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Manager  Joined: 10 Jul 2009
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If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer
(2) y + xz is an odd integer

Originally posted by Aleehsgonji on 28 Jul 2009, 22:25.
Last edited by Bunuel on 17 Jun 2019, 08:35, edited 2 times in total.
Edited the question and added the OA.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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Mountain14 wrote:
jlgdr wrote:
Aleehsgonji wrote:
If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer
(2) y + xz is an odd integer

Odd/Even questions can be usually solved quite easily if one tries some operations with the statements

We want to know if x is even integer

We are given that xy+z is odd

Statement 1

xq + xz is even

Subtracting

z(x+1) is odd

Therefore, x+1 should be odd and x should be even

Sufficient

Statement 2

Not sufficient

Just my 2c

Cheers
J

I am not clear with the red part.

When you subtract $$xy + z=odd$$ from $$xy+xz=even$$ you'll get: $$xz-z=even-odd=odd$$ --> $$z(x-1)=odd$$. For the product of two integers to be odd, both of them must be odd --> $$z$$ and $$x-1$$ are odd. If $$x-1=odd$$, then x must be even: $$x-1=x-odd=odd$$ --> $$x=odd+odd=even$$.

Hope it's clear.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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IMO A

1. xy + xz is an even integer - SUFFICIENT
Given:
xy + z is odd ...(i)
xy + xz is even ...(ii)

subtracting (ii) from (i)
we get xz - z, which should be odd (* since odd - even = odd)
=> z(x-1) is odd
=> both z and (x-1) is odd
=> since (x-1) is odd, x must be even.

2. y + xz is an odd integer -INSUFFICIENT
Given:
xy + z is odd ...(i)
y + xz is odd ...(ii)

subtracting (ii) from (i)
we get xy + z - y - xz
= (x-1)(y-z) , which should be even
=> either (x-1) is even or (y-z) is even ....insufficient to determine
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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You can look at Statement 1 conceptually: when we add z to xy, we get something odd. However, when we add xz to xy, we get something even. So certainly one of z or xz is odd, the other even. Now if xz is different from z, then multiplying by x must have changed z, and that could only happen if x is even and z odd. That's a bit tricky to explain, but I hope that's clear.

Or you can proceed algebraically - notice the similarity between the expression in the question and in Statement 1. We know that xy + xz is even, and xy + z is odd. When you subtract this second expression from the first, you're subtracting an odd from an even, so must get an odd: xy + xz - (xy + z) = xz - z = z(x-1) is odd. Since this is a product, z must be odd, and x-1 must be odd, so x is even. Sufficient.

For Statement 2, all the letters could be odd, so not sufficient.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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IanStewart wrote:
You can look at Statement 1 conceptually: when we add z to xy, we get something odd. However, when we add xz to xy, we get something even. So certainly one of z or xz is odd, the other even. Now if xz is different from z, then multiplying by x must have changed z, and that could only happen if x is even and z odd. That's a bit tricky to explain, but I hope that's clear.

Or you can proceed algebraically - notice the similarity between the expression in the question and in Statement 1. We know that xy + xz is even, and xy + z is odd. When you subtract this second expression from the first, you're subtracting an odd from an even, so must get an odd: xy + xz - (xy + z) = xz - z = z(x-1) is odd. Since this is a product, z must be odd, and x-1 must be odd, so x is even. Sufficient.

For Statement 2, all the letters could be odd, so not sufficient.

Fr St2,
y+xz odd
xy+z odd
=> y+zx+xy+z even
=> y(x+1)+z(x+1) even
=> (y+z)(x+1) event
x+1 can be odd or even means that x can be even or odd, insuff
Manager  Joined: 21 May 2009
Posts: 77
Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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1
Took me less than a min

if XY+Z is odd
these are the possiblities

X Y Z

E E O
O E O
E O O
O O E

so substituting these values in Option A gives the solution write away as Even.
in B we have 2 diff results so

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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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Aleehsgonji wrote:
If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer
(2) y + xz is an odd integer

Odd/Even questions can be usually solved quite easily if one tries some operations with the statements

We want to know if x is even integer

We are given that xy+z is odd

Statement 1

xq + xz is even

Subtracting

z(x+1) is odd

Therefore, x+1 should be odd and x should be even

Sufficient

Statement 2

Not sufficient

Just my 2c

Cheers
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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1
Odd(O) Even (E)
given:
x,y,z integers
xy+z=O
so only the following scenarios can fulfill the constraints
a) EO+O
b) EE+O
c) OE+O
d) OO+E

question:
x=E?

