If x, y, and z are positive integers..............

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

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

Source: OG12

OA is C.
Last edited by chendawg on Wed Mar 09, 2011 8:35 am, edited 1 time in total.

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by [email protected] » Sun Mar 06, 2011 1:38 pm
chendawg wrote:If x, y, and z are positive integers, is x - y odd ?

(1) x = z²
(2) y = (z - 1)²
Whether (x - y) is odd or even, depends upon both x and y.

Statement 1: Not sufficient as no information about y.

Statement 2: Not sufficient as no information about x.

1 & 2 Together: (x - y) = z² - (z - 1)² = z² - (z² -2z + 1) = (2z - 1) = Even - 1 = odd

Sufficient

The correct answer is C.
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by GMATGuruNY » Sun Mar 06, 2011 3:37 pm
chendawg wrote:If x, y, and z are positive integers, is x-y odd ?

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

Source: OG12

OA after some discussion.
Statement 1: x = z^2
No information about y.
Insufficient.

Statement 2: y = (z-1)^2
No information about x.
Insufficient.

Statements 1 and 2 combined:
This is an even vs. odd DS question. Since both x and y are in terms of z, all we have to do is plug in one even value for z and one odd value for z.

If z = 2:
x = 2^2 = 4.
y = (2-1)^2 = 1.
x-y = 4-1 = 3.

If z = 3:
x = 3^2 = 9.
y = (3-1)^2 = 4.
x-y = 9-4 = 5.

Since x-y is odd in both cases, sufficient.

The correct answer is C.
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by [email protected] » Sun Mar 06, 2011 3:55 pm
Additionally, you could also notice that z-1 and z are consecutive integers. Therefore, it's a given that one is even and one is odd.

Squaring a number doesn't change its parity (i.e. even/odd), so z^2 and (z-1)^2 (and thus x and y) are some combination of even and odd.

As a result, the difference x-y must be odd.
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by mensanumber » Sat Dec 05, 2015 10:34 am
I think most of the gmat problems could have elegant solutions like the one given by Rich 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


[email protected] wrote:Additionally, you could also notice that z-1 and z are consecutive integers. Therefore, it's a given that one is even and one is odd.

Squaring a number doesn't change its parity (i.e. even/odd), so z^2 and (z-1)^2 (and thus x and y) are some combination of even and odd.

As a result, the difference x-y must be odd.

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by [email protected] » Mon Dec 07, 2015 8:49 am
If x, y, and z are positive integers, is x-y odd?

1) x = z²
2) y = (z-1)²
Here's an algebraic approach:

Target question: Is x-y odd?

Given: x, y, and z are positive integers

Statement 1: x = z²
There's no information about y, so there's no way to determine whether or not x-y is odd.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: y = (z-1)²
There's no information about x, so there's no way to determine whether or not x-y is odd.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1: x = z²
Statement 2: y = (z-1)²
Subtract equations to get: x-y = z² - (z-1)²
Expand to get: x-y = z² - [z² - 2z + 1]
Simplify to get: x-y = 2z - 1
Since z is a positive integer, we know that 2z is EVEN, which means 2z-1 is ODD.
If 2z-1 is ODD, we can conclude that x-y is definitely ODD
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer = C

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