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

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

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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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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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

Answer is A

Video solution from Quant Reasoning starts at 0:26:
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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

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

Answer is A

Just my 2c

Cheers
J

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

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.

(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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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.

Please find solution as attached.

Answer: Option A

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.


_________________

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

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Is x an even integer?

(1) \(xy + xz is a even integer\)
\(xy + z = odd\)

\(xy + xz - xy - z = xz - z\)

\(xz - z = odd\)
Therefore \(z(x-1) = odd\)
z and x - 1 must be odd; x must be even. SUFFICIENT.

(2) \(y + xz = odd\)
\(xy + z = odd\)
\(y + xz + xy + z = odd\)
\(y(1 + x) + z(1 + x)\)
\((y + z ) (1 + x) = even\)

We can have one term = even or two terms = even. INSUFFICIENT.
_________________

I have a question regarding one of the Methods used above several times to analyze Statement 2. Hopefully someone could provide me with some feedback.

chetan2u EMPOWERgmatRichC


Is X an Even Integer?

Given: XY + Z = ODD

Statement 2: Y + XZ = ODD


If we look at either the Given Info. or Statement 2 Alone, the following 2 Cases are NOT Possible:

Case 1: All 3 Variables are EVEN: E * E + E = Even, NOT Odd

or

Case 2: All 3 Variables are ODD: O * O + O = Even, NOT Odd


However, some methods above have analyzed Statement 2 by Combining the Given Information with Statement 2's Information:

XY + Z = Odd Integer

+ (Y + XZ = Odd Integer)
____________________

XY + Z + Y + XZ = EVEN Integer


then by Factoring:

XY + Y + XZ + Z = EVEN Integer

Y (X + 1) + Z (X + 1) = Even Integer

(X + 1) (Y + Z) = Even Integer


if we just look at this combined information alone, it seems as if all 3 Variables could be ODD or all 3 Variables could be EVEN

X, Y, Z = ODD
(Odd + 1) (Odd + Odd) = (Even) (Even) = Even Integer

X, Y, Z = EVEN
(Even + 1) (Even + Even) = (Odd) (Even) = Even Integer


My question is the following:

How can we avoid making such invalid inferences after combining the Information? Should we note the available cases (such as EMPOWERgmatRichC did at the outset) or avoid any combination of information that involves Multiplication and/or Division in questions where you have to determine the Even or Odd nature of a Variable?

Hi

When you get an equation, where product of two terms is odd, it limits your possibilities : Only one Odd*Odd.
The moment you convert it into even, the possibilities increase : Even*Even or Even*Odd or Odd*Even. So always try to keep it to product as odd unless the question demands.

Example x*y=odd.....Clearly, if x and y are integers, both are odd.
But I multiply by 2 => 2*x*y=even. Now I cannot even say that xy is odd or even.

Whatever method you follow : Look at the possibilities initially, and do not do something to the equation that increases the possibilities.
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Thank you much for the reply. Will remember it.

I was looking at a couple of the older posts and they seemed to gloss over that fact.



Posted from my mobile device

Hi Fdambro294,

As a general rule, going through a math 'step' is supposed to make the remaining work EASIER to deal with - and the examples that you are referring to do not appear to do that. By combining the information in the way described, you would end up inadvertently creating new possibilities that do NOT fit the original information.

Here's a much simpler example of that concept:
If X = 2....
we could square both sides and end up with X^2 = 4....
However, this creates a new option that does not fit the original 'limitations' (in this "new" equation, X could also equal -2).

By dealing with the information one 'piece' at a time (as I did in my post), you should be able to avoid those types of potential errors in logic.

GMAT assassins aren't born, they're made,
Rich