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Intern  Joined: 15 Oct 2009
Posts: 10
If y and z are integers, is y*(z + 1) odd? (1) y is odd (2)  [#permalink]

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3 00:00

Difficulty:   5% (low)

Question Stats: 81% (00:49) correct 19% (01:00) wrong based on 285 sessions

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If y and z are integers, is y*(z + 1) odd?

(1) y is odd
(2) z is even

OA: C

Why can't i plugin either of them as zero ? In that case, OA should be E. Any inputs to correct my thought process would be great ?
Manager  Joined: 09 Jun 2010
Posts: 98
Re: Odd or not  [#permalink]

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The answer should be E

Q: is XY(Z+1) odd??

(1) y is odd => value of XY(Z+1) depends on Y & (Z+1). Y may be even or (Z+1) may be even. Insufficient
(2) z is even => value of XY(Z+1) depends on Y & X either of them can be even. Insufficient

Combine: We don't know whether X is odd or even. If X = odd then product is odd & if X = even then product is even. Insufficient

Math Expert V
Joined: 02 Sep 2009
Posts: 58288
Re: Odd or not  [#permalink]

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2
anticipation wrote:
If y and z are integers, is y*(z + 1) odd?

(1) y is odd
(2) z is even

OA: C

Why can't i plugin either of them as zero ? In that case, OA should be E. Any inputs to correct my thought process would be great ?

Hi, and welcome to Gmat Club. Below is a solution for your problem.

The product of two integers (in our case $$y$$ and $$z+1$$) to be odd both of them must be odd.

Question: is $$y*(z + 1)=odd$$ --> so basically the question asks is $$y=odd$$ and $$z+1=odd$$, or $$z=even$$?

(1) $$y$$ is odd, we don't know whether $$z=even$$. Not sufficient.

(2) $$z$$ is even, we don't know whether $$y=odd$$. Not sufficient.

(1)+(2) $$y=odd$$ and $$z=even$$, so both necessary conditions are satisfied. Sufficient.

As for your question: first of all, $$y$$ can not be zero as $$y=odd$$ and zero is an even integer, next if $$z=0=even$$, then $$z+1=0+1=even+odd=odd$$ and as from (1) $$y=odd$$ we still have the product of two odd numbers which is odd.

For more on this issues please check Number Theory chapter of Math Book (link in my signature).

Hope it helps.
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Manager  Joined: 14 Jun 2010
Posts: 164
Re: Odd or not  [#permalink]

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The answer should be C...
the Q asks if y(z+1) is odd , in other words it asks if Y is odd and Z is even ( Z+1) is odd ONLY when Z is even.
Thus we have C, points 1 and 2 give the info required!
Intern  Joined: 06 Sep 2010
Posts: 10
Re: Odd or not  [#permalink]

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For y*(z + 1) to be Odd..the product of y and (z+1) has to be odd..It could happen in only one way when both the terms are odd(like 3*3=9).
1.) When y is odd - This will only give us one part of the answer since nature of z is unknown.Hence,INSUFFICIENT
2.) When z is even - This is also INSUFFICIENT coz nature of y is unknown.

On combining both cases,we can conclude that that when z is added to 1,it will become an odd number and when multiplied by an odd number y,it will always give an odd value.

Hence,answer should be C
Manager  Joined: 04 Aug 2010
Posts: 79
Re: Odd or not  [#permalink]

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1
If y and z are integers, is y*(z + 1) odd?

(1) y is odd
(2) z is even

Basically there are two conditions where you can answer if a product is odd:
either (a) both terms are odd - THEN product would be odd
or (b) one of the terms are even - THEN product would be even

Evaluate (1) y is odd
- According to condition (a), we don't know if (z + 1) is odd too.
- INSUFFICIENT

Evaluate (2) z is even
- Which means term (z + 1) is odd
- According to condition (a), we don't know if y is odd too.
- INSUFFICIENT

Evaluate (1) + (2)
- (z + 1) is odd AND y is odd
- SUFFICIENT to conclude that the product of the 2 terms is odd

Ans C
Intern  Joined: 15 Oct 2009
Posts: 10
Re: Odd or not  [#permalink]

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anticipation wrote:
If y and z are integers, is y*(z + 1) odd?

(1) y is odd
(2) z is even

OA: C

Why can't i plugin either of them as zero ? In that case, OA should be E. Any inputs to correct my thought process would be great ?

Thanks everyone for the solution. The key here is to know that zero is even.
Pays to know the parity of zero.
Manager  Joined: 17 Nov 2009
Posts: 215
Re: Odd or not  [#permalink]

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anticipation wrote:
If y and z are integers, is y*(z + 1) odd?

(1) y is odd
(2) z is even

OA: C

Why can't i plugin either of them as zero ? In that case, OA should be E. Any inputs to correct my thought process would be great ?

Rephrase as:

Is Y odd and Z even?

Logic: O*O=Odd.
since z+1 should be odd Z should be even.

st 1: answers only 1 part of what is needed the Y part
st 2: answers only 1 part of what is needed the Z part

Together I have all I need on Y and Z
Manager  Joined: 17 Nov 2009
Posts: 215
Re: Odd or not  [#permalink]

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Could this be a 700 level? Does not seem like the regular 700 questions
Manager  Joined: 19 Sep 2010
Posts: 124
Re: Odd or not  [#permalink]

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No..even i don't rate it as a 700 level question . I gave few MGMAT tests . The 700 level there was pretty tough , more
calculation oriented and time consuming.
VP  Status: mission completed!
Joined: 02 Jul 2009
Posts: 1214
GPA: 3.77
Re: Odd or not  [#permalink]

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1
agnok wrote:
Could this be a 700 level? Does not seem like the regular 700 questions

no way. It is max of 600 level.

To solve just plug in even numbers including 0 {0;2} and few odd {-1;1;3} you will see that in any of cases the result will be odd.

No chances for us after Bunuel's excellent explanations.
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Location: United States (CA)
Re: If y and z are integers, is y*(z + 1) odd? (1) y is odd (2)  [#permalink]

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With DS always spend 10 second stop and write down what is being asked.
Here we are asked y*(z + 1) odd?
Only way we can say this is by know whether y and z are odd/even.

1. NS - we don't know about z.
2. NS - we don't know about y.

1&2: S - have all the info to answer.

anticipation wrote:
If y and z are integers, is y*(z + 1) odd?

(1) y is odd
(2) z is even

OA: C

Why can't i plugin either of them as zero ? In that case, OA should be E. Any inputs to correct my thought process would be great ?

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Non-Human User Joined: 09 Sep 2013
Posts: 13087
Re: If y and z are integers, is y*(z + 1) odd? (1) y is odd (2)  [#permalink]

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_________________ Re: If y and z are integers, is y*(z + 1) odd? (1) y is odd (2)   [#permalink] 30 Sep 2018, 09:57
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