If p, q and r are integers, and pq + r is an odd integer, is p an even

Question

If p, q and r are integers, and pq + r is an odd integer, is p an even integer?

(1) pq + pr is an even integer.
(2) p + qr is an odd integer.

chetan2u · Accepted Answer

Hi..
pq is common in statement I and the main statement..
So try and remove it to get a fresh equation..
pq+pr is EVEN and pq+r is odd..
Subtract them so pq+pr-(pq+r)=EVEN-ODD=ODD..
So pq+pr-pq-r=pr-r=(p-1)r is ODD
So both p-1 and r are ODD..
If p-1 is odd, p will be even 
Suff

niks18 · Answer

Given  pq+r=Odd

Statement 1 :  pq+pr=Even , add  r  to both sides

pq+r+pr=Even+r => Odd+pr=Even+r

or  pr-r=Even-Odd => r(p-1)=Odd

1  is odd so  p  has to be even because if  p  is odd then  p-1  is even and  r(p-1)  will be even which is not possible.  Sufficient

Statement 2 :  p+qr=Odd

so if  p=even  &  r=Odd  or  p=odd ,  q=odd  and  r=even . So no unique solution.  Insufficient

Option  A

-------------------------------------------

Another method. Plug in some values and test

itisSheldon · Answer

chetan2u  
Even Though i got the answer right, I took more than 2 minutes here. Is there any way to solve this question within a minute or two. I arrived at the correct answer by assuming 2 cases for each statement. Like
pq+r = odd, So  case 1  pq=odd; r= even
 Case 2  pq=even; r=odd

EgmatQuantExpert · Answer

Solution

We are given: 
 •  p ,  q , and  r  are  3  integers.
•  (p*q + r)  is an odd integer.

(p*q + r)  can be odd for different cases, let us see:

https://e-gmat.com/blogs/wp-content/uploads/2018/03/10.png

Statement-1  “  (p * q + p * r)  is an even integer .”

• We can conclude  p * (q + r)  is an even integer. 
•  p * (q + r)  is an even integer for different cases among the above table.
 1.  p = Even,  q = Even,  r = Odd
2.  p = Even,  q = Odd,  r = Odd

In both the cases,  p  is an even integer. 
Hence,  Statement 1 alone is sufficient to answer the question.

Statement-2  “  (p + q * r)  is an odd integer ”.

•  p + (q * r)  is an odd integer for different cases among the above table.
 1.  p  = Odd,  q = Even,  r =Odd
2.  p  = Odd,  q = odd,  r =Even
3.  p  = Even,  q = Odd,  r =Odd

p  can be even or odd.  
Hence,  Statement 2 alone is not sufficient to answer the question.

Answer: A

sambitspm · Answer

It honestly took me 2-3 attempts to get to the right answer.  We need to find the cases for p,q,r for pq+r as odd. And that's it. That was the trick .

deepalidmoon · Answer

Hi Payal Mam,

Why can't odd(odd+odd) which is even be considered in this case? P then becomes odd as well.

I am not able to understand this why p is not considered odd in this problem

Anshu0012 · Answer

I have a doubt in Statement 1,Please clarify
If p(q+r)
If p=odd(3)
q+r=even(5+5)or (6+6) 
then p(q+r) is also even.
so how could we so sure that P is even.

Need Experts advice

Thanks

chetan2u · Answer

Then it will not satisfy pq+r to be an odd integer, a condition given in main statement. 
3*5+5=20=even
3*6+6=24=even

bumpbot · Answer

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If p, q and r are integers, and pq + r is an odd integer, is p an even

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If p, q and r are integers, and pq + r is an odd integer, is p an even

Prep Toolkit

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My Rewards