Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 08 Oct 2010
Posts: 1

How many different prime numbers are factors of the positive
[#permalink]
Show Tags
Updated on: 08 Sep 2015, 09:37
Question Stats:
59% (00:42) correct 41% (01:02) wrong based on 1753 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by xyzgmat on 08 Oct 2010, 17:50.
Last edited by ENGRTOMBA2018 on 08 Sep 2015, 09:37, edited 1 time in total.
Merged identical topics and updated the source tag to GMATPREP




Math Expert
Joined: 02 Sep 2009
Posts: 49381

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
08 Oct 2010, 19:11




Senior Manager
Status: Not afraid of failures, disappointments, and falls.
Joined: 20 Jan 2010
Posts: 275
Concentration: Technology, Entrepreneurship
WE: Operations (Telecommunications)

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
08 Oct 2010, 18:50
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=61. 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 2n1=61=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
_________________
"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??



Senior Manager
Status: Not afraid of failures, disappointments, and falls.
Joined: 20 Jan 2010
Posts: 275
Concentration: Technology, Entrepreneurship
WE: Operations (Telecommunications)

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
08 Oct 2010, 22:55
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)
_________________
"I choose to rise after every fall" Target=770 http://challengemba.blogspot.com Kudos??



Intern
Joined: 24 Dec 2011
Posts: 5

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
28 Dec 2011, 02:22
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



Senior Manager
Joined: 28 Jul 2011
Posts: 389
Location: United States
Concentration: International Business, General Management
GPA: 3.86
WE: Accounting (Commercial Banking)

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
28 Dec 2011, 08:42
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
Joined: 24 Dec 2011
Posts: 5

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
28 Dec 2011, 08:57
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
Joined: 26 Jun 2011
Posts: 242
Location: India

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
28 Dec 2011, 09:00
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 ...



Manager
Joined: 29 Jul 2011
Posts: 94
Location: United States

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
11 Jan 2012, 14:59
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 = 34 min



Math Expert
Joined: 02 Sep 2009
Posts: 49381

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
29 Jan 2012, 18:42



Manager
Status: MBA Aspirant
Joined: 12 Jun 2010
Posts: 143
Location: India
Concentration: Finance, International Business
WE: Information Technology (Investment Banking)

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
29 Jan 2012, 20:57
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. Answer: B. Similar question: differentprimefactorsds95585.htmlHope 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
Concentration: Leadership, Nonprofit
WE: Education (Education)

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
29 Jan 2012, 21:16
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. Answer: B. Similar question: differentprimefactorsds95585.htmlHope 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: 49381

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
30 Jan 2012, 02:50
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. Answer: B. Similar question: differentprimefactorsds95585.htmlHope 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  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? Extrahard Quant Tests with Brilliant Analytics



Verbal Forum Moderator
Joined: 10 Oct 2012
Posts: 614

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
24 Apr 2013, 23:27
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.
_________________
All that is equal and notDeep Dive Inequality
Hit and Trial for Integral Solutions



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8292
Location: Pune, India

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
25 Apr 2013, 01:52
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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Status: *Lost and found*
Joined: 25 Feb 2013
Posts: 122
Location: India
Concentration: General Management, Technology
GPA: 3.5
WE: Web Development (Computer Software)

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
13 May 2013, 10:02
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: 74

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
13 May 2013, 10:46
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 x1 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: 52
Concentration: Strategy, General Management
WE: Consulting (Telecommunications)

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
13 May 2013, 10:50
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.



Math Expert
Joined: 02 Sep 2009
Posts: 49381

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
12 Jun 2013, 04:27



Intern
Joined: 21 Feb 2015
Posts: 26

Re: How many different prime numbers are factors of the positive
[#permalink]
Show Tags
24 Feb 2015, 18:18
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.




Re: How many different prime numbers are factors of the positive &nbs
[#permalink]
24 Feb 2015, 18:18



Go to page
1 2
Next
[ 34 posts ]