1) x(y+z)=E
i. (E)(O+O) --> fits scenario a -->yes, x can be even
ii. (O)(E+E) --> n/a - doesn't fit any scenarios
iii. (O)(O+O) --> n/a - doesn't fit any scenarios

stop testing, x can't be odd, sufficient

2) y+xz = O
i. E+(O)(O) --> fits scenario a -->yes, x can be even
ii. O+(E)(E) --> n/a - doesn't fit any scenarios
iii. O+(O)(E) --> fits scenario d -->no, x can be odd

stop testing, x can be either even or odd

insufficient

A
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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1
rahsin wrote:
If x, y, z are integers and xy+z is an odd integer, is x an even integer.

1. xy +xz is an even integer

2. y+ xz is an odd integer

Statement 1

xy+xz is even
xy+z is odd

Subtract the two. xy+xz-xy-z = xz-z

xz-z will be odd (even - odd will always result in odd)

z(x-1) will be odd

This is only possible when both z and (x-1) are odd.

x-1 is odd. This means that x is even.

Sufficient.

Statement 2

y+xz is odd
xy+z is odd

Add the two. y+xz+xy+z = x(y+z)+1 (y+z) = (x+1) (y+z).

(x+1)(y+z) will be even (odd+odd is even)

Now here x+1 can be even and y+z can be odd. (Even * odd will result in an even result)
or x+1 can be odd and y+z can be even (Even * odd will result in an even result)
or both (x+1) and (y+z) can be even (Even * even will result in an even result)

So x+1 can be both odd and even. In other words, x can be both even and odd.

This is only possible when both z and (x-1) are odd.

x-1 is odd. This means that x is even.

NOT Sufficient.

A is the answer.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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(1) Since xy + z is odd and xy +xz is even, subtracting we can deduce that z-xz = z(1-x) is odd. Thus 1 - x is odd, so x must be even. SUFF

(2) SUbtracting, we see that xy - y + z - xz = y(x - 1) - z(x - 1) =(y - z)(x - 1) is even. If y - z is odd, x - 1 is even and thus x is odd. On the other hand, if y - z is even, x is even. NOT SUFF

Hence A.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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Aleehsgonji wrote:
If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer
(2) y + xz is an odd integer

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables (x and y) and 1 equations, C is most likely to be the answer and so we should consider 1) & 2) first.

Conditions 1) & 2)
xy + xz : even ⇒ xy:even, xz=even | xy:odd, xz:odd
y + xz : odd ⇒ y: odd, xz=even | y: even, xz: odd
xy + z: odd ⇒ xy: even, z:odd | xy:odd, z:even
⇒ x:even, y:odd, z:odd
Both conditions together are sufficient.

Since this is an integer question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1)
xy + xz : even ⇒ xy:even, xz=even | xy:odd, xz:odd
xy + z: odd ⇒ xy: even, z:odd | xy:odd, z:even
⇒ x:even, y:?, z:odd
The condition 1) is sufficient since we can determine x is even.

Condition 2)
y + xz : odd ⇒ y: odd, xz=even | y: even, xz: odd
xy + z: odd ⇒ xy:odd, z:even | xy: even, z:odd
x = 2, y = 1, z = 1 | x = 1, y = 2, z = 1
The condition 2) is not sufficient.

Therefore, A is the answer.

Normally, in problems which require 2 equations, such as those in which the original conditions include 2 variables, or 3 variables and 1 equation, or 4 variables and 2 equations, each of conditions 1) and 2) provide an additional equation. In these problems, the two key possibilities are that C is the answer (with probability 70%), and E is the answer (with probability 25%). Thus, there is only a 5% chance that A, B or D is the answer. This occurs in common mistake types 3 and 4. Since C (both conditions together are sufficient) is the most likely answer, we save time by first checking whether conditions 1) and 2) are sufficient, when taken together. Obviously, there may be cases in which the answer is A, B, D or E, but if conditions 1) and 2) are NOT sufficient when taken together, the answer must be E.
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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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Aleehsgonji wrote:
If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer
(2) y + xz is an odd integer

Please find solution as attached.

Attachments Screen Shot 2018-07-13 at 9.12.45 AM.png [ 360.57 KiB | Viewed 23588 times ]

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Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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Aleehsgonji wrote:
If x, y and z are integers and xy + z is an odd integer, is x an even integer?

(1) xy + xz is an even integer
(2) y + xz is an odd integer

Statement 1:
Test whether it's possible that x is odd.
Let x=1.
Substituting x=1 into xy+z = ODD, we get:
y+z = ODD
Substituting x=1 into xy+xz = EVEN, we get:
y+z = EVEN
The equations in red contradict each other, implying that it is not possible that x is odd.
Thus, x must be even.
SUFFICIENT.

Statement 2:
Test whether it's possible that x is odd.
Let x=1.
Substituting x=1 into xy+z = ODD, we get:
y+z = ODD
Substituting x=1 into y+xz = ODD, we get:
y+z = ODD
Since the equations in green are the same, it's possible that x is odd.

