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# If r and s are integers and rs + r is odd, which of the following must

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Intern
Joined: 06 Apr 2007
Posts: 9
If r and s are integers and rs + r is odd, which of the following must [#permalink]

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01 Jun 2007, 23:57
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If r and s are integers and rs + r is odd, which of the following must be even ?

A. R
B. s
C. r + s
D. rs - r
E. r^2 + s
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Oct 2014, 14:08, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
VP
Joined: 08 Jun 2005
Posts: 1141
Re: If r and s are integers and rs + r is odd, which of the following must [#permalink]

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02 Jun 2007, 11:55
Himalayan is right !

according to OG , zero is even

R*S+R =

e=even
o=odd

we have two options:

(1) e*o+e = e
(2) o*e+o = o

since the stem states that R*S+R =odd then we will choose (2) and S must be even.
CEO
Joined: 17 May 2007
Posts: 2945
Re: If r and s are integers and rs + r is odd, which of the following must [#permalink]

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02 Jun 2007, 15:43
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RS+R = R(S+1) --> which is ODD
Now this implies that R is odd AND S+1 is odd. which means S is even

Ans B.
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Concentration: Marketing, General Management
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Re: If r and s are integers and rs + r is odd, which of the following must [#permalink]

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04 Oct 2014, 13:02
bsd_lover wrote:
RS+R = R(S+1) --> which is ODD
Now this implies that R is odd AND S+1 is odd. which means S is even

Ans B.

I am not sure if i got this one ..

i mean if r is odd what is the possibility of s being odd or even ??
Math Expert
Joined: 02 Sep 2009
Posts: 43361
If r and s are integers and rs + r is odd, which of the following must [#permalink]

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04 Oct 2014, 14:18
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minakshigurani wrote:
bsd_lover wrote:
RS+R = R(S+1) --> which is ODD
Now this implies that R is odd AND S+1 is odd. which means S is even

Ans B.

I am not sure if i got this one ..

i mean if r is odd what is the possibility of s being odd or even ??

If r and s are integers and rs + r is odd, which of the following must be even ?

A. r
B. s
C. r + s
D. rs - r
E. r^2 + s

Given that rs + r is odd --> $$rs + r =r*(s+1)=odd$$. For the product of two integers, r and s+1, to be odd both must be odd. Theretofore, r and s+1 are odd, which means that r is odd and s is even.

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Intern
Joined: 21 Dec 2014
Posts: 11
Re: If r and s are integers and rs + r is odd, which of the following must [#permalink]

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04 Jun 2015, 05:02
Great explanation, Bunuel.
Love your approach in every answer you put in the threads.
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Re: If r and s are integers and rs + r is odd, which of the following must [#permalink]

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04 Jun 2015, 16:46
Hi All,

This question is perfect for TESTing VALUES.

We're told that R and S are integers and that (R)(S)+R = ODD. We're asked which of the 5 answers MUST be even...

IF....
R = 1
S = 2
(1)(2)+1 = 3

Answer A: R = 1 NOT even
Answer B: S = 2 This IS EVEN
Answer C: R+S = 3 NOT even
Answer D: RS-R = 2-1 = 1 NOT even
Answer E: R^2+S = 1+2 = 3 NOT even

[Reveal] Spoiler:
B

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SVP
Joined: 06 Nov 2014
Posts: 1904
Re: If r and s are integers and rs + r is odd, which of the following must [#permalink]

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10 Jun 2015, 22:41
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If r and s are integers and rs + r is odd, which of the following must be even ?

If r is factored out to rewrite this as r (s + 1), we can see it as a product of two parts, r and s + 1. For their product to be odd, r must be odd and s + 1 must be odd. We also know that even + odd = odd. For s+1 to be odd, s must be even since 1 is odd.
A. R
B. s
C. r + s
D. rs - r
E. r^2 + s
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Posts: 14171
Re: If r and s are integers and rs + r is odd, which of the following must [#permalink]

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29 Jul 2016, 23:56
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Re: If r and s are integers and rs + r is odd, which of the following must [#permalink]

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09 Sep 2017, 06:13
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If r and s are integers and rs + r is odd, which of the following must   [#permalink] 09 Sep 2017, 06:13
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# If r and s are integers and rs + r is odd, which of the following must

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