Last visit was: 27 Apr 2026, 00:32 It is currently 27 Apr 2026, 00:32
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
chetan2u
User avatar
GMAT Expert
Joined: 02 Aug 2009
Last visit: 26 Apr 2026
Posts: 11,229
Own Kudos:
45,025
 [4]
Given Kudos: 335
Status:Math and DI Expert
Location: India
Concentration: Human Resources, General Management
GMAT Focus 1: 735 Q90 V89 DI81
Products:
Expert
Expert reply
GMAT Focus 1: 735 Q90 V89 DI81
Posts: 11,229
Kudos: 45,025
 [4]
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
User avatar
broall
User avatar
Retired Moderator
Joined: 10 Oct 2016
Last visit: 07 Apr 2021
Posts: 1,133
Own Kudos:
Given Kudos: 65
Status:Long way to go!
Location: Viet Nam
Posts: 1,133
Kudos: 7,377
Kudos
Add Kudos
Bookmarks
Bookmark this Post
avatar
vs224
Joined: 25 Jan 2013
Last visit: 05 Feb 2022
Posts: 29
Own Kudos:
Given Kudos: 5,909
Location: United States
Concentration: General Management, Entrepreneurship
Schools: Johnson '21
Schools: Johnson '21
Posts: 29
Kudos: 15
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
asthagupta
Joined: 10 Sep 2015
Last visit: 13 May 2021
Posts: 49
Own Kudos:
Given Kudos: 76
Location: India
Concentration: Finance, Human Resources
GMAT 1: 640 Q47 V31
GMAT 2: 660 Q47 V35
GMAT 3: 700 Q49 V36
GPA: 4
Products:
GMAT 3: 700 Q49 V36
Posts: 49
Kudos: 130
Kudos
Add Kudos
Bookmarks
Bookmark this Post
A??
r+s = 2(s+1)
st1> r is prime, so s is also odd as r cannot be 2(if r=2 then s=0). Other than 2 for every prime R, S will be odd. Odd + 1 is even so S+1 is even.
Even * 2 will be a multiple of 4.
suff
St2> If s is prime, S+1 will be even,
Even * 2 will be a multiple of 4.

But if S=2, then r=4.
in-suff

hence A
Please tell OA, my exam is on 15th of this month
User avatar
niks18
User avatar
Retired Moderator
Joined: 25 Feb 2013
Last visit: 30 Jun 2021
Posts: 862
Own Kudos:
1,806
 [1]
Given Kudos: 54
Location: India
GPA: 3.82
Products:
Posts: 862
Kudos: 1,806
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
chetan2u
If r and s are positive integers and r=s+2, is r+s divisible by 4?
(1) r is a prime number.
(2) s is a prime number.


A new tricky DS Question
\(r+s = r+r-2\) (as \(r=s+2\), so \(s=r-2\))
This implies: \(2r-2=2(r-1)\).
hence \(\frac{(r+s)}{4} = \frac{2(r-1)}{4} = \frac{(r-1)}{2}\)-------------\((1)\)
So for \((r+s)\) to be divisible by \(4\), \(r\) needs to be ODD

Statement 1: this implies that \(r\) is ODD. Hence sufficient
(note: \(r= s+2\) and \(s\) is a positive integer, hence \(s\) cannot be \(0\), so \(r\) cannot be \(2\), the only even prime no)

Statement 2: if \(s =2\), then \(r=4 = Even\), hence equation \((1)\) is not divisible by \(4\)
but if \(s =\) odd prime no for eg. \(3\), then \(r = 5 = Odd\), hence equation \((1)\) is divisible by \(4\). Hence the statement is Insufficient

