GMATPrep DS: Even and Odd

[Champion]

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

OA: A (highlight text to see OA)

Whats the quickest way to solve this?

-Champion-

[RonPurewal]

NOTE
the word "parity", as used in the following discussion, means "whether something is even or odd", in the same way in which "sign" means whether a number is positive or negative.

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

OA: A (highlight text to see OA)

Whats the quickest way to solve this?

-Champion-

the best way to approach things like this is to break down the compound statements into statements about the parity of the individual variables.
to do that, you'll almost certainly have to split the statements into cases.

"xy + z is odd"
two numbers can add to give an odd sum only if they have opposite parity. hence:
case 1: xy is odd, z is even
there's only one way this can happen:
x = odd, y = odd, z = even. (1)
case 2: xy is even, z is odd
there are 3 ways in which this can happen:
x = even, y = even, z = odd (2a)
x = odd, y = even, z = odd (2b)
x = even, y = odd, z = odd (2c)

this is a bit awkward, but, once you've divided the question prompt up into cases, all you have to do is look at your results, check the cases, and you'll have an answer.

--

statement (1)
the easiest way to handle expressions like this is to factor out common terms. you can handle the statement without doing so, but it's more work that way.
pull out x:
x(y + z) is even.
this means that at least one of x and (y + z) is even.
* if x is even, regardless of the parity of (y + z), then the answer to the prompt question is "yes" and we're done.
* the other possibility would be x = odd and (y + z) = even. this is impossible, though, as it doesn't satisfy any of the cases above.
therefore, the answer must be "yes".
sufficient.

--

statement (2)
this means that y and xz have opposite parity.
* y = even, xz = odd --> this means x = odd, y = even, z = odd. that's case (2b), which gives "no" to the question.
at this point you're done, because STATEMENTS CAN'T CONTRADICT EACH OTHER, se you know that "yes" MUST be a possibility with this statement (as statement #1 gives exclusively "yes" answers).
if you use this statement first, you'll have to keep going through the cases.
insufficient.

ans = a

[Guest]

NOTE
the word "parity", as used in the following discussion, means "whether something is even or odd", in the same way in which "sign" means whether a number is positive or negative.

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

OA: A (highlight text to see OA)

Whats the quickest way to solve this?

-Champion-

the best way to approach things like this is to break down the compound statements into statements about the parity of the individual variables.
to do that, you'll almost certainly have to split the statements into cases.

"xy + z is odd"
two numbers can add to give an odd sum only if they have opposite parity. hence:
case 1: xy is odd, z is even
there's only one way this can happen:
x = odd, y = odd, z = even. (1)
case 2: xy is even, z is odd
there are 3 ways in which this can happen:
x = even, y = even, z = odd (2a)
x = odd, y = even, z = odd (2b)
x = even, y = odd, z = odd (2c)

this is a bit awkward, but, once you've divided the question prompt up into cases, all you have to do is look at your results, check the cases, and you'll have an answer.

--

statement (1)
the easiest way to handle expressions like this is to factor out common terms. you can handle the statement without doing so, but it's more work that way.
pull out x:
x(y + z) is even.
this means that at least one of x and (y + z) is even.
* if x is even, regardless of the parity of (y + z), then the answer to the prompt question is "yes" and we're done.
* the other possibility would be x = odd and (y + z) = even. this is impossible, though, as it doesn't satisfy any of the cases above.
therefore, the answer must be "yes".
sufficient.

--

statement (2)
this means that y and xz have opposite parity.
* y = even, xz = odd --> this means x = odd, y = even, z = odd. that's case (2b), which gives "no" to the question.
at this point you're done, because STATEMENTS CAN'T CONTRADICT EACH OTHER, se you know that "yes" MUST be a possibility with this statement (as statement #1 gives exclusively "yes" answers).
if you use this statement first, you'll have to keep going through the cases.
insufficient.

ans = a

wow. thats a realy quick way.

[JonathanSchneider]

: )

[jitenderjain065]

NOTE
the word "parity", as used in the following discussion, means "whether something is even or odd", in the same way in which "sign" means whether a number is positive or negative.

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

OA: A (highlight text to see OA)

Whats the quickest way to solve this?

-Champion-

the best way to approach things like this is to break down the compound statements into statements about the parity of the individual variables.
to do that, you'll almost certainly have to split the statements into cases.

"xy + z is odd"
two numbers can add to give an odd sum only if they have opposite parity. hence:
case 1: xy is odd, z is even
there's only one way this can happen:
x = odd, y = odd, z = even. (1)
case 2: xy is even, z is odd
there are 3 ways in which this can happen:
x = even, y = even, z = odd (2a)
x = odd, y = even, z = odd (2b)
x = even, y = odd, z = odd (2c)

this is a bit awkward, but, once you've divided the question prompt up into cases, all you have to do is look at your results, check the cases, and you'll have an answer.

