If x, y, and z are positive integers, is xy odd ?
(1) x = z^2
(2) y = (z  1)^2
Source: OG12
OA is C.
If x, y, and z are positive integers..............
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Whether (x  y) is odd or even, depends upon both x and y.chendawg wrote:If x, y, and z are positive integers, is x  y odd ?
(1) x = zÂ²
(2) y = (z  1)Â²
Statement 1: Not sufficient as no information about y.
Statement 2: Not sufficient as no information about x.
1 & 2 Together: (x  y) = zÂ²  (z  1)Â² = zÂ²  (zÂ² 2z + 1) = (2z  1) = Even  1 = odd
Sufficient
The correct answer is C.
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Statement 1: x = z^2chendawg wrote:If x, y, and z are positive integers, is xy odd ?
(1) x = z^2
(2) y = (z  1)^2
Source: OG12
OA after some discussion.
No information about y.
Insufficient.
Statement 2: y = (z1)^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 = (21)^2 = 1.
xy = 41 = 3.
If z = 3:
x = 3^2 = 9.
y = (31)^2 = 4.
xy = 94 = 5.
Since xy is odd in both cases, sufficient.
The correct answer is C.
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Additionally, you could also notice that z1 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 (z1)^2 (and thus x and y) are some combination of even and odd.
As a result, the difference xy must be odd.
Squaring a number doesn't change its parity (i.e. even/odd), so z^2 and (z1)^2 (and thus x and y) are some combination of even and odd.
As a result, the difference xy must be odd.
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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=(z1)^q...where p and q are pos int
or
1) x=z^p & 2)y=(z3)^q...where p and q are pos int
or
1) x=z^p & 2)y=(zm)^q...where m=odd integer
1) x=z^p & 2)y=(z1)^q...where p and q are pos int
or
1) x=z^p & 2)y=(z3)^q...where p and q are pos int
or
1) x=z^p & 2)y=(zm)^q...where m=odd integer
[email protected] wrote:Additionally, you could also notice that z1 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 (z1)^2 (and thus x and y) are some combination of even and odd.
As a result, the difference xy must be odd.
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Here's an algebraic approach:If x, y, and z are positive integers, is xy odd?
1) x = zÂ²
2) y = (z1)Â²
Target question: Is xy odd?
Given: x, y, and z are positive integers
Statement 1: x = zÂ²
There's no information about y, so there's no way to determine whether or not xy is odd.
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: y = (z1)Â²
There's no information about x, so there's no way to determine whether or not xy is odd.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
Statements 1 and 2 combined
Statement 1: x = zÂ²
Statement 2: y = (z1)Â²
Subtract equations to get: xy = zÂ²  (z1)Â²
Expand to get: xy = zÂ²  [zÂ²  2z + 1]
Simplify to get: xy = 2z  1
Since z is a positive integer, we know that 2z is EVEN, which means 2z1 is ODD.
If 2z1 is ODD, we can conclude that xy is definitely ODD
Since we can answer the target question with certainty, the combined statements are SUFFICIENT
Answer = C
Cheers,
Brent