BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

p is a positive odd integer,

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 200
Joined: Thu Aug 07, 2008 5:14 pm
Thanked: 1 times
by ska7945 » Fri Oct 03, 2008 7:39 pm
If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(1) When p is divided by 8, the remainder is 5.

(2) p is the sum of the squares of two positive integers.
let's beat GMAT.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
Master | Next Rank: 500 Posts
Posts: 133
Joined: Sat Dec 29, 2007 2:43 am
Thanked: 12 times

by raju232007 » Fri Oct 03, 2008 11:22 pm
the first statement tells that when p is divided by 8,the remainder is 5...

so p must be 8n+5 (where n=0,1,2,3,4..etc)
p= 5,21,29,37.....etc
Now when p is divided by 4 a remainder of 1 is obtained...

therefore statement 1 is sufficient..

Statement 2 tells that p is the sum of squares of two positive integers..

p=1^2+2^2=5...5/4==remainder:1
p=3^2+5^2=34...34/4==remainder:2

But remember it is given in the question that p is a positive odd integer so the second case should not be considered....

Statement 2 is also sufficient...

Hence the ans is D...
Let me know if you still have any doubts..
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 135
Joined: Mon Jun 23, 2008 3:56 am
Thanked: 7 times

by sethids » Sat Oct 04, 2008 3:36 pm
raju232007 wrote:the first statement tells that when p is divided by 8,the remainder is 5...

so p must be 8n+5 (where n=0,1,2,3,4..etc)
p= 5,21,29,37.....etc
Now when p is divided by 4 a remainder of 1 is obtained...

therefore statement 1 is sufficient..

Statement 2 tells that p is the sum of squares of two positive integers..

p=1^2+2^2=5...5/4==remainder:1
p=3^2+5^2=34...34/4==remainder:2

But remember it is given in the question that p is a positive odd integer so the second case should not be considered....

Statement 2 is also sufficient...

Hence the ans is D...
Let me know if you still have any doubts..
I agree with your answer but its not that the second case is not to be considered. The second case you have mentioned needs some modification.
The question says that p is a positive odd number. According to statement 2 we cannot rule out p being the sum of even and odd squares.
However each such sum when divided by 4 yields a remainder of 1.
e.g. p could be 100 + 81 = 181, 4 + 9 = 13, 36 + 49 = 85 ...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 133
Joined: Sat Dec 29, 2007 2:43 am
Thanked: 12 times

by raju232007 » Sat Oct 04, 2008 11:17 pm
You have considered the squares of consecutive integers...
9^2+10^2=181............181/4..remainder=1
2^2+3^2=13................13/4..remainder=1
6^2+7^2=85.................85/4..remainder=1

But as i have already pointed out in my previous post the condition in the question is that p should be odd...so you should never consider the the case of sum of two numbers which results in an even number...

In other words...you should never consider the sum of squares of two consecutive odd integers which results in a even number

1^2+3^2=10........10/4..remainder=2
3^2+5^2=34........34/4..remainder=2
5^2+7^2=74........74/4..remainder=2

So all the above cases should be neglected as the condition in the question states that p should be an odd integer...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member
Thanked: 27 times
GMAT Score:660(

by 4meonly » Sun Oct 05, 2008 12:19 am
I agree with D

I had another approach to (1) because i hate to pick numbers

Q: p=4q+R, R=?

(1)
p=8q+5
p>0, so min value of p is 5
8q is divisible by 4 with R=0
5 is divisible by 4 with R=1
so R is always =1
SUFF

(2)
I used mentioned above approach
SUFF

D

What do you think about my approach to (1)?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 133
Joined: Sat Dec 29, 2007 2:43 am
Thanked: 12 times

by raju232007 » Sun Oct 05, 2008 1:56 am
yeah...even the above mentioned method works perfectly fine..and its much quicker than assuming numbers
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 198
Joined: Fri Mar 09, 2007 8:40 pm
Thanked: 7 times

by kris610 » Sat Oct 11, 2008 4:56 pm
4meonly wrote:I agree with D

I had another approach to (1) because i hate to pick numbers

Q: p=4q+R, R=?

(1)
p=8q+5
p>0, so min value of p is 5
8q is divisible by 4 with R=0
5 is divisible by 4 with R=1
so R is always =1
SUFF

(2)
I used mentioned above approach
SUFF

D

I agree with your approach to 1.

But for 2) picking numbers to confirm may be a better idea.

9 + 16 = 25 remainder is 1

9 + 25 = 34 remainder is 2

What do you think about my approach to (1)?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 891
Joined: Sat Aug 16, 2008 4:21 amFollowed by: 1 member
Thanked: 27 times
GMAT Score:660(

by 4meonly » Sat Oct 11, 2008 10:42 pm
I think you arfe asking about your approach to (2) because you agree with those of (1)

p is a positive odd integer (main stem) so this e.g. is correct
9 + 16 = 25 remainder is 1

but this is not
9 + 25 = 34 remainder is 2
because 34 is even. we need odd p

HTH
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Tue Feb 03, 2009 2:01 am
Thanked: 1 times

by fleurdelisse » Mon Apr 20, 2009 9:49 am
How can you more generally prove that (2) is sufficient??
You can't go and sum all possible squares that yield to something odd, what if there is some pair of squares that yield to a remainder of 3?? How do we prove that the remainer can only be 1 and not 3??
Please help.
Thanks.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members
Location: Montreal
Thanked: 1090 times
GMAT Score:780

Re: p is a positive odd integer,

by Ian Stewart » Mon Apr 20, 2009 3:23 pm
ska7945 wrote:If p is a positive odd integer, what is the remainder when p is divided by 4 ?

(2) p is the sum of the squares of two positive integers.
If you want an algebraic proof, using Statement 2 alone:

p = x^2 + y^2, where x and y are integers. Since p is odd, one of the squares must be even, one odd - let's say x is even, y is odd. Then x = 2k for some integer k, and y = 2m + 1, for some integer m. So,

p = x^2 + y^2
p = (2k)^2 + (2m + 1)^2
p = 4k^2 + 4m^2 + 4m + 1
p = 4(k^2 + m^2 + m) + 1

So p is 1 more than a multiple of 4, and the remainder will be 1 when p is divided by 4. Sufficient.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 44
Joined: Tue Feb 03, 2009 2:01 am
Thanked: 1 times

by fleurdelisse » Tue Apr 21, 2009 1:05 am
Great! Thanks a lot!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1