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

 It is currently 04 Jul 2015, 06:13

### 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

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

# 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:
Director
Status: Finally Done. Admitted in Kellogg for 2015 intake
Joined: 25 Jun 2011
Posts: 552
Location: United Kingdom
Concentration: International Business, Strategy
GMAT 1: 730 Q49 V40
GPA: 2.9
WE: Information Technology (Consulting)
Followers: 29

Kudos [?]: 1166 [4] , given: 217

How many different prime numbers are factors of the positive [#permalink]  29 Jan 2012, 17:31
4
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

58% (01:38) correct 42% (01:04) wrong based on 531 sessions
How many different prime numbers are factors of the positive integer n?

(1) Four different prime numbers are factors of 2n.
(2) Four different prime numbers are factors of n^2.
[Reveal] Spoiler: OA

_________________

Best Regards,
E.

MGMAT 1 --> 530
MGMAT 2--> 640
MGMAT 3 ---> 610
GMAT ==> 730

Math Expert
Joined: 02 Sep 2009
Posts: 28272
Followers: 4471

Kudos [?]: 45161 [14] , given: 6646

Re: Factors of positive integer n [#permalink]  29 Jan 2012, 17:42
14
KUDOS
Expert's post
11
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.

Similar question: different-prime-factors-ds-95585.html

Hope it helps.
_________________
Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 178
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)
Followers: 3

Kudos [?]: 39 [0], given: 1

Re: Factors of positive integer n [#permalink]  29 Jan 2012, 19:57
Bunuel wrote:
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.

Similar question: different-prime-factors-ds-95585.html

Hope it helps.

Hi Bunnel,
for statement 1: if n has 2 as a factor(e.g.: 2*3*4*5) then 2n becomes 2*2*3*4*5 then n has 4 factors, but if n dosn't has 2 as a factor (e.g.: 3*4*5) then 2 n becomes 2*3*4*5 then also n has 4 factors.

so both statement alone are sufficient.
Intern
Joined: 08 Dec 2011
Posts: 39
Location: United States
WE: Education (Education)
Followers: 0

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

Re: Factors of positive integer n [#permalink]  29 Jan 2012, 20:16
subhajeet wrote:
Bunuel wrote:
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.

Similar question: different-prime-factors-ds-95585.html

Hope it helps.

Hi Bunnel,
for statement 1: if n has 2 as a factor(e.g.: 2*3*4*5) then 2n becomes 2*2*3*4*5 then n has 4 factors, but if n dosn't has 2 as a factor (e.g.: 3*4*5) then 2 n becomes 2*3*4*5 then also n has 4 factors.

so both statement alone are sufficient.

1) 4 different numbers are prime factors of 2n.
2n could equal 2*3*5*7 in which case n would equal 3*5*7. So in this case n would have only 3 prime factors.
But it is also possible that 2n = 2*2*3*5*7. In this case n=2*3*5*7 and so n has 4 prime factors.
So Bunuel is right. Statement 1 is not sufficient
Math Expert
Joined: 02 Sep 2009
Posts: 28272
Followers: 4471

Kudos [?]: 45161 [2] , given: 6646

Re: Factors of positive integer n [#permalink]  30 Jan 2012, 01:50
2
KUDOS
Expert's post
subhajeet wrote:
Bunuel wrote:
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.

Similar question: different-prime-factors-ds-95585.html

Hope it helps.

Hi Bunnel,
for statement 1: if n has 2 as a factor(e.g.: 2*3*4*5) then 2n becomes 2*2*3*4*5 then n has 4 factors, but if n dosn't has 2 as a factor (e.g.: 3*4*5) then 2n becomes 2*3*4*5 then also n has 4 factors.

so both statement alone are sufficient.

In your own example "if n dosn't has 2 as a factor (e.g.: 3*4*5)" n has only 3 factors not 4.

(1) says: 4 different prime numbers are factors of 2n. Now, if 2 is not a prime factor of n then 2n would have one more prime than n (this same exact 2), thus n has 3 prime factors. But if 2 is already a prime factor of n then 2n has the same number of prime factors as n.

Hope it's clear.
_________________
Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 629
Followers: 57

Kudos [?]: 723 [0], given: 135

Re: Number Properties [#permalink]  24 Apr 2013, 22:27
Expert's post
1
This post was
BOOKMARKED
mannava189 wrote:
How many different prime numbers are factors of the positive integer n ?
(1) Four different prime numbers are factors of 2n.
(2) Four different prime numbers are factors of n^2.