Test whether it's possible that x is even.
Let x=2.
Substituting x=2 into xy+z = ODD, we get:
2y+z = ODD
Substituting x=2 into y+xz = ODD, we get:
y+2z = ODD
The equations in green are both viable if y=1 and z=1, implying that it's possible that x is even.

Since the answer to the question stem is NO in the first case but YES in the second case, INSUFFICIENT.

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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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Hi All,

We're told that X, Y and Z are INTEGERS and (X)(Y) + Z is an ODD integer. We're asked if X is an EVEN integer. This is a YES/NO question and can be solved by either TESTing VALUES or using Number Properties. While it certainly appears more complex than a typical DS prompt, the basic Number Property rules involved are just about multiplication and addition, so you might find it easiest to think in terms of those patterns (although this question is step-heavy and you do have to be thorough with your work and not miss any of the options).

To start, since (X)(Y) + Z is an ODD, we can 'map out' the limited number of possibilities for the 3 variables. We could have...

(X)(Y) = Even, Z = Odd
(X)(Y) = Odd, Z = Even

From here, we then need to sub-divide (X)(Y):
-IF (X)(Y) is Even, then either both are Even OR one is Odd and the other is Even (this includes the option in which one is 0 and the other could be any integer).
-IF (X)(Y) is Odd, then both MUST be Odd

This ultimately means that these are the only 4 options for the three variables:
X = Even, Y = Even, Z = Odd
X = Even, Y = Odd, Z = Odd
X = Odd, Y = Even, Z = Odd
X = Odd, Y = Odd, Z = Even

Having this list should make working through the two Facts a lot easier.

(1) (X)(Y) + (X)(Z) is an EVEN integer

Using the 4 options above, (X)(Y) + (X)(Z) would be....

(E)(E) + (E)(O) = Even... This IS a match
(E)(O) + (E)(O) = Even... This IS a match
(O)(E) + (O)(O) = Odd... This is NOT a match
(O)(O) + (O)(E) = Odd... This is NOT a match

Two of the options fit Fact 1 - and they BOTH require that X be EVEN, so the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

(2) Y + (X)(Z) is an ODD integer

Using the 4 options above, (Y) + (X)(Z) would be....

(E) + (E)(O) = Even... This is NOT a match
(O) + (E)(O) = Odd... This IS a match
(E) + (O)(O) = Odd... This IS a match
(O) + (O)(E) = Odd... This IS a match

In the three options that fit Fact 2, X is sometimes Even (a "Yes" answer) but sometimes Odd (a "No" Answer).
Fact 2 is INSUFFICIENT

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GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 If x, y and z are integers and xy + z is an odd integer, is x an even  [#permalink]

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rahsin wrote:
If x, y, z are integers and xy+z is an odd integer, is x an even integer.

1. xy +xz is an even integer

2. y+ xz is an odd integer

Statement 1

xy+xz is even
xy+z is odd

Subtract the two. xy+xz-xy-z = xz-z

xz-z will be odd (even - odd will always result in odd)

z(x-1) will be odd

This is only possible when both z and (x-1) are odd.

x-1 is odd. This means that x is even.

Sufficient.

Statement 2

y+xz is odd
xy+z is odd

Add the two. y+xz+xy+z = x(y+z)+1 (y+z) = (x+1) (y+z).

(x+1)(y+z) will be even (odd+odd is even)

Now here x+1 can be even and y+z can be odd. (Even * odd will result in an even result)
or x+1 can be odd and y+z can be even (Even * odd will result in an even result)
or both (x+1) and (y+z) can be even (Even * even will result in an even result)

So x+1 can be both odd and even. In other words, x can be both even and odd.

This is only possible when both z and (x-1) are odd.

x-1 is odd. This means that x is even.

NOT Sufficient.

A is the answer.

Both addition or subtraction of equations can work. Order to equations (Eq 2 - Eq 1 or Eq 1 - Eq 2) for subtraction doesn't matter.

For statement 1: Instead of subtracting, if we add then:
xy+z = odd
xy +xz = even

xy + z + xy + xz = odd + even => 2xy + z(1+x) = odd => even + z(1+x) = odd => z(1+x) = odd => 1+x = odd => odd + x = odd => x = even => Sufficient

For statement 1: Instead of adding, if we subtract then:
xy+z = odd
y+ xz = odd

xy + z - y - xz = odd - odd => y(x-1) + z(1-x) = even => y(x-1) -z(x-1) = even => (x-1)(y-z) = even => (x-1) = even or odd => If (x-1) = even, then x = odd. If (x-1) = odd, then x = even. => Not Sufficient
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