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

It is currently 27 Aug 2015, 21:35
GMAT Club Tests

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

How many different prime numbers are factors of the positive

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
1 KUDOS received
Intern
Intern
avatar
Joined: 08 Oct 2010
Posts: 1
Followers: 0

Kudos [?]: 4 [1] , given: 0

How many different prime numbers are factors of the positive [#permalink] New post 08 Oct 2010, 16:50
1
This post received
KUDOS
5
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

55% (01:48) correct 45% (01:14) wrong based on 273 sessions
How many different prime numbers are factors of positive integer n?

(1) 4 different prime numbers are factors of 2n
(2) 4 different prime numbers are factors of n^2
[Reveal] Spoiler: OA
1 KUDOS received
Senior Manager
Senior Manager
User avatar
Joined: 20 Jan 2010
Posts: 278
Schools: HBS, Stanford, Haas, Ross, Cornell, LBS, INSEAD, Oxford, IESE/IE
Followers: 14

Kudos [?]: 165 [1] , given: 117

Re: DS question : need help [#permalink] New post 08 Oct 2010, 17:50
1
This post received
KUDOS
Well, my answer is D. I did reveal OA after solving.
Here's my explanation (Please! do correct me if I am wrong):

Statement I: Let's suppose if 2,3,5,7 are 4 different prime numbers of 2n (2 would be always a factor as it is 2n), then there are total of 6 factors of 2n=4 different prime numbers+1+the number itself (in this case 210). Divide 2n with 2 (one of the 4 different prime factors) and we get "n" and now the factors would be (excluded 2) 1,3,5,7,210=5 factors=number of factors of 2n -1=6-1. Same goes for the case when we choose 2,5,7,11 as 4 different prime numbers, then we'll also get--> Factors of n=factors of 2n-1=6-1=5 (as 2 always would be divided and we'll be left with one factors less than those of 2n). Statement 1 SUFFICIENT.
If, Statement I is sufficient then answer can be either A or D, so eliminate B,C, and E.

Statement II:Let's suppose 2,3,5,7 are different prime number factors of n^2, then 2*3*5*7=210=n^2 and for n=sqrt (210)= not an whole number or not an integer= It's factors would be 1 & the number itself. There are 2 factors of Sqrt (210) and ultimately of "n". We can't take "Sqrt" of any prime number or products of different prime numbers (no repetition of same prime number) because prime number is only divisible by 1 & itself. So, factors of "n" would always be 1 & itself-->only 2 factors of "n". Same is the case if we take other 4 different prime numbers.
Statement II is SUFFICIENT

That's why Answer should be D in my opinion.
Would love to see Bunuel's opinion :)
_________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so."
Target=780
http://challengemba.blogspot.com
Kudos??

Expert Post
3 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 29090
Followers: 4720

Kudos [?]: 49587 [3] , given: 7395

Re: DS question : need help [#permalink] New post 08 Oct 2010, 18:11
3
This post received
KUDOS
Expert's post
3
This post was
BOOKMARKED
How many different prime numbers are factors of the positive integer n?

(1) 4 different prime numbers are factors of 2n --> if \(n\) itself has 2 as a factor (eg \(n=2*3*5*7\)) than its total # of primes is 4 but if \(n\) doesn't have 2 as a factor (eg \(n=3*5*7\)) than its total # of primes is 3. Not sufficient.

(2) 4 different prime numbers are factors of n^2 --> \(n^x\) (where \(x\) is an integer \(\geq{1}\)) will have as many different prime factors as integer \(n\), exponentiation doesn't "produce" primes. So, 4 different prime numbers are factors of \(n\). Sufficient.

Answer: B.

Similar question: different-prime-factors-ds-95585.html?hilit=produce%20prime#p735955

Hope it helps.
_________________

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

PLEASE READ AND FOLLOW: 12 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 ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. 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?
25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings

Senior Manager
Senior Manager
User avatar
Joined: 20 Jan 2010
Posts: 278
Schools: HBS, Stanford, Haas, Ross, Cornell, LBS, INSEAD, Oxford, IESE/IE
Followers: 14

Kudos [?]: 165 [0], given: 117

Re: DS question : need help [#permalink] New post 08 Oct 2010, 21:55
Thanks! Bunuel for clarification while doing this question, I knew in my heart that I am doing something wrong that's why I couldn't ask a better person but you. My bad, I didn't read clearly that it is "how many different prime numbers are factors" not "how many factors".
@arundas, totally right. Thanks buddy :). Looks like gotta pay more attention & work on questions related to factors (doing MGMAT Number Properties)
_________________

"Don't be afraid of the space between your dreams and reality. If you can dream it, you can make it so."
Target=780
http://challengemba.blogspot.com
Kudos??