--

statement (1)
the easiest way to handle expressions like this is to factor out common terms. you can handle the statement without doing so, but it's more work that way.
pull out x:
x(y + z) is even.
this means that at least one of x and (y + z) is even.
* if x is even, regardless of the parity of (y + z), then the answer to the prompt question is "yes" and we're done.
* the other possibility would be x = odd and (y + z) = even. this is impossible, though, as it doesn't satisfy any of the cases above.
therefore, the answer must be "yes".
sufficient.

--

statement (2)
this means that y and xz have opposite parity.
* y = even, xz = odd --> this means x = odd, y = even, z = odd. that's case (2b), which gives "no" to the question.
at this point you're done, because STATEMENTS CAN'T CONTRADICT EACH OTHER, se you know that "yes" MUST be a possibility with this statement (as statement #1 gives exclusively "yes" answers).
if you use this statement first, you'll have to keep going through the cases.
insufficient.

ans = a

hey Ron
does not we have to check for y=odd, xz=even in 2) statement
i think, i am not able to understand the "STATEMENTS CAN'T CONTRADICT EACH OTHER"

[RonPurewal]

hey Ron
does not we have to check for y=odd, xz=even in 2) statement
i think, i am not able to understand the "STATEMENTS CAN'T CONTRADICT EACH OTHER"

the two statements can't have interpretations that contradict each other. i.e., statement (1) and statement (2) must have at least some overlap.

for instance:
(1) x is positive
(2) x is negative
this is impossible, since x can't be both positive and negative. these are contradictory statements.

(1) x is nonnegative
(2) x is not positive
this is ok, since x = 0 satisfies (1) and (2) together. therefore, the statements are not contradictory.

--

in this case:

statement (1) gives ALL "YES" answers.
therefore, it's IMPOSSIBLE for statement (2) to give all "NO" answers, since that would mean that the statements are contradictory.
so, if you get a "no", then you can instantly conclude that the overall answer is "maybe" and therefore insufficient.

by the way, if this sort of reasoning doesn't make immediate sense to you, then don't use it on the test. you'll never, ever need to use this sort of reasoning, although it can save you time here and there if you understand it (since it can lower the # of cases you have to test).

--

and yes, you could just go through the cases for statement (2). there aren't that many of them.

[imanemekouar]

Can you please help.
I did like Ron: set up all the possibilities made distinction between odd and even numbers but after that I was staring at them .How did he elimimate the answer choice ,how did he compare them with possibilities he set up.

[RonPurewal]

Can you please help.
I did like Ron: set up all the possibilities made distinction between odd and even numbers but after that I was staring at them .How did he elimimate the answer choice ,how did he compare them with possibilities he set up.

to help you out here, we'll need more details. what did you have set up? how was it organized? what did you try to do with it?

without these specifics, we'll basically just wind up repeating everything we posted above. thanks

[vijaykumar.kondepudi]

Hi,
Can we solve the problem this way?

Problem Stem: XY + Z = ODD ----> EQ 1

Statement 1: XY + XZ = EVEN ----> EQ 2

Subtract EQ 2 - EQ 1

XY cancels in both the Equations and we get:

Z * (X-1) = ODD
This implies, (X-1) is ODD and therefore, X= EVEN (Sufficient)

Statement 2: Y + XZ = ODD --> EQ 3
ADD EQ 3 and EQ 1
(X+1) * (Y+Z) = EVEN
This doesn't confirm the parity of X. Therefore insufficinet.

Ans: A

[RonPurewal]

Hi,
Can we solve the problem this way?

Problem Stem: XY + Z = ODD ----> EQ 1

Statement 1: XY + XZ = EVEN ----> EQ 2

Subtract EQ 2 - EQ 1

XY cancels in both the Equations and we get:

Z * (X-1) = ODD
This implies, (X-1) is ODD and therefore, X= EVEN (Sufficient)

yeah, that works. nicely done.

Statement 2: Y + XZ = ODD --> EQ 3
ADD EQ 3 and EQ 1
(X+1) * (Y+Z) = EVEN
This doesn't confirm the parity of X. Therefore insufficinet.

