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

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If x, y and z are positive integers, is x - y odd? [#permalink] New post 09 Mar 2011, 10:29
00:00
A
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C
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E

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80% (01:59) correct 19% (01:24) wrong based on 202 sessions
If x, y and z are positive integers, is x - y odd?

(1) x=z^2
(2) y=(z-1)^2
[Reveal] Spoiler: OA

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Re: 172 OG DS [#permalink] New post 09 Mar 2011, 10:40
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Baten80 wrote:
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?


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

(1) x=z^2. No info about y. Not sufficient.

(2) y=(z-1)^2. No info about x. Not sufficient.

(1)+(2) Subtract (2) from (1): x-y=z^2-(z^2-2z+1)=2z-1=even-odd=odd. Sufficient.

Answer: C.
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Re: 172 OG DS [#permalink] New post 09 Mar 2011, 20:59
(1) and (2) are clearly not enough on their own.

So from (1) and (2) together:

z^2 - z^2 + 2z - 1 = 2z-1 which is odd, so answer is C.
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Re: 172 OG DS [#permalink] New post 10 Mar 2011, 06:45
Thank u guys.
These solutions are also OG solution. Right?
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Re: 172 OG DS [#permalink] New post 10 Mar 2011, 10:32
Another way to solve this problem:

(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.
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Re: 172 OG DS [#permalink] New post 10 Mar 2011, 20:34
Expert's post
Baten80 wrote:
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

Answer (C)
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Re: 172 OG DS [#permalink] New post 11 Mar 2011, 17:09
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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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Re: 172 OG DS [#permalink] New post 12 Mar 2011, 17:39
1. Not sufficient.

x = z^2 , no information about y

x-y can be even or odd

lets say x y z are 16 3 4 respectively => x-y = 13 odd
x y z are 16 6 4 respectively => x-y = 12 even

2. Not sufficient

no x , similar to the above example , depending upon x , x-y can be odd or even . Not sufficient.

together

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

=> x-y = 2z-1 .i.e is odd.

Hence answer is C.
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Re: If x, y and z are positive integers, is x - y odd? [#permalink] New post 07 Jan 2014, 04:31
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Re: If x, y and z are positive integers, is x - y odd?   [#permalink] 07 Jan 2014, 04:31
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