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Re: DS question : need help [#permalink]
08 Oct 2010, 17:50

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=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
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Re: DS question : need help [#permalink]
08 Oct 2010, 18:11

4

This post received KUDOS

Expert's post

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.

Re: DS question : need help [#permalink]
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)
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"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??

Re: How many different prime numbers are factors of +ve integer [#permalink]
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

Re: How many different prime numbers are factors of +ve integer [#permalink]
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...........

Re: How many different prime numbers are factors of +ve integer [#permalink]
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.

Re: How many different prime numbers are factors of +ve integer [#permalink]
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.
_________________

Re: How many different prime numbers are factors of +ve integer [#permalink]
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.

Re: How many different prime numbers are factors of +ve integer [#permalink]
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.
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