Find all School-related info fast with the new School-Specific MBA Forum

It is currently 30 Sep 2014, 06:11

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

The integers m and p are such that 2<m<p, and m is not a

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Intern
Intern
avatar
Joined: 12 Sep 2010
Posts: 10
Followers: 0

Kudos [?]: 5 [1] , given: 0

The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 19 Sep 2010, 23:56
1
This post received
KUDOS
3
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

45% (02:38) correct 55% (01:53) wrong based on 348 sessions
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

(1) the greatest common factor of m and p is 2
(2) the least common multiple of m and p is 30

--------------------------------------------------------------
A small doubt! Are the concepts GCD,LCM applicable to negative numbers?If so,please explain and tel me why -ve numbers are not considered in the Fact1 of the following problem?

--------------------------------------------------------------
I understood that Fact2 is not sufficient but am not sure about Fact1. If negative numbers are taken,then,for example, GCD(-8,6) is 2 and gives new remainder.Am I going wrong somewhere?
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Jun 2013, 03:29, edited 2 times in total.
Edited the question and added the OA.
Expert Post
6 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23055
Followers: 3535

Kudos [?]: 27206 [6] , given: 2725

Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 03:07
6
This post received
KUDOS
Expert's post
ravitejapandiri wrote:
A small doubt!Are the concepts GCD,LCM applicable to negative numbers?If so,please explain and tel me why -ve numbers are not considered in the Fact1 of the following problem?

--------------------------------------------------------------
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

(1) the greatest common factor of m and p is 2
(2) the least common multiple of m and p is 30
--------------------------------------------------------------

I understood that Fact2 is not sufficient but am not sure about Fact1.If negative numbers are taken,then,for example, GCD(-8,6) is 2 and gives new remainder.Am I going wrong somewhere?


SOME NOTES:

1. GCD and LCM
The greatest common divisor (GCD), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.

So GCD can only be positive integer. It should be obvious as greatest factor of two integers can not be negative. For example if -3 is a factor of two integer then 3 is also a factor of these two integers.

The lowest common multiple (LCM), of two integers a and b is the smallest positive integer that is a multiple both of a and of b.

So LCM can only be positive integer. It's also quite obvious as if we don not limit LCM to positive integer then LCM won't make sense any more. For example what would be the lowest common multiple of 2 and 3 if LCM could be negative? There is no answer to this question.

2. DIVISIBILITY QUESTIONS ON GMAT

EVERY GMAT divisibility question will tell you in advance that any unknowns represent positive integers.

3. REMAINDER

GMAT Prep definition of the remainder:
If a and d are positive integers, there exists unique integers q and r, such that a = qd + r and 0\leq{r}<d. q is called a quotient and r is called a remainder.

Moreover many GMAT books say factor is a "positive divisor", d>0.

I've never seen GMAT question asking the remainder when dividend (a) is negative, but if we'll cancel this restriction (and consider dividend=a<0), but leave the other restriction (0\leq{r}<d), then division of negative integer by positive integer could be calculated as follow:

-8 divided by 6 will result: 0\leq{r}<d, a=qd + r --> 0\leq{r}<6, -8=(-2)*6+4. Hence remainder=r=4.

TO SUMMARIZE, DON'T WORRY ABOUT NEGATIVE DIVIDENDS, DIVISORS OR REMAINDERS ON GMAT.


BACK TO THE ORIGINAL QUESTION:

The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

Given: 2<m<p and \frac{p}{m}\neq{integer}, which means that r>0. Note here that as 0<2<m<p then your example of -8 and 6 is not valid as both m and p are positive.
Question: r=?

(1) the greatest common factor of m and p is 2 --> both p and m are even (as both have 2 as a factor) --> even divided by even can give only even remainder (0, 2, 4, ...), since remainder is not zero (as \frac{p}{m}\neq{integer}), then remainder must be more than 1: 2, 4, ... Sufficient.

(2) the least common multiple of m and p is 30 --> if m=5 and p=6, remainder=1=1, answer to the question would be NO. BUT if m=10 and p=15 remainder=5>1 answer to the question would be YES. Two different answers. Not sufficient.

Answer: A.

Hope it helps.

Also:
Please post PS questions in the PS subforum: gmat-problem-solving-ps-140/
Please post DS questions in the DS subforum: gmat-data-sufficiency-ds-141/
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

3 KUDOS received
Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 76

Kudos [?]: 487 [3] , given: 25

GMAT ToolKit User Reviews Badge
Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 01:20
3
This post received
KUDOS
ravitejapandiri wrote:
A small doubt!Are the concepts GCD,LCM applicable to negative numbers?If so,please explain and tel me why -ve numbers are not considered in the Fact1 of the following problem?

