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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Remainder

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 458
Joined: Sun Aug 31, 2008 10:44 amFollowed by: 1 member
Thanked: 3 times
by gmat009 » Thu Oct 09, 2008 1:07 pm
If p and n are positive integers p >n , what is the remainder when p^2-n^2 is divided by 15

The remainder when p+q is divided by 5 is 1
The remainder when p-q is divided by 3 is 1
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
User avatar
Master | Next Rank: 500 Posts
Posts: 246
Joined: Mon May 19, 2008 7:34 am
Location: Texaco Gas Station
Thanked: 7 times

by cubicle_bound_misfit » Thu Oct 09, 2008 2:01 pm
IMO answer is E.

is the q) p^2 - q^2 's remainder when divided by 15?
Cubicle Bound Misfit
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 496
Joined: Sun May 04, 2008 11:01 pm
Location: mumbai
Thanked: 7 times
GMAT Score:640

by stubbornp » Thu Oct 09, 2008 7:37 pm
imo E...

For statement1,The remainder when p+q is divided by 5 is 1p=5 q=1---remainder is 9

p=4 q=2---remainder is 12

p=7 q=4----remainder is 3--------Insufficient

For stmt 2,
The remainder when p-q is divided by 3 is 1

p=5 q=1 -----remainder is 12

p=7 q=1-----remainder is 3--insufficient


By combining both,again we have two different remainders.So insufficient..

....
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 » Thu Oct 09, 2008 8:03 pm
statement 1:

It states that when p+q is divided by 5 the remainder is 1

so lets assume the no to be 5n+1 ( here n=0,1,2,3...etc)

p+q=5n+1

Insufficient

statement 2:

It states that when p-q is divided by 3 the remainder is 1

so lets assume the no to be 3n+1 ( here n=0,1,2,3...etc)

p-q = 3n+1

Insufficient

Now,lets combine both the statements and see whether we can deduce a solution

for n=1
p+q=6
p-q=4
2p=10
p=5 and q=1

(p^2-n^2)/15=24/15..........remainder=9

for n=2
p+q=11
p-q=7
2p=18
p=9 and q=2

(p^2-n^2)/15=77/15.........remainder=2

IT is observed that the remainder varies in both the cases..

Hence E is the ans...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 246
Joined: Mon May 19, 2008 7:34 am
Location: Texaco Gas Station
Thanked: 7 times

by cubicle_bound_misfit » Thu Oct 09, 2008 8:31 pm
well, picking no may not be the best option as in real GMAT you may have to find the .004% of 1/0.125th growth of fat level in Jay Leno's potbelly.
So better to use

p-q = 5x +1

p+q = 3Y+1 etc.
hope that helps.
Cubicle Bound Misfit
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 833
Joined: Mon Aug 04, 2008 1:56 am
Thanked: 13 times

by vivek.kapoor83 » Thu Oct 09, 2008 10:22 pm
imo E
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 29
Joined: Mon Sep 01, 2008 3:24 pm
Thanked: 3 times

Re: Remainder

by gmat2010 » Fri Oct 10, 2008 3:16 pm
gmat009 wrote:If p and n are positive integers p >n , what is the remainder when p^2-n^2 is divided by 15

The remainder when p+q is divided by 5 is 1
The remainder when p-q is divided by 3 is 1

Quick question, wouldn't this be obviously E since both of the statements are not mentioning n at all?
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(

Re: Remainder

by 4meonly » Sat Oct 11, 2008 9:04 am
gmat2010 wrote:
gmat009 wrote:If p and n are positive integers p >n , what is the remainder when p^2-n^2 is divided by 15

The remainder when p+q is divided by 5 is 1
The remainder when p-q is divided by 3 is 1

Quick question, wouldn't this be obviously E since both of the statements are not mentioning n at all?
It hink it is a typo
IMHO it should be

If p and n are positive integers p >n , what is the remainder when p^2-n^2 is divided by 15

The remainder when p+n is divided by 5 is 1
The remainder when p-n is divided by 3 is 1
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 43
Joined: Sat May 31, 2008 11:31 am

by anb » Fri Nov 21, 2008 5:23 pm
I understand how the equations were derived, but how does knowing that p+q= 5z + 1 sufficient to know that statement 1 (or 2) is insufficient?

Don't you still need to pick numbers and contradict the Y-N question?

Thanks!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2467
Joined: Thu Aug 28, 2008 6:14 pmFollowed by: 11 members
Thanked: 331 times

by cramya » Fri Nov 21, 2008 5:52 pm
anb,

Refer here https://www.beatthegmat.com/remainders-t22704.html#95683

The same problem has been discussed and I have picked numbers.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 43
Joined: Sat May 31, 2008 11:31 am

by anb » Fri Nov 21, 2008 7:21 pm
Thanks!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1