From F.S 1, for n = 3*5*7, we have 3 different prime factors for n. However, for n = 2*3*5*7, we would have 4 different prime factors for n. Insufficient.

From F.S 2, we know that $$n^2 = a*b*c*d$$,where a,b,c,d are primes.Thus, as n is also a positive integer, n = $$\sqrt{a*b*c*d}$$ = integer. As square root of any prime number is not an integer, thus, for n to be an integer, all the primes have to be in some even power form.

Thus, the number of prime factors for n = 4.Sufficient.

B.
_________________
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 5682
Location: Pune, India
Followers: 1413

Kudos [?]: 7336 [9] , given: 186

Re: Number Properties [#permalink]  25 Apr 2013, 00:52
9
KUDOS
Expert's post
2
This post was
BOOKMARKED
mannava189 wrote:
How many different prime numbers are factors of the positive integer n ?
(1) Four different prime numbers are factors of 2n.
(2) Four different prime numbers are factors of n2.

The question will be very straight forward if you just understand the prime factorization of a number.

When we say that n has 4 distinct prime factors, it means that

$$n = 2^a * 3^b * 5^c * 7^d$$ or $$n = 2^a * 5^b * 11^c * 17^d$$ or $$n = 23^a * 31^b * 59^c * 17^d$$ etc

What if we square n?
We get $$n^2 = 2^{2a} * 3^{2b} * 5^{2c} * 7^{2d}$$ etc
Notice that the number of prime factors will not change.
So if we know that n^2 has 4 prime factors, we can say that n MUST have 4 prime factors too. So statement 2 alone is sufficient.

(2) Four different prime numbers are factors of 2n.

There is a complication here. 2n introduces a new prime number 2. We don't know whether n had 2 before or not.

2n has 4 different prime factors implies n can have either 3 different prime factors or 4 different prime factors. Hence this statement alone is not sufficient. Let's look at examples:

Case 1:
n has 3 different prime factors.
Say, $$n = 3*5*7$$ (3 prime factors)
$$2n = 2*3*5*7$$ (4 prime factors)

Case 2:
n has 4 different prime factors
Say,$$n = 2*3*5*7$$
$$2n = 2^2*3*5*7$$ (still 4 prime factors)
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Manager
Status: *Lost and found*
Joined: 25 Feb 2013
Posts: 123
Location: India
Concentration: General Management, Technology
GMAT 1: 640 Q42 V37
GPA: 3.5
WE: Web Development (Computer Software)
Followers: 15

Kudos [?]: 84 [1] , given: 14

Re: Number Properties [#permalink]  13 May 2013, 09:02
1
KUDOS
coolpintu wrote:
How many different prime numbers are factors of the positive integer n?
(1) four different prime numbers are factors of 2n
(2) four different prime numbers are factors of n^2.

I feel the answer should be [B]. Please find below my explanation:

Statement 1. Since its mentioned, there are 4 different prime factors under 2n, 2 must be one of them, so the question arises can be simply subtract 1 from 4 and claim that n has 3 different prime factors?! But in case n already has 2 as a factor, simply subtracting would not give you the exact count. For example is 2*n = 2*2*3, the total different prime factors of 2n would be 2 but when it comes to n, the value is still 2!

Hence, since we are not sure if n is even or not, we cannot say for sure that n has 3 prime factors or not. Hence, Insufficient.

Statement 2. Since its n^2, all the prime factors under n must be doubled. No exponent will be 1. Hence if the term n^2 has 4 different prime factors, it can be assumed that n too has 4 different prime factors. Hence, sufficient!

Therefore the answer should be [B]. Hope I am correct!

Regards,
Arpan
_________________

Feed me some KUDOS! *always hungry*

My Thread : Recommendation Letters

Manager
Joined: 11 Jun 2010
Posts: 84
Followers: 0

Kudos [?]: 11 [1] , given: 17

Re: Number Properties [#permalink]  13 May 2013, 09:46
1
KUDOS
How many different prime numbers are factors of the positive integer n?
(1) four different prime numbers are factors of 2n
(2) four different prime numbers are factors of n^2

St1: If 2 is one of the primes, then factors of 2N = {2,3,5,7} or 2N = 210. N then becomes 105 and N has 3 distinct prime factors {3,5,7}
But if 2 is not one of the primes, and if {3,5,7,11} are 4 distinct prime factors of 2N, then 2N = 1155. N becomes 577.5 and N has distinct 4 prime factors {3,5,7,11}.
Hence INSUFFICIENT

TAkeAwAY: If 2 is one of the prime factors of 2N, then N has x-1 primes (excluding 2) but if 2 is not one of the prime factors of 2N, then N has as many distinct primes factors as 2N itself.

