|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 13 May 2010
Posts: 108
|
Does the integer k have a factor p such that 1<p<k? [#permalink]
Show Tags
 19 Jun 2010, 03:49
3
This post received
KUDOS
30
This post was
BOOKMARKED
D
E
Question Stats:
61% (01:54) correct
39% (01:18) wrong based on 1313 sessions
HideShow timer Statistics
i dont have any idea to solve the attached problem. Please provide methodology to solve such problems. Does the integer k have a factor p such that 1<p<k? (1) k > 4! (2) 13! + 2<= k <= 13!+13
Official Answer and Stats are available only to registered users. Register/ Login.
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 39598
|
Re: gmat prep ds [#permalink]
Show Tags
19 Jun 2010, 04:51
20
This post received
KUDOS
Expert's post
21
This post was
BOOKMARKED
Does the integer k have a factor p such that 1<p<k? Question basically asks whether \(k\) is a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number). (1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient. (2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\), the same way as we did for 8). Sufficient. Answer: B. Check similar question: factor-factorials-100670.htmlHope it's clear.
_________________
New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!
Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.
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?
Extra-hard Quant Tests with Brilliant Analytics
|
|
|
|
|
|
Manager
Joined: 13 May 2010
Posts: 108
|
Re: gmat prep ds [#permalink]
Show Tags
19 Jun 2010, 05:46
Thanks..can not be clearer than this.
|
|
|
|
|
|
Retired Moderator
Status: 2000 posts! I don't know whether I should feel great or sad about it! LOL
Joined: 04 Oct 2009
Posts: 1662
Location: Peru
Schools: Harvard, Stanford, Wharton, MIT & HKS (Government)
WE 1: Economic research
WE 2: Banking
WE 3: Government: Foreign Trade and SMEs
|
Re: gmat prep ds [#permalink]
Show Tags
03 Nov 2010, 15:20
Bunuel wrote: Does the integer k have a factor p such that 1<p<k? Question basically asks is \(k\) a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number). (1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient. (2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\)). Sufficient. Answer: B. Check similar question: factor-factorials-100670.htmlHope it's clear. Thanks Bunuel. I have a question, what would happen if the second statement said this? (2) \(13!+ 1\leq{k}\leq{13!+13}\) I have seen some similar problems, in which they add 1. Could you post some links about it? Thanks!
_________________
"Life’s battle doesn’t always go to stronger or faster men; but sooner or later the man who wins is the one who thinks he can."
My Integrated Reasoning Logbook / Diary: http://gmatclub.com/forum/my-ir-logbook-diary-133264.html
GMAT Club Premium Membership - big benefits and savings
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 39598
|
Does the integer k have a factor p such that 1<p<k? [#permalink]
Show Tags
03 Nov 2010, 15:51
metallicafan wrote: Bunuel wrote: Does the integer k have a factor p such that 1<p<k? Question basically asks is \(k\) a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number). (1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient. (2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\)). Sufficient. Answer: B. Check similar question: factor-factorials-100670.htmlHope it's clear. Thanks Bunuel. I have a question, what would happen if the second statement said this? (2) \(13!+ 1\leq{k}\leq{13!+13}\) I have seen some similar problems, in which they add 1. Could you post some links about it? Thanks! So basically we should determine whether \(13!+1\) is a prime number (as shown above all other possible values of k are not are not prime), which cannot be done without a computer. There are some particular values of \(k=n!+1\) for which we can say whether it's a prime or not with help of Wilson's theorem, but again it's out of the scope of GMAT. By the way: \(13!+1\) is not a prime number, it has two distinct prime factors: \(13!+1=83*75,024,347=6,227,020,801\), so the answer still will be B.
_________________
New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!
Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.
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?
Extra-hard Quant Tests with Brilliant Analytics
|
|
|
|
|
|
Intern
Joined: 12 Nov 2010
Posts: 22
|
Re: gmat prep ds [#permalink]
Show Tags
10 Feb 2011, 18:04
Bunuel wrote: Does the integer k have a factor p such that 1<p<k? Question basically asks whether \(k\) is a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number). (1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient. (2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\)). Sufficient. Answer: B. Check similar question: factor-factorials-100670.htmlHope it's clear. Sorry for the question, but I am just not understanding how you get 8*(2*4*5*6*7*9*10*11*12*13+1) from 13!+8 then how do you know that k is a multiple of 8...I obviously have some deficiencies when it comes to number properties. So then how is k never a prime between those parameters? I would think (1*2*3*4*5*6*7*8*9*10*11*12*13) +8 ...
