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



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Re: How many different prime numbers are factors of the positive [#permalink]
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24 Feb 2015, 18:06



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Re: How many different prime numbers are factors of the positive [#permalink]
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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.
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Re: How many different prime numbers are factors of the positive [#permalink]
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29 Apr 2015, 05:47
Here's another similar question for you to practice: How many different prime factors does positive integer N have?
(1) \(\frac{N}{2}\) has 2 different odd prime factors
(2) \(\frac{N}{3}\) has 2 different prime factorsWill post the OA and OE in a couple of days in the same thread. Please post your analysis below. Happy Solving! Best Regards Japinder
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Re: How many different prime numbers are factors of the positive [#permalink]
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29 Apr 2015, 09:30
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Nice question! As for (1) We are told that 2n contains 4 different primes. if n contains the factor 2, then it would be repeated and n would still include 4 primes. If n does not contain the factor 2, then n contains only 3 different primes. This gives us the two possibilities 3 and 4. As for (2) We are told that n^2 has 4 different primes. If we prime factorize n and square the factorization(e.g. (2*3*5*7)^2) we would simply be repeating the same factors(in our example(2*2*3*3*5*5*7*7) and no new primes would be introduced by the manipulation. Thus, B is sufficient.
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Re: How many different prime numbers are factors of the positive [#permalink]
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29 Apr 2015, 09:35
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EgmatQuantExpert wrote: Here's another similar question for you to practice: How many different prime factors does positive integer N have?
(1) \(\frac{N}{2}\) has 2 different odd prime factors
(2) \(\frac{N}{3}\) has 2 different prime factorsWill post the OA and OE in a couple of days in the same thread. Please post your analysis below. Happy Solving! Best Regards Japinder I'll go with A. The following is my reasoning: (1) N/2 Has 2 odd primes. 2 is a prime factor of N as we could divide by 2 and end up with an integer. So N = 2*o1*o2 where o1 and o2 are two different odd prime numbers. (2) N/3 has 2 different prime factors. We now have two possible options. 3 is obviously one of the prime factors, but it could be repeated if e.g. N=3*3*5 then N has 2 different primes. If N = 2*3*5 then N/3=2*5, resulting in 3 different primes. Thus A should be correct.
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Re: How many different prime numbers are factors of the positive [#permalink]
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07 May 2015, 03:37
EgmatQuantExpert wrote: Here's another similar question for you to practice: How many different prime factors does positive integer N have?
(1) \(\frac{N}{2}\) has 2 different odd prime factors
(2) \(\frac{N}{3}\) has 2 different prime factorsWill post the OA and OE in a couple of days in the same thread. Please post your analysis below. Happy Solving! Best Regards Japinder The correct answer is Option A.Statement 1: \(\frac{N}{2}\) has 2 different odd prime factorsLet the 2 different odd prime factors of \(\frac{N}{2}\) be OP1 and OP2. This means, we can write: \(N = 2*OP1^{a}*OP2^{b}\), where a and b are positive integers From here, it's clear that N has 3 prime factors, 1 of which is even and 2 are odd. Thus, Statement 1 is sufficient. Statement 2: \(\frac{N}{3}\) has 2 different prime factorsLet the 2 different prime factors of \(\frac{N}{3}\) be P1 and P2 This means, we can write: \(N = 3*P1^{a}*P2^{b}\), where a and b are positive integers If P1 = 3, then N has 2 prime factors But if P1 or P2 are different from 3, then N has 3 prime factors Therefore, Statement 2 is not sufficient to ascertain the exact number of prime factors of N. Hope this was helpful! Japinder
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Re: How many different prime numbers are factors of the positive [#permalink]
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08 Sep 2015, 08:20
3. 196!item!187;#058&000141 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.
The answer to this question is B (2) is sufficient, but (1) is insufficient. I got this wrong as I put C (both together are sufficient).
This is my understanding of why B is the correct answer: (1) is insufficient since if 2n is only divisible by 2 and not divisible by 4, then n would have one less prime factor than 2n. However, if 2n is divisible by 4, then 2n could have 4 different prime factors.
(2) is sufficient since if those 4 distinct prime numbers are factors of n^2, and given that n is an integer, then those 4 prime numbers should also be factors of n since the square root of any of those prime numbers would not be an integer. Therefore the answer is B.
Is this understanding correct? Please feel free to add any further interpretations that I might need.
Thanks.



