1
z/3 = 2*k+1 where k is an integer
<=> z = 3*(2*k + 1) = ODD*ODD = ODD

SUFF.

From 2
3*z = 2*k+1
<=> z= (2*k+1)/3

So,
o If 2*k+1 = 3*n (n an ODD integer), then:
z = 3*n / 3 and z is an ODD

o If 2*k+1 is not a multiple of 3, then:
z is even not an integer.

INSUFF.
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No, odds and evens are only for integers. While the question asks about whether Z is odd, statements 1 and 2 gives clue some fractions that contain z.

Statemnet 1 confirms that Z is indeed an integer, while statement 2 does not neccessarily indicate that Z is an integer, regardless whether odd or even.

My updated Answer: A
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The Alternative approach to this question is to try to find specific numbers that would eliminate certain answers because they lead to more than one option.
Statement (1) works for 3,9,15,21... all odd. Since z/3 equals odd, z = 3 times odd, so it must be odd. Statement (1) is sufficient, so (B), (C) and (E) are eliminated.
Statement (2) works for 1,3,5,7... but also for 1/3, so there's more than one option. Answer choice (D) is eliminated. The correct answer is (A).

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Is z an odd integer?

a) z/3 is odd integer
z=3*odd = odd*odd = odd {sufficient, cancel (BCE)}

b)3z= odd integer
z= odd integer /3
it may or may not be integer. Not sufficient

Answer A
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the logic which people are using to negate B is the same logic which can be used to negate A, as if Z is something like a 5.1 now divide by 3 it will give you a odd figure. which is 1.7
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the logic which people are using to negate B is the same logic which can be used to negate A, as if Z is something like a 5.1 now divide by 3 it will give you a odd figure. which is 1.7
­
No. Only integers can be even or odd. 1.7 is not odd becasue it's not an integer.
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