Intern
Intern
avatar
Joined: 24 Dec 2011
Posts: 5
GMAT 1: 760 Q50 V42
Followers: 0

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

Re: How many different prime numbers are factors of +ve integer [#permalink] New post 28 Dec 2011, 01:22
Well, from statement 1 you get that 2n has 4 different prime factors. This implies that n has at least 3 different prime factors (if n is not even) and 4 different prime factors (if n is even).
Two possibilities - A and D out.

From statement 2 we get that n^2 has 4 different prime factors. This would mean that n will also have the same 4 different prime factors as n^2 {why? because any additional factors that n^2 has over n will be composite and not prime. We can understand this better with an example. The smallest number with 4 different prime factors is 210 (2*3*5*7). 210^2 = (2^2)*(3^2)*(5^2)*(7^2). Thus, no new prime factors come into play.}
Answer: B
Director
Director
avatar
Joined: 28 Jul 2011
Posts: 563
Location: United States
Concentration: International Business, General Management
GPA: 3.86
WE: Accounting (Commercial Banking)
Followers: 2

Kudos [?]: 102 [0], given: 16

Re: How many different prime numbers are factors of +ve integer [#permalink] New post 28 Dec 2011, 07:42
vjalan wrote:
Well, from statement 1 you get that 2n has 4 different prime factors. This implies that n has at least 3 different prime factors (if n is not even) and 4 different prime factors (if n is even).
Two possibilities - A and D out.

From statement 2 we get that n^2 has 4 different prime factors. This would mean that n will also have the same 4 different prime factors as n^2 {why? because any additional factors that n^2 has over n will be composite and not prime. We can understand this better with an example. The smallest number with 4 different prime factors is 210 (2*3*5*7). 210^2 = (2^2)*(3^2)*(5^2)*(7^2). Thus, no new prime factors come into play.}
Answer: B


Sorry i didn't understand your explanation

From 1 how did you say if "n" is even then it has 4 factors...because when you take n=2 then 2 has only 2 factors and when n is odd take n=1 then it has only 1 factor...........

From 2 i didn't get anything

Can you please be more clear?????

Thanks in advance
_________________

+1 Kudos If found helpful..

Intern
Intern
avatar
Joined: 24 Dec 2011
Posts: 5
GMAT 1: 760 Q50 V42
Followers: 0

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

Re: How many different prime numbers are factors of +ve integer [#permalink] New post 28 Dec 2011, 07:57
Well, ur example is wrong. Take n=such a value that 2n has 4 different prime factors. Smallest number with 4 different prime factors = 210 (2x3x5x7). So, let 2n = 210. Then n = 105 (odd, has 3 prime factors). Again, for 2n = 420 (2x2x3x5x7), which again has 4 different prime factors, n = 210 (even, 4 prime factors).
Is it clearer now?? Don't hesitate if u still have doubts..

Now, coming to statement 2, n^2 has 4 prime factors. Let these be a,b,c,d. Now, for n to a positive integer, n^2 has to be a perfect square. Thus n^2 = (a^2) * (b^2) * (c^2) * (d^2). So, n = a*b*c*d. Number of prime factors of n = number of prime factors of n^2 = 4.

Answer is B.
Manager
Manager
avatar
Joined: 26 Jun 2011
Posts: 249
Location: India
GMAT 1: 760 Q51 V41
Followers: 11

Kudos [?]: 47 [0], given: 26

Re: How many different prime numbers are factors of +ve integer [#permalink] New post 28 Dec 2011, 08:00
kotela wrote:
vjalan wrote:
Well, from statement 1 you get that 2n has 4 different prime factors. This implies that n has at least 3 different prime factors (if n is not even) and 4 different prime factors (if n is even).
Two possibilities - A and D out.

From statement 2 we get that n^2 has 4 different prime factors. This would mean that n will also have the same 4 different prime factors as n^2 {why? because any additional factors that n^2 has over n will be composite and not prime. We can understand this better with an example. The smallest number with 4 different prime factors is 210 (2*3*5*7). 210^2 = (2^2)*(3^2)*(5^2)*(7^2). Thus, no new prime factors come into play.}
Answer: B


Sorry i didn't understand your explanation

From 1 how did you say if "n" is even then it has 4 factors...because when you take n=2 then 2 has only 2 factors and when n is odd take n=1 then it has only 1 factor...........