--------------------------------------------------------------
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

(1) the greatest common factor of m and p is 2
(2) the least common multiple of m and p is 30
--------------------------------------------------------------

I understood that Fact2 is not sufficient but am not sure about Fact1.If negative numbers are taken,then,for example, GCD(-8,6) is 2 and gives new remainder.Am I going wrong somewhere?


(1) This is sufficient. We know gcd(m,p)=2, this means both m & p are even.
Let q be the quotient and r the remainder when p is divided by m
p = q*m + r
Note that r cannot be 0, since if r=0, then m divides p and is therefore a factor of p, which we know is not true
Also note r cannot be 1, since if r=1, them q*m (even number) + 1 (odd) is an odd number, which is not possible since we know that p is even
Hence r>1.

(2) Not sufficient. eg. m=6,p=15,lcm=30, remainder=3; m=5,p=6,lcm=30, remainder=1

Ans is (A)

Note on negative numbers
I am not entirely sure what the doubt above is, but atleast in this question we are given 2<m<p, so neither can be negative.
In general when talking about division using negative numbers, there can be two values of r, and we have to decide on definition before hand, so I dont think this something the GMAT can test on. Eg, consider division of 10 by -7. 10 = 1*(-7) + (-3) & 10=2*(-7) + (4). Both -3 and 4 are well defined remainders as both satisfy the condition 0<=Abs(remainder)<Abs(divisor). I have never seen GMAT questions deal with negative divisions & remainders because of this ambiguity.

_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Get the best GMAT Prep Resources with GMAT Club Premium Membership


Last edited by shrouded1 on 20 Sep 2010, 01:40, edited 1 time in total.
Expert Post
1 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23055
Followers: 3535

Kudos [?]: 27206 [1] , given: 2725

Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 03:17
1
This post received
KUDOS
Expert's post
1 KUDOS received
Manager
Manager
avatar
Joined: 28 Feb 2012
Posts: 115
Concentration: Strategy, International Business
Schools: INSEAD Jan '13
GPA: 3.9
WE: Marketing (Other)
Followers: 0

Kudos [?]: 18 [1] , given: 17

GMAT ToolKit User
Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 06 Jun 2013, 22:17
1
This post received
KUDOS
ravitejapandiri wrote:
The integers m and p are such that 2<m<p, and m is not a factor of p. if r is the remainder when p is divided by m, is r>1?

(1) the greatest common factor of m and p is 2
(2) the least common multiple of m and p is 30




(1) from this statement we know that both m and n are even integers, whenever we divide even integers to each other the remainder will be 0 or more than 1. However since m and n must be greater than 2 and the greatest common factor is 2, each integer contains some distinct from each other factors. For example, 4(2*2) and 6(3*2), 6(3*2) and 8(2*2*2). So the remainder will be greater than 1 in any case. Sufficient.

(2) here we can have several integers that fit into the 2nd statement's condition and have different remainders. For example, 5 and 6, their LCM is 30 the remainder is 1; and 15 and 6, their LCM is again 30 but the remainder is greater than 1. Not sufficient.
_________________

If you found my post useful and/or interesting - you are welcome to give kudos!

CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2793
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 180

Kudos [?]: 956 [0], given: 235

Reviews Badge
Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 01:34
Shrouded1 :

m=10,p=30,lcm=30, remainder=0 --> is wrong , it is given that m is not a factor of P

you may take m =6 and p =10
_________________

Fight for your dreams :For all those who fear from Verbal- lets give it a fight

Money Saved is the Money Earned :)

Jo Bole So Nihaal , Sat Shri Akaal

:thanks Support GMAT Club by putting a GMAT Club badge on your blog/Facebook :thanks

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Gmat test review :
670-to-710-a-long-journey-without-destination-still-happy-141642.html

Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 76

Kudos [?]: 487 [0], given: 25

GMAT ToolKit User Reviews Badge
Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 01:39
gurpreetsingh wrote:
Shrouded1 :

m=10,p=30,lcm=30, remainder=0 --> is wrong , it is given that m is not a factor of P

you may take m =6 and p =10


Oops, you are right, I should take m=5, p=6, then remainder=1. Still works for the question

Editing my solution
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Get the best GMAT Prep Resources with GMAT Club Premium Membership

GMAT Forum Moderator
avatar
Status: Accepting donations for the mohater MBA fund
Joined: 05 Feb 2008
Posts: 1860
Location: United States
Concentration: Healthcare, Economics
Schools: Ross '14 (M)
GMAT 1: 710 Q48 V38
GMAT 2: Q V
GPA: 3.54
WE: Accounting (Manufacturing)
Followers: 55

Kudos [?]: 404 [0], given: 234

GMAT ToolKit User Premium Member Reviews Badge
Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 03:08
Already posted:
new-set-of-good-ds-85441.html

Also, posted in the wrong forum. This belongs in the quant forum.

