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(1) Two cases: (i) If x = odd => z = odd (ii) If x = even => z = even

(2) Two cases: (i) if y = odd => z = even (and vice versa) (ii) if y = even => Z = odd (and vice versa)

(1) & (2) combined, again two cases: (i) If x = odd => y = even (ii) If x = even => y = odd

As subtracting an odd number from an even number and subtracting an even number from an odd number always results in an odd number it follows that C is the correct solution.

If x, y and z are positive integers, is x - y odd? (1) x=z^2 (2) y=(z-1)^2

Can this problem be solve by plunging number?

Yes, you can plug in numbers. Generally, in even odd questions, plugging numbers works. (mind you, generally, not always)

Break it down in the following way:

If x, y and z are positive integers, is x - y odd? Question: Is one of x and y even and one odd? (because x - y will be odd only if one of them is even and one is odd) (1) x=z^2 If z is even, x is even. If z is odd, x is odd. No info about y. or if z = 2, x is 4. If z = 1, x is 1.

(2) y=(z-1)^2 If z is even, y is odd. If z is odd, y is even. No info about x. or If z = 2, y = 1. If z = 1, y is 0 (even number).

Together: If z is even, x is even and y is odd. If z is odd, x is odd and y is even. One is always odd, other is always even. or If z is 2, x = 4 and y = 1 If z = 1, x = 1 and y = 0

Re: If x, y and z are positive integers, is x - y odd? [#permalink]

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07 Jan 2014, 04:31

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Re: If x, y and z are positive integers, is x - y odd? [#permalink]

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22 Feb 2015, 06:42

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If x, y and z are positive integers, is x - y odd? [#permalink]

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08 Mar 2015, 11:51

Another approach might be the following which doesn't involve calculating:

(1) x = z^2 Not suff. (as explained) (2) y = (z-1)^2 Not suff. (as explained)

(1) & (2) x = z^2 states that x is some number, which for instance could be even. An even number to any power stays even. So if z^2 is even, x is even.

Now (2) states that (z-1)^2. Hence, as z^2 is even, (z-1)^2 has to be odd. Again, an odd number to any power stays odd. Therefore, x has to be even (in the example) and y has to be odd.

If x, y and z are positive integers, is x - y odd? [#permalink]

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05 Dec 2015, 10:21

I think most of the gmat problems could have elegant solutions like the one given below. Though this one is not a very difficult question, i think it has implications. Test writers could make it into a difficult question by changing the answer choices as follows:

1) x=z^p & 2)y=(z-1)^q...where p and q are pos int

or 1) x=z^p & 2)y=(z-3)^q...where p and q are pos int

or 1) x=z^p & 2)y=(z-m)^q...where m=odd integer

What do math experts think about it?

gmat1220 wrote:

Rephrasing the question - Is x odd and y even or x even and y odd? Or Is x and y squares of consecutive integers respectively?

1. Insufficient. y is unknown 2. Insufficient. x is unknown

combine 1) + 2) z and z-1 are consecutive integers. Hence sufficient.

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x, y and z are positive integers, is x - y odd?

(1) x=z^2 (2) y=(z-1)^2

There are 3 variables (x,y,z) but only 2 equations are given by the 2 conditions, so there is high chance (E) will become the answer. Looking at the conditions together, x=z^2, y=(z-1)^2, and as x=odd and y=even or x=even and y=odd, the answer to the question always becomes 'yes' and the answer seems like (C) However, this is a question with commonly made mistakes, so just to make sure, if the conditions are examined separately, we cannot know the value of y from condition 1, and condition 2 is not sufficient as well, so the answer becomes (C).

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.
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