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How many different prime numbers are factors of the positive [#permalink]
 29 Jan 2012, 18:31
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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.
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Re: Factors of positive integer n [#permalink]
29 Jan 2012, 18:42
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Re: Factors of positive integer n [#permalink]
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: different-prime-factors-ds-95585.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.
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Re: Factors of positive integer n [#permalink]
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: different-prime-factors-ds-95585.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
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Re: Factors of positive integer n [#permalink]
30 Jan 2012, 02:50
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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: different-prime-factors-ds-95585.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.
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Re: Number Properties [#permalink]
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.
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Re: Number Properties [#permalink]
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*72n = 2^2*3*5*7 (still 4 prime factors)
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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.
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Re: Number Properties [#permalink]
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!  Regards, Arpan
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Re: Number Properties [#permalink]
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.
Ans B
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Re: Number Properties [#permalink]
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.
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Re: Number Properties [#permalink]
13 May 2013, 23:25
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Re: Number Properties
[#permalink]
13 May 2013, 23:25
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