Please search in the future before posting.

Locked.
_________________

Strategy Discussion Thread | Strategy Master | GMAT Debrief| Please discuss strategies in discussion thread. Master thread will be updated accordingly. | GC Member Write Ups

Get the best GMAT Prep Resources with GMAT Club Premium Membership

GMAT Forum Moderator
avatar
Status: Accepting donations for the mohater MBA fund
Joined: 05 Feb 2008
Posts: 1860
Location: United States
Concentration: Healthcare, Economics
Schools: Ross '14 (M)
GMAT 1: 710 Q48 V38
GMAT 2: Q V
GPA: 3.54
WE: Accounting (Manufacturing)
Followers: 55

Kudos [?]: 404 [0], given: 234

GMAT ToolKit User Premium Member Reviews Badge
Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 04:34
Bunuel wrote:
Moved to the proper subforum, so now it's unlocked for further discussion.


booooo you stole my thunder :)
_________________

Strategy Discussion Thread | Strategy Master | GMAT Debrief| Please discuss strategies in discussion thread. Master thread will be updated accordingly. | GC Member Write Ups

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 04 Aug 2010
Posts: 158
Followers: 2

Kudos [?]: 24 [0], given: 15

Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 18:05
When evaluating statement (2) LCD = 30, is there a way to get the conclusion of insufficient by looking at the prime factors of 30 rather than sub-ing in numbers?
Retired Moderator
User avatar
Joined: 02 Sep 2010
Posts: 807
Location: London
Followers: 76

Kudos [?]: 487 [0], given: 25

GMAT ToolKit User Reviews Badge
Re: DS:GCD of numbers. [#permalink] New post 20 Sep 2010, 19:13
FQ wrote:
When evaluating statement (2) LCD = 30, is there a way to get the conclusion of insufficient by looking at the prime factors of 30 rather than sub-ing in numbers?


Its one and the same thing. You are looking for 2 factors of 30, such that one is a multiple of the other OR p=km+1 and at the same time their LCM is 30.
_________________

Math write-ups
1) Algebra-101 2) Sequences 3) Set combinatorics 4) 3-D geometry

My GMAT story

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23055
Followers: 3535

Kudos [?]: 27206 [0], given: 2725

Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 06 Jun 2013, 05:08
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

Theory on remainders problems: remainders-144665.html

All DS remainders problems to practice: search.php?search_id=tag&tag_id=198
All PS remainders problems to practice: search.php?search_id=tag&tag_id=199

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Director
Director
avatar
Joined: 29 Nov 2012
Posts: 929
Followers: 12

Kudos [?]: 285 [0], given: 543

Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 10 Oct 2013, 05:52
What would be a good numerical example to satisfy statement 1
m=8, p=14?
_________________

Click +1 Kudos if my post helped...

Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/

GMAT Prep software What if scenarios gmat-prep-software-analysis-and-what-if-scenarios-146146.html

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23055
Followers: 3535

Kudos [?]: 27206 [0], given: 2725

Re: The integers m and p are such that 2<m<p, and m is not a [#permalink] New post 10 Oct 2013, 06:22
Expert's post
Re: The integers m and p are such that 2<m<p, and m is not a   [#permalink] 10 Oct 2013, 06:22
    Similar topics Author Replies Last post
Similar
Topics:
19 Experts publish their posts in the topic Is |p|^2<|p| ? Zarrolou 9 18 Apr 2013, 12:32
The integers m and p are such that 2 < m <p. m is not Caas 4 08 Jun 2007, 09:00
5 Experts publish their posts in the topic The integers m and p are such that 2<m<p, and m is not a fac mrmikec 13 16 Jun 2006, 17:05
The integers m and p are such that 2<m<p and m is not ellisje22 7 06 Jan 2006, 08:05
The integers m and p are such that 2 < m < p and m is TeHCM 6 26 Dec 2005, 17:16
Display posts from previous: Sort by

The integers m and p are such that 2<m<p, and m is not a

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.