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 39598
|
Re: gmat prep ds [#permalink]
Show Tags
10 Feb 2011, 18:19
2
This post received
KUDOS
Expert's post
1
This post was
BOOKMARKED
ChenggongMAS wrote: Bunuel wrote: Does the integer k have a factor p such that 1<p<k? Question basically asks whether \(k\) is a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number). (1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient. (2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\)). Sufficient. Answer: B. Check similar question: factor-factorials-100670.htmlHope it's clear. Sorry for the question, but I am just not understanding how you get 8*(2*4*5*6*7*9*10*11*12*13+1) from 13!+8 then how do you know that k is a multiple of 8...I obviously have some deficiencies when it comes to number properties. So then how is k never a prime between those parameters? I would think (1*2*3*4*5*6*7*8*9*10*11*12*13) +8 ... k=13!+8 means k=2*3*4*5*6*7* 8*9*10*11*12*13+ 8, now factor out 8: k=8*(2*4*5*6*7*9*10*11*12*13+1) --> k is a multiple of 8 as k=8*something --> as k is a multiple of 8 it can not be a prime number. You can have any number from 2 to 13 inclusive instead of 8, and you'll be able to factor out this number the same way as you did with 8, so any number of a type 13!+x, wher x is from 2 to 13 inclusive will be a multiple of x, thus not a prime number. Check the link in my first post for similar problem. Also check Number Theory chapter of Math Book: math-number-theory-88376.htmlHope it's clear.
_________________
New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!
Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.
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?
Extra-hard Quant Tests with Brilliant Analytics
|
|
|
|
|
|
Manager
Joined: 29 Jul 2011
Posts: 107
Location: United States
|
Re: OG 11 [#permalink]
Show Tags
13 Dec 2011, 15:37
Answer can't be E, it has to be B. I don't have OG with me, but I need to check. 1. By itself not sufficient. K could be prime. 2. 13! + 2 <= K <= 13! + 13 ensures that you have ever number from 2 - 13 dividing k. That is: 13! + 2 -> div by 2 13! + 3 -> div by 3 ... so on till 13. So, answer should be B.
_________________
I am the master of my fate. I am the captain of my soul.
Please consider giving +1 Kudos if deserved!
DS - If negative answer only, still sufficient. No need to find exact solution.
PS - Always look at the answers first
CR - Read the question stem first, hunt for conclusion
SC - Meaning first, Grammar second
RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min
|
|
|
|
|
|
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 537
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)
|
Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
29 Jan 2012, 16:15
4
This post received
KUDOS
34
This post was
BOOKMARKED
Does the integer k have a factor p such that 1 < p < k ? (1) k > 4! (2) 13! + 2 <= k <= 13! + 13 So considering statement 1
K > 24
Lets say k = 25, so the factors are 1, 5 and 25. P will be 1 as 1 is the factor of every number. Therefore, insufficient.
Considering Statement 2
13!+2<= k<=13!+13
simplifying this expression will give
2<=K<=13, so can be anything from 2...............13. As p can and cannot be greater than 1 this statement is insufficient.
Combining the two, I think will not give an answer too. Therefore, for me its E. Any thoughts anyone as I don't have an OA? _________________
Best Regards,
E.
MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
Last edited by Bunuel on 16 Sep 2014, 13:56, edited 4 times in total.
Added the OA
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 39598
|
Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
29 Jan 2012, 16:21
28
This post received
KUDOS
Expert's post
20
This post was
BOOKMARKED
Does the integer k have a factor p such that 1<p<k? Question basically asks whether \(k\) is a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number). (1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient. (2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*3*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\), the same way as we did for 8). Sufficient. Answer: B. Check similar question: if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.htmlHope it's clear.
_________________
New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!
Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.
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?
Extra-hard Quant Tests with Brilliant Analytics
|
|
|
|
|
|
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 537
Location: United Kingdom
Concentration: International Business, Strategy
GPA: 2.9
WE: Information Technology (Consulting)
|
Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
29 Jan 2012, 16:22
Thanks for the great explanation.