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Re: How many different prime numbers are factors of the positive [#permalink]
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08 Sep 2015, 08:30
alexk91 wrote: 3. 196!item!187;#058&000141 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.
The answer to this question is B (2) is sufficient, but (1) is insufficient. I got this wrong as I put C (both together are sufficient).
This is my understanding of why B is the correct answer: (1) is insufficient since if 2n is only divisible by 2 and not divisible by 4, then n would have one less prime factor than 2n. However, if 2n is divisible by 4, then 2n could have 4 different prime factors.
(2) is sufficient since if those 4 distinct prime numbers are factors of n^2, and given that n is an integer, then those 4 prime numbers should also be factors of n since the square root of any of those prime numbers would not be an integer. Therefore the answer is B.
Is this understanding correct? Please feel free to add any further interpretations that I might need.
Thanks. Please search for a question before posting. This has already been discussed above.
Topics merged.Statement 1 is not sufficient for obvious reasons. Statement 2 is sufficient as prime factors of \(n^k\) (n, k \(\in\) positive integer) = prime factors of n . Remember this as a rule.When you square any integer, you are only multiplying the same integer with itself and as such not adding to the number of different prime factors. Your interpretation is also consistent with this. Case in point, \(n^2 = n*n\) , this means that \(n^2\) CAN NOT have different prime factors from those of 'n'. Hope this helps.



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Re: How many different prime numbers are factors of the positive [#permalink]
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28 Jan 2017, 12:51
1) If 2n has 4 different prime numbers as factors, n might either have 4 or 3 prime numbers as factors depending on whether 2 is one of the prime factors of n or not. If n is even, and 2 is one of its prime factors, then n has 4 different prime numbers as factors. If n is odd, and 2 is not one of its prime factors, then n has three different prime factors. 2) Perfect square has two identical sets of prime factors. n contains one of these sets, and thus has the same number of unique prime factors as n^2 does.
Statement 2 alone is sufficient, and statement 1 alone is not sufficient



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Re: How many different prime numbers are factors of the positive [#permalink]
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24 Feb 2017, 02:42
Superset The value of the number of prime numbers will be a positive integer
Translation In order to find the answer, we need: 1# exact value of n 2# exact number of prime factors on n 3# factorization of n
Statement analysis St 1: case1# if n is even. Then n will be having 4 prime factors. case2# if n is odd. Then n will be having 3 prime factors. INSUFFICIENT
St 2: prime factors of n^2 = prime factors of n. Therefore number of prime factors for n is 4. ANSWER
Option B



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Re: How many different prime numbers are factors of the positive [#permalink]
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27 Aug 2017, 09:28
How do we know if we are talking of the prime bases or the powers of those prime bases when doing prime factorisation in this question? I got confused between them. VeritasPrepKarishma wrote: 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)



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Re: How many different prime numbers are factors of the positive [#permalink]
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25 Oct 2017, 19:03
xyzgmat wrote: 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 Interesting question there is a subtle trick in statement 1 Statement 1 The trick is that 2 itself could be counted as one of the prime factors, or not , not enough info to tell Statement 2 Certainly B



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How many different prime numbers are factors of the positive [#permalink]
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30 Oct 2017, 13:18
First step is to write \(n\) in its prime factorized form. \(n = (P_1)^m * (P_2)^n * (P_3)^o * (P_4)^p...\) ; where \(P_1, P_2, P_3\) and \(P_4\) and so on are the prime factors. We don't know the number of prime factors or their powers. Statement 1: 4 different prime numbers are factors of 2nThis can be reworded as: 2n has 4 different prime factors. \(2*n =2* (P_1)^m * (P_2)^n * (P_3)^o ...\) ; One of the prime factors of n is 2 but we still don't know the other three. Insufficient.AD BCE Statement 2: 4 different prime numbers are factors of n^2A handy rule to remember is that \(n^x\) has the same number of prime factors has \(n\). Therefore, if \(n^2\) has 4 different prime factors, \(n\) also has 4 different prime factors. Sufficient.AD B CEAnswer is B
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How many different prime numbers are factors of the positive [#permalink]
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22 Nov 2017, 13:17
How many different prime numbers are factors of positive integer n? In other words, find the prime factorisation of n.
(1) 4 different prime numbers are factors of 2n Scenario 1: n originally had 3 prime factors and (2) was not one of them. E.g. n = (3)(5)(7)(8). 2n would give us 4 prime numbers (satisfying the statement) and n would have 3 prime factors. Scenario 2: n originally had 4 prime factors and (2) was one of those prime factors. E.g. n = (2)(3)(5)(7). 2n would give us 4 prime numbers (satisfying the statement) and n would have 4 prime factors. As there are at least 2 possible scenarios, this statement is insufficient.
(2) 4 different prime numbers are factors of n^2 (n)^2 would just mean that all of the prime numbers in n's originally prime factorisation are multiplied twice. [E.g.: (25^2) = (5*5)^2 = (5*5*5*5)]. That means the number of different prime factors in n^2 would be the same number of different factors in n. Sufficient.




How many different prime numbers are factors of the positive
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