Ans: A

this, on the other hand, does not work, although it gives the correct answer in this instance by what is more or less pure luck.
the problem is this: after you add together your equations #1 and #3, you are completely neglecting those individual equations, and looking ONLY at the combined equation.
if you combine two or more equations/inequalities in a data sufficiency problem, you must still take into account the equations/inequalities that you originally combined!

here's a stupid example, but one that should serve to get the point across:
If x > 2, is x = 3.5?
(1) x > 4

if we took your approach above, we would add together x > 2 and x > 4 to produce 2x > 6, which simplifies to x > 3.
if we reason as you did above -- only using this new inequality, and neglecting the original two inequalities -- then we would declare that this statement is insufficient, since knowing that x > 3 doesn't determine whether or not x = 3.5.
however, that's the wrong answer -- one of the INDIVIDUAL statements, x > 4, is enough by itself to draw the conclusion that x is not 3.5.

so, in other words, if you combine two equations/inequalities in a DS problem, then you must take into account
* the newly combined equation/inequality
BUT ALSO
* the original equations/inequalities, in the state in which they existed before you combined them.

[vijaykumar.kondepudi]

Hi Ron,
I get what you wanted to say in the above post, But how do I use this info to conclude that statement 2 was not sufficient.

Basically, how Do I apply:

* the original equations/inequalities, in the state in which they existed before you combined them.

to the 2nd statement of the problem ?

Thanks a lot !!

[mschwrtz]

One point Ron made here,

"statement (2)
this means that y and xz have opposite parity.
* y = even, xz = odd --> this means x = odd, y = even, z = odd. that's case (2b), which gives "no" to the question.
at this point you're done, because STATEMENTS CAN'T CONTRADICT EACH OTHER, se you know that "yes" MUST be a possibility with this statement (as statement #1 gives exclusively "yes" answers).
if you use this statement first, you'll have to keep going through the cases.
insufficient."

is that S2 must allow "yes," or it would contradict S1. He didn't combine the statements in the way that we do when choosing between C and E.

However, let's consider how we might determine that S2 allows "yes" if we don't have something like S1 to which to refer.

y+xz is an odd integer

y=odd, xz=even

this allows that x=even.

[me.parashar]

Hi,
Can we solve the problem this way?

Problem Stem: XY + Z = ODD ----> EQ 1

Statement 1: XY + XZ = EVEN ----> EQ 2

Subtract EQ 2 - EQ 1

XY cancels in both the Equations and we get:

Z * (X-1) = ODD
This implies, (X-1) is ODD and therefore, X= EVEN (Sufficient)

yeah, that works. nicely done.

Statement 2: Y + XZ = ODD --> EQ 3
ADD EQ 3 and EQ 1
(X+1) * (Y+Z) = EVEN
This doesn't confirm the parity of X. Therefore insufficinet.

Ans: A

this, on the other hand, does not work, although it gives the correct answer in this instance by what is more or less pure luck.
the problem is this: after you add together your equations #1 and #3, you are completely neglecting those individual equations, and looking ONLY at the combined equation.
if you combine two or more equations/inequalities in a data sufficiency problem, you must still take into account the equations/inequalities that you originally combined!

here's a stupid example, but one that should serve to get the point across:
If x > 2, is x = 3.5?
(1) x > 4

if we took your approach above, we would add together x > 2 and x > 4 to produce 2x > 6, which simplifies to x > 3.
if we reason as you did above -- only using this new inequality, and neglecting the original two inequalities -- then we would declare that this statement is insufficient, since knowing that x > 3 doesn't determine whether or not x = 3.5.
however, that's the wrong answer -- one of the INDIVIDUAL statements, x > 4, is enough by itself to draw the conclusion that x is not 3.5.

so, in other words, if you combine two equations/inequalities in a DS problem, then you must take into account
* the newly combined equation/inequality
BUT ALSO
* the original equations/inequalities, in the state in which they existed before you combined them.

Hi Ron,
I do understand the inequality concept that you explained above but I'm having hard time finding an example of the same concept with an EQUALITY. Can you please provide one such example. I believe that the thing that you explained above should apply on inequalities but not on equations (EQUALITIES).

[RonPurewal]

Hi Ron,
I do understand the inequality concept that you explained above but I'm having hard time finding an example of the same concept with an EQUALITY. Can you please provide one such example. I believe that the thing that you explained above should apply on inequalities but not on equations (EQUALITIES).

hi,
i don't recall any official problems with that feature, but there are certainly problems in which combining 2 equations leads to loss of information.
here's one example. granted, it's a somewhat stupid/trivial example, but it's an example nonetheless:
Is x > 4?
(1) x = 5
(2) y = 2
ok, so the answer is clearly (a). but, if you add the equations together, you get x + y = 7, which (by itself) no longer guarantees that x > 4.
this isn't a perfect example (because you wouldn't ever combine these statements in the first place, since the first one is already sufficient), but you get the idea.

this is all somewhat inconsequential, anyway. the more general lesson here is simply that you should always notice all the available information in these problems. that includes any equations or inequalities that hold in a particular instance.