_________________
Best Regards,
E.
MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730
|
|
|
|
|
|
Intern
Joined: 06 Nov 2011
Posts: 37
Location: Germany
Concentration: Entrepreneurship, General Management
GMAT Date: 03-10-2012
GPA: 3
|
Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
14 Feb 2012, 14:27
Hi bunuel I don't understand the definition of a prime ( 1<p<k ). I know what a prime is but this drives me nuts. For an example I take 10 for k. I have factors 2x5 that are greater than 1 and smaller than 10.  Or is it asking for one factor? I see "a factor" in the question, which is singular. Then I understand this. It is really tricky.  Posted from GMAT ToolKit
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 39598
|
Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
14 Feb 2012, 16:15
3
This post received
KUDOS
Expert's post
4
This post was
BOOKMARKED
M3tm4n wrote: Hi bunuel I don't understand the definition of a prime ( 1<p<k ). I know what a prime is but this drives me nuts. For an example I take 10 for k. I have factors 2x5 that are greater than 1 and smaller than 10. Or is it asking for one factor? I see "a factor" in the question, which is singular. Then I understand this. It is really tricky. Posted from GMAT ToolKitQuestion asks whether some number k has a factor p which is more than 1 but less than k. For example if k=10 then the answer is yes, since both 2 and 5 are factors of 10 and are more than 1 and less than 10. But if for example k=7=prime then the answer is no, since 7 has no factor which is more than 1 and less than 7. Now, look at the definition of a prime number: a prime number is a positive integer with exactly two factors: 1 and itself. So, we can say that the questions asks whether k is a prime number, because if it is then it won't have a factor which is more than 1 and less than k. Hope it's clear.
_________________
New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!
Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.
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?
Extra-hard Quant Tests with Brilliant Analytics
|
|
|
|
|
|
Manager
Joined: 03 Jul 2012
Posts: 133
GPA: 3.9
WE: Programming (Computer Software)
|
Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
01 Sep 2012, 08:32
Got this one wrong. Can somebody explain how to solve this?
Attachments
DBtful DS.png [ 7.39 KiB | Viewed 30334 times ]
|
|
|
|
|
|
Intern
Joined: 30 Oct 2011
Posts: 46
|
Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
18 Nov 2012, 03:18
Bunuel wrote: Does the integer k have a factor p such that 1<p<k? Question basically asks whether \(k\) is a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number). (1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient. (2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\), the same way as we did for 8). Sufficient. Answer: B. Check similar question: if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.htmlHope it's clear. Hi Bunuel, I have a small query here- Had the question been : 4!+2<k<4!+6, if I follow the approach explained above I may conclude that k is not a prime no. reason being all the factors of 4! are the factors of 4!+?. Since its easy to calculate here the values between 4!+2 and 4!+6, I already know that it includes 29 which is a prime no., where as in case of 13!+2<k<13!+13, it is not easy to calculate 13!, we may be missing some value which is a prime no. How do we make sure that we are not missing anything? (Assuming we can not cross check by calculating the values because of the time constraint at the exam time)
|
|
|
|
|
|
Math Expert
Joined: 02 Sep 2009
Posts: 39598
|
Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
18 Nov 2012, 05:01
mneeti wrote: Bunuel wrote: Does the integer k have a factor p such that 1<p<k? Question basically asks whether \(k\) is a prime number. If it is, then it won't have a factor \(p\) such that \(1<p<k\) (definition of a prime number). (1) \(k>4!\) --> \(k\) is more than some number (\(4!=24\)). \(k\) may or may not be a prime. Not sufficient. (2) \(13!+2\leq{k}\leq{13!+13}\) --> \(k\) can not be a prime. For instance if \(k=13!+8=8*(2*4*5*6*7*9*10*11*12*13+1)\), then \(k\) is a multiple of 8, so not a prime. Same for all other numbers in this range. So, \(k=13!+x\), where \(2\leq{x}\leq{13}\) will definitely be a multiple of \(x\) (as we would be able to factor out \(x\) out of \(13!+x\), the same way as we did for 8). Sufficient. Answer: B. Check similar question: if-x-is-an-integer-does-x-have-a-factor-n-such-that-100670.htmlHope it's clear. Hi Bunuel, I have a small query here- Had the question been : 4!+2<k<4!+6, if I follow the approach explained above I may conclude that k is not a prime no. reason being all the factors of 4! are the factors of 4!+?. Since its easy to calculate here the values between 4!+2 and 4!+6, I already know that it includes 29 which is a prime no., where as in case of 13!+2<k<13!+13, it is not easy to calculate 13!, we may be missing some value which is a prime no. How do we make sure that we are not missing anything? (Assuming we can not cross check by calculating the values because of the time constraint at the exam time) If \(13!+2\leq{k}\leq{13!+13}\), then we can factor out \(x\) out of \(13!+x\) (where \(2\leq{x}\leq{13}\)), which means that k is not a prime. But you cannot apply the same logic with 4!+2<k<4!+6, since if k=4!+5, then you cannot factor out 5 out of it.