Option A
avatar
AlexGmat2017
Joined: 13 Jan 2017
Last visit: 20 Apr 2021
Posts: 29
Own Kudos:
Given Kudos: 92
Posts: 29
Kudos: 5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
B.
r + s : 4 is equivalent to (r + s) : 4 =(s+s+2):4=2*(s+2):4=(s+2):2. Hence, s must be an even number for residual to be equal to 0.
User avatar
niks18
User avatar
Retired Moderator
Joined: 25 Feb 2013
Last visit: 30 Jun 2021
Posts: 862
Own Kudos:
Given Kudos: 54
Location: India
GPA: 3.82
Products:
Posts: 862
Kudos: 1,806
Kudos
Add Kudos
Bookmarks
Bookmark this Post
AlexGmat2017
B.
r + s : 4 is equivalent to (r + s) : 4 =(s+s+2):4=2*(s+2):4=(s+2):2. Hence, s must be an even number for residual to be equal to 0.

Hi AlexGmat2017

as per statement 2 "s" is prime no so it can be even for eg. 2 or it can be odd for eg. 3. the highlighted section is not always true
avatar
kaulshanx
Joined: 19 Sep 2016
Last visit: 17 Oct 2018
Posts: 27
Own Kudos:
Given Kudos: 3
Posts: 27
Kudos: 3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
A . R=s+2, statement 1 - r is prime which means s can take values of 1,3,5,9 etc so thay r would be 3,5,7,11 and so on and adding r+s would be 4,8,12,20,24 which are all multiples of 4 and hence is sufficient.
Statement 2- s is prime means s can take values of 2,3,5,7,11 etc and r cam take values of 4,5,7,9,13 and r+s takes values of 6,8,12,16,24 and so on and 6 is not divisible by 4 hence 8,12,16,24 are, hence insufficient. So answer A

Sent from my SM-N920G using GMAT Club Forum mobile app
User avatar
hellosanthosh2k2
Joined: 02 Apr 2014
Last visit: 07 Dec 2020
Posts: 360
Own Kudos:
Given Kudos: 1,227
Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34
GPA: 3.5
Schools: XLRI"20
GMAT 1: 700 Q50 V34
Posts: 360
Kudos: 620
Kudos
Add Kudos
Bookmarks
Bookmark this Post
\(r = s + 2\) => \(r - s =2\)

Statement 1:
\(r\) is prime number, \(r\) can't be 2, if it were, then \(s = 0\) but \(s\) is positive
so \(r\) is odd prime number,
any odd number can be written as \((2p + 1)\) or \((2k + 3)\) or \((2l + 5)\).. where \(p,k,l\)are integers
Let \(r = 2k + 3\), then \(s = r - 2\) => \(2k + 1\)
so \(r + s = (2k + 3 + 2k + 1) = (4k + 4)\) => which is clearly divisible by 4 => suff

Statement 2:
\(s\) is prime number.
if \(s = 2 => r = 4 => (r+s) = 6\) not divisible by 4
if \(s\) is odd prime number, we can use above method in statement 1, to prove that it is divisible by 4
so two diff possiblities => Not Sufficient

Answer (A)
User avatar
gurmukh
Joined: 18 Dec 2017
Last visit: 30 Dec 2025
Posts: 258
Own Kudos:
Given Kudos: 20
Posts: 258
Kudos: 269
Kudos
Add Kudos
Bookmarks
Bookmark this Post
From first statement
R-S =2
Where R is a prime number so R is either 6k +1 or 6k-1 form where k is an integer.
Now, if R is 6k+1 form then S is 6k -1 form, so their difference is 2
Therefore R+ S = 12k
Hence divisible by 4
Now, if R is 6k-1 form S is 6k+1 form, so their difference is 2
Therefore R + S =12k
Hence given statement is sufficient
From second statement
S is a prime number
R =3 +2 =5
R + S = 5+3=8
Divisible by 4
R = 2+2 =4
R +S = 4+2 =6
Not divisible by 4
Hence not sufficient
Therefore answer is option A

Posted from my mobile device
Moderators:
Math Expert
109921 posts
498 posts
212 posts