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Re: How many different prime numbers are factors of the positive [#permalink]

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08 Oct 2010, 18:50

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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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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: How many different prime numbers are factors of the positive [#permalink]

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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)
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Re: How many different prime numbers are factors of the positive [#permalink]

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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

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Re: How many different prime numbers are factors of the positive [#permalink]

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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...........

Re: How many different prime numbers are factors of the positive [#permalink]

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28 Dec 2011, 08:57

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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 the positive [#permalink]

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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.
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Re: How many different prime numbers are factors of the positive [#permalink]

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11 Jan 2012, 14:59

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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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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: How many different prime numbers are factors of the positive [#permalink]

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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.

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.

Re: How many different prime numbers are factors of the positive [#permalink]

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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.

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

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.

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.

Re: How many different prime numbers are factors of the positive [#permalink]

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24 Apr 2013, 23:27

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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.

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

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)
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Re: How many different prime numbers are factors of the positive [#permalink]

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13 May 2013, 10:02

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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!

Re: How many different prime numbers are factors of the positive [#permalink]

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13 May 2013, 10:46

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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.

Re: How many different prime numbers are factors of the positive [#permalink]

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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.