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.
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If x, y, and z are positive integers..............
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Anurag@Gurome
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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 = z²
(2) y = (z - 1)²
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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GMATGuruNY
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Statement 1: x = z^2chendawg 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.
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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Rich@VeritasPrep
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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.
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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mensanumber
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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=(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
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
Rich@VeritasPrep 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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Here's an algebraic approach:If x, y, and z are positive integers, is x-y odd?
1) x = z²
2) y = (z-1)²
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
Cheers,
Brent
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