_________________
New to the Math Forum?
Please read this: All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!
Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.
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?
Extra-hard Quant Tests with Brilliant Analytics
|
|
|
|
|
|
Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GPA: 3.23
|
Re: Does the integer k have a factor p such that 1<p<k? [#permalink]
Show Tags
22 Jan 2013, 03:40
gmatcracker2010 wrote: i dont have any idea to solve the attached problem. Please provide methodology to solve such problems. Does the integer k have a factor p such that 1<p<k? (1) k > 4! (2) 13! + 2<= k <= 13!+13 1. if k=29 p=29= k NO! if k =35 p could be 7 or 5 YES! INSUFFICIENT! 2. 13! + 2 = 2(13!/2 + 1) = 2*odd 13! + 3 = 3(13!/3 + 1) = 3*odd ... 13! + 13 = 13 (12! +1) = 13*odd Always there is a 1< p < 13! + n Answer: B
_________________
Impossible is nothing to God.
|
|
|
|
|
|
Director
Joined: 29 Nov 2012
Posts: 878
|
Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
27 Jan 2013, 21:25
So basically in the second statement we can factor out a number from 2-13 but if its higher than that we can't say for sure it would be a prime or not?
_________________
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 http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html
|
|
|
|
|
|
AGSM Thread Master
Joined: 19 Jul 2012
Posts: 169
Location: India
Concentration: Marketing, International Business
GPA: 3.3
|
Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
29 May 2013, 23:06
1
This post received
KUDOS
crazypriya wrote: Does the integer k have a factor p such that 1<p<k? 1.k>4! 2.13!+2<=k<=13!+13 The question is basically asking whether K is prime or not. If it is prime then P can be 1 or k else it will follow 1<p<k. Statement 1. k>4!
k>24;
let's say k=25, then p=5; we get a Yes to the question.
if k=29; then p=1 or 29; we get a No to the question.
Thus, insufficient.Statement 2: 13!+2<=k<=13!+13
k can take any value from 13!+2 to 13!+13.
All the values in this range has one thing in common. They all can have one common value:
for e.g. 13!+2= 2[(13*12*...except 2)+1]
13!+10=10[(13*12*...(except 10)+1]
No matter whether the value in the bracket is prime or not; the common value will make it non prime and will always give a yes to the question.
Thus, sufficient.
|
|
|
|
|
|
Intern
Joined: 21 Apr 2013
Posts: 9
|
Re: Does the integer k have a factor p such that 1 < p < k ? [#permalink]
Show Tags
29 May 2013, 23:17
crazypriya wrote: Does the integer k have a factor p such that 1<p<k? 1.k>4! 2.13!+2<=k<=13!+13 Essentially this is about prime and non-prime. Option 1: k>4! Now assume k=5. then is there any factor of 5 which lies between 1 and 5? NO. Take k=6. then there is 2,3 which lies between 1 and 6. So 1 alone insufficient. Option 2: 13!+2<=k<=13!+13 Take say 13!+3. now it can be rewritten as 3*{(13.12.11.....4.2.1)+1} So there is 3 as factor between 1 and k. Similarly you can take the common factor out of all nos. in option 2. Therefore option 2 is sufficient! Add Kudos if this helps!
|
|
|
|
|
|
|
Re: Does the integer k have a factor p such that 1 < p < k ?
[#permalink]
29 May 2013, 23:17
|
|
|
|
|
Go to page
1 2 3
Next
[ 43 posts ]
|
|
|
|
|
|
|
|
13