ST2: 4 different prime numbers are factors of n^2
N^2 = (2*3*5*7)^2 then n is a multiple of (2*3*5*7)
thus # of primes in N will be the same as # of primes in n^2
Another example if n^2 = (3*5*13*19)^2, then n must be a multiple of (3*5*13*19)
Hence sufficient. There will be 4 different prime factors of N.

Ans B
Manager
Joined: 26 Feb 2013
Posts: 53
Concentration: Strategy, General Management
GMAT 1: 660 Q50 V30
WE: Consulting (Telecommunications)
Followers: 0

Kudos [?]: 6 [0], given: 16

Re: Number Properties [#permalink]  13 May 2013, 09:50
My answer is B.

from Stmt 1.
four different prime numbers are factors of 2n.
Let 2n = 2*3*5*7. in this case n= 3*5*7..meaning n has three prime factors.
let 2n = 2^3 *3*5*7. in this case n= 2^2 *3*5*7..meaning n has 4 prime factors.

hence its insufficient.

from stmt 2:
four different prime numbers are factors of n^2

let n^2 = a^2p * b^2q*c^2r*d^2s..where a,b,c,d are prime numbers and p,q,r,s are integers(as it is given that n is an integer).. taking square root on both sides
n = a^p * b^q*c^r*d^s...meaning n will have 4 prime factors.

sufficient.
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 5368
Followers: 310

Kudos [?]: 60 [0], given: 0

Re: How many different prime numbers are factors of the positive [#permalink]  03 Aug 2014, 00:13
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.
_________________
Intern
Joined: 21 Feb 2015
Posts: 27
Followers: 0

Kudos [?]: 2 [0], given: 11

Re: How many different prime numbers are factors of the positive [#permalink]  24 Feb 2015, 17:18
I do not understand your explanation for statement one. Also 1 is a prime number too therefore 2n can have the following prime factors 1, 3, 5, 7. Please correct me if I am wrong. I just don't understand what makes you assumed that 2n can have less than four prime factors when the question states that 2n has four prime factors.
Math Expert
Joined: 02 Sep 2009
Posts: 28272
Followers: 4471

Kudos [?]: 45161 [0], given: 6646

Re: How many different prime numbers are factors of the positive [#permalink]  24 Feb 2015, 18:06
Expert's post
mawus wrote:
I do not understand your explanation for statement one. Also 1 is a prime number too therefore 2n can have the following prime factors 1, 3, 5, 7. Please correct me if I am wrong. I just don't understand what makes you assumed that 2n can have less than four prime factors when the question states that 2n has four prime factors.

1 is most definitely NOT a prime number. As for your other question please refer to the example in my solution above.
_________________
Chat Moderator
Joined: 19 Apr 2013
Posts: 508
Concentration: Strategy, Entrepreneurship
GPA: 4
Followers: 7

Kudos [?]: 48 [0], given: 327

Re: How many different prime numbers are factors of the positive [#permalink]  02 Mar 2015, 08:12
mawus wrote:
I do not understand your explanation for statement one. Also 1 is a prime number too therefore 2n can have the following prime factors 1, 3, 5, 7. Please correct me if I am wrong. I just don't understand what makes you assumed that 2n can have less than four prime factors when the question states that 2n has four prime factors.

If it is difficult for you to understand or remember why 1 is not a prime number, remember that 1 can be deleted only to one number. However, prime numbers must be able to be deleted only by two numbers.
_________________

If my post was helpful, press Kudos. If not, then just press Kudos !!!

Re: How many different prime numbers are factors of the positive   [#permalink] 02 Mar 2015, 08:12
Similar topics Replies Last post
Similar
Topics:
11 How many different prime numbers are factors of the positive 16 08 Oct 2010, 16:50
25 If K is a positive integer, how many different prime numbers 7 10 Jun 2010, 00:33
How many different prime numbers are factors of the positive 4 22 Oct 2006, 22:37
How many different prime nos are factors of the positive 5 30 Jul 2006, 04:22
Display posts from previous: Sort by

# How many different prime numbers are factors of the positive

 Question banks Downloads My Bookmarks Reviews Important topics

 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®.