From 2 i didn't get anything

Can you please be more clear?????

Thanks in advance


1) 2n has four diff. prime factors. So, n can be 3*5*7 (odd) or 2*3*5*7 (even) or their multiples. Not sufficient.
2) n will have the same prime factors as its square n^2. This is valid for every integer n. Sufficient.
_________________

The chase begins ...

Director
Director
avatar
Joined: 28 Jul 2011
Posts: 563
Location: United States
Concentration: International Business, General Management
GPA: 3.86
WE: Accounting (Commercial Banking)
Followers: 2

Kudos [?]: 102 [0], given: 16

Re: How many different prime numbers are factors of +ve integer [#permalink] New post 28 Dec 2011, 15:56
vailad wrote:
kotela wrote:
vjalan wrote:
Well, from statement 1 you get that 2n has 4 different prime factors. This implies that n has at least 3 different prime factors (if n is not even) and 4 different prime factors (if n is even).
Two possibilities - A and D out.

From statement 2 we get that n^2 has 4 different prime factors. This would mean that n will also have the same 4 different prime factors as n^2 {why? because any additional factors that n^2 has over n will be composite and not prime. We can understand this better with an example. The smallest number with 4 different prime factors is 210 (2*3*5*7). 210^2 = (2^2)*(3^2)*(5^2)*(7^2). Thus, no new prime factors come into play.}
Answer: B


Sorry i didn't understand your explanation

From 1 how did you say if "n" is even then it has 4 factors...because when you take n=2 then 2 has only 2 factors and when n is odd take n=1 then it has only 1 factor...........

From 2 i didn't get anything

Can you please be more clear?????

Thanks in advance


1) 2n has four diff. prime factors. So, n can be 3*5*7 (odd) or 2*3*5*7 (even) or their multiples. Not sufficient.
2) n will have the same prime factors as its square n^2. This is valid for every integer n. Sufficient.



Thanks a lot I got it....
_________________

+1 Kudos If found helpful..

Manager
Manager
User avatar
Joined: 29 Jul 2011
Posts: 108
Location: United States
Followers: 5

Kudos [?]: 45 [0], given: 6

Re: How many different prime numbers are factors of +ve integer [#permalink] New post 11 Jan 2012, 13:59
Easy one.

1. Here, say n=3x5x7. Then, 2n will have 4 primes, but n will have 3 primes. If n=2x3x5x7, then 2n and n will have same primes. insuff.
2. n^2 will always have same number of primes as n, suff.

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

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 29090
Followers: 4720

Kudos [?]: 49587 [0], given: 7395

Re: How many different prime numbers are factors of the positive [#permalink] New post 12 Jun 2013, 03:27
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 Number Properties: math-number-theory-88376.html

All DS Number Properties Problems to practice: search.php?search_id=tag&tag_id=38
All PS Number Properties Problems to practice: search.php?search_id=tag&tag_id=59

_________________

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

PLEASE READ AND FOLLOW: 12 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 ; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) ; 12. 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?
25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings

GMAT Club Legend
GMAT Club Legend
User avatar
Joined: 09 Sep 2013
Posts: 6079
Followers: 340

Kudos [?]: 68 [0], given: 0

Premium Member
Re: How many different prime numbers are factors of the positive [#permalink] New post 17 Jul 2014, 17:19
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Director
Director
avatar
Joined: 08 Jun 2010
Posts: 750
Followers: 0

Kudos [?]: 52 [0], given: 175

Re: How many different prime numbers are factors of the positive [#permalink] New post 28 Apr 2015, 02:56
xyzgmat wrote:
How many different prime numbers are factors of positive integer n?

(1) 4 different prime numbers are factors of 2n
(2) 4 different prime numbers are factors of n^2


very nice, I miss this one

very basic but fatal
Expert Post
e-GMAT Representative
User avatar
Joined: 04 Jan 2015
Posts: 338
Followers: 65

Kudos [?]: 471 [0], given: 83

Re: How many different prime numbers are factors of the positive [#permalink] New post 29 Apr 2015, 05:47
Expert's post
Here's another similar question for you to practice:

How many different prime factors does positive integer N have?

(1) \(\frac{N}{2}\) has 2 different odd prime factors

(2) \(\frac{N}{3}\) has 2 different prime factors


Will post the OA and OE in a couple of days in the same thread. Please post your analysis below. Happy Solving! :-D

Best Regards

Japinder
_________________

https://e-gmat.com/courses/quant-live-prep/

Image

Intern
Intern
User avatar
Joined: 15 Mar 2015
Posts: 46
Followers: 1

Kudos [?]: 15 [0], given: 6

Re: How many different prime numbers are factors of the positive [#permalink] New post 29 Apr 2015, 09:30
Nice question!

