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# If x, y and z are positive integers, is xz even?

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If x, y and z are positive integers, is xz even?  [#permalink]

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01 Oct 2018, 03:55
00:00

Difficulty:

55% (hard)

Question Stats:

65% (01:37) correct 35% (02:17) wrong based on 42 sessions

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If x, y and z are positive integers, is xz even?

(1) (2xy − x) is even
(2) (x^2 + xz) is eve

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

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01 Oct 2018, 04:06
Bunuel wrote:
If x, y and z are positive integers, is xz even?

(1) (2xy − x) is even
(2) (x^2 + xz) is eve

Given: If x, y and z are positive integers

Question: is xz even?

Statement 1: (2xy − x) is even
i.e. $$x(2y-1) = Even$$
But (2y-1) = Even - 1 = Odd therefore,
x = Even
i.e. xz = even
SUFFICIENT

Statement 2: $$(x^2 + xz)$$ is even

$$(x^2 + xz) = x(x+z) = even$$
Case 1: x = even and z is odd i.e. xz = even
Case 2: x = even and z is Even
Case 3: x = Odd and z is Odd i.e. xz = Odd
NOT SUFFICIENT

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If x, y and z are positive integers, is xz even?  [#permalink]

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01 Oct 2018, 05:50
Top Contributor
Bunuel wrote:
If x, y and z are positive integers, is xz even?

(1) (2xy − x) is even
(2) (x² + xz) is even

Some important rules:
#1. ODD +/- ODD = EVEN
#2. ODD +/- EVEN = ODD
#3. EVEN +/- EVEN = EVEN

#4. (ODD)(ODD) = ODD
#5. (ODD)(EVEN) = EVEN
#6. (EVEN)(EVEN) = EVEN

Target question: Is xz even?

Given: x, y and z are positive integers

Statement 1: (2xy − x) is even
Notice that 2xy will be EVEN for all values of x and y.
So, statement 1 is really telling us that EVEN - x = EVEN, which we can rewrite as: EVEN + EVEN = x
By rule #3, we can conclude that x is EVEN
If x is EVEN, them we can be certain (from rules #5 and #6) that xz is EVEN
So, the answer to the target question is YES, xz is even
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: (x² + xz) is even
There are several values of x and z that satisfy statement 2. Here are two:
Case a: x = 2 and z = 1. Notice that x² + xz = 2² + (2)(1) = 6, which is even. In this case, xz = (2)(1) = 2. So, the answer to the target question is YES, xz IS even
Case b: x = 1 and z = 1. Notice that x² + xz = 1² + (1)(1) = 2, which is even. In this case, xz = (1)(1) = 1. So, the answer to the target question is NO, xz is NOT even
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

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Brent
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Re: If x, y and z are positive integers, is xz even?  [#permalink]

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01 Oct 2018, 06:14
Bunuel wrote:
If x, y and z are positive integers, is xz even?

(1) (2xy − x) is even
(2) (x^2 + xz) is even

even - even = even

odd + odd = even.

Statement 1:

2xy = even. So x has to be even. Thus xz must be even. Sufficient.

Statement 2:

case 1:

let x be 2 and

$$x^2 = even$$

xz = even.

case 2:

x = 3 and z = 3

$$3^2 + 3*3 = 18$$

but 3*3 = 9 , which is odd.

Thus not sufficient .

Re: If x, y and z are positive integers, is xz even?   [#permalink] 01 Oct 2018, 06:14
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# If x, y and z are positive integers, is xz even?

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