As for (1) We are told that 2n contains 4 different primes. if n contains the factor 2, then it would be repeated and n would still include 4 primes. If n does not contain the factor 2, then n contains only 3 different primes. This gives us the two possibilities 3 and 4.

As for (2) We are told that n^2 has 4 different primes. If we prime factorize n and square the factorization(e.g. (2*3*5*7)^2) we would simply be repeating the same factors(in our example(2*2*3*3*5*5*7*7) and no new primes would be introduced by the manipulation.

Thus, B is sufficient.
1 KUDOS received
Intern
Intern
User avatar
Joined: 15 Mar 2015
Posts: 46
Followers: 1

Kudos [?]: 15 [1] , given: 6

Re: How many different prime numbers are factors of the positive [#permalink] New post 29 Apr 2015, 09:35
1
This post received
KUDOS
EgmatQuantExpert wrote:
Here's another similar question for you to practice:

How many different prime factors does positive integer N have?

(1) \(\frac{N}{2}\) has 2 different odd prime factors

(2) \(\frac{N}{3}\) has 2 different prime factors


Will post the OA and OE in a couple of days in the same thread. Please post your analysis below. Happy Solving! :-D

Best Regards

Japinder



I'll go with A.

The following is my reasoning:

(1) N/2 Has 2 odd primes.

2 is a prime factor of N as we could divide by 2 and end up with an integer. So N = 2*o1*o2 where o1 and o2 are two different odd prime numbers.

(2) N/3 has 2 different prime factors.

We now have two possible options. 3 is obviously one of the prime factors, but it could be repeated if e.g. N=3*3*5 then N has 2 different primes.
If N = 2*3*5 then N/3=2*5, resulting in 3 different primes.

Thus A should be correct.
Expert Post
e-GMAT Representative
User avatar
Joined: 04 Jan 2015
Posts: 338
Followers: 65

Kudos [?]: 471 [0], given: 83

Re: How many different prime numbers are factors of the positive [#permalink] New post 07 May 2015, 03:37
Expert's post
EgmatQuantExpert wrote:
Here's another similar question for you to practice:

How many different prime factors does positive integer N have?

(1) \(\frac{N}{2}\) has 2 different odd prime factors

(2) \(\frac{N}{3}\) has 2 different prime factors


Will post the OA and OE in a couple of days in the same thread. Please post your analysis below. Happy Solving! :-D

Best Regards

Japinder



The correct answer is Option A.

Statement 1:
\(\frac{N}{2}\) has 2 different odd prime factors


Let the 2 different odd prime factors of \(\frac{N}{2}\) be OP1 and OP2.

This means, we can write:

\(N = 2*OP1^{a}*OP2^{b}\), where a and b are positive integers

From here, it's clear that N has 3 prime factors, 1 of which is even and 2 are odd.

Thus, Statement 1 is sufficient.

Statement 2:
\(\frac{N}{3}\) has 2 different prime factors


Let the 2 different prime factors of \(\frac{N}{3}\) be P1 and P2

This means, we can write:

\(N = 3*P1^{a}*P2^{b}\), where a and b are positive integers

If P1 = 3, then N has 2 prime factors
But if P1 or P2 are different from 3, then N has 3 prime factors

Therefore, Statement 2 is not sufficient to ascertain the exact number of prime factors of N.

Hope this was helpful! :)

Japinder
_________________

https://e-gmat.com/courses/quant-live-prep/

Image

Re: How many different prime numbers are factors of the positive   [#permalink] 07 May 2015, 03:37
    Similar topics Author Replies Last post
Similar
Topics:
4 Experts publish their posts in the topic If m and n are different positive integers, then how many prime number Bunuel 7 16 Jul 2015, 00:27
10 Experts publish their posts in the topic How many different prime factors does positive integer n have? Bunuel 9 14 May 2015, 04:25
3 Experts publish their posts in the topic How many different prime factors does x have? Harley1980 2 12 Apr 2015, 23:57
51 Experts publish their posts in the topic How many different prime numbers are factors of the positive enigma123 13 29 Jan 2012, 17:31
27 Experts publish their posts in the topic If K is a positive integer, how many different prime numbers ykaiim 7 10 Jun 2010, 00:33
Display posts from previous: Sort by

How many different prime numbers are factors of the positive

  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®.