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How many different prime numbers are factors of the positive integer n

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

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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.

Originally posted by xyzgmat on 08 Oct 2010, 17:50.
Last edited by Bunuel on 05 Jun 2019, 03:04, edited 2 times in total.
Renamed the topic and edited the question.
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Re: How many different prime numbers are factors of the positive integer n  [#permalink]

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New post 08 Oct 2010, 19:11
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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.

Answer: B.

Similar question: different-prime-factors-ds-95585.html?hilit=produce%20prime#p735955

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

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New post 28 Dec 2011, 02:22
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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 integer n  [#permalink]

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

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New post 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 integer n  [#permalink]

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New post 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.

Answer is B.
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Re: How many different prime numbers are factors of the positive integer n  [#permalink]

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

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New post 29 Jan 2012, 18:42
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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.

Answer: B.

Similar question: different-prime-factors-ds-95585.html

Hope it helps.
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Re: How many different prime numbers are factors of the positive integer n  [#permalink]

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New post 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.html

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

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New post 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.html

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

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New post 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.

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

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New post 25 Apr 2013, 01:52
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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 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 integer n  [#permalink]

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

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New post 24 Feb 2015, 19:06
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.


1 is most definitely NOT a prime number. As for your other question please refer to the example in my solution above.
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Re: How many different prime numbers are factors of the positive integer n  [#permalink]

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New post 07 May 2015, 04:37
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 factors


Will post the OA and OE in a couple of days in the same thread. Please post your analysis below. Happy Solving! :-D

Best Regards

Japinder



The correct answer is Option A.

Statement 1:
\(\frac{N}{2}\) has 2 different odd prime factors


Let 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 factors


Let 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! :)

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

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New post 24 Feb 2017, 03:42
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 integer n  [#permalink]

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New post 30 Oct 2017, 14:18
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 2n

This 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^2

A 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 BCE

Answer is B
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Re: How many different prime numbers are factors of the positive integer n  [#permalink]

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New post 22 Nov 2017, 14:17
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.
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Re: How many different prime numbers are factors of the positive integer n  [#permalink]

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New post 07 Aug 2019, 10:05
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xyzgmat 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.


Target question: How many different prime numbers are factors of the positive integer n?

IMPORTANT: A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k (or a factor of k), then k is "hiding" within the prime factorization of N

Examples:
24 is divisible by 3 <--> 24 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7

Given all of this, we can phrase the target question as: How many different prime numbers are "hiding" in the prime factorization of n?

Statement 1: 4 different prime numbers are factors of 2n
There are several value of n that meet this condition. Here are two:
Case a: n = (3)(5)(7), in which case there are 3 different prime numbers "hiding" in the prime factorization of n
Case b: n = (2)(3)(5)(7), in which case there are 4 different prime numbers "hiding" in the prime factorization of n
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statement 2: 4 different prime numbers are factors of n²
IMPORTANT: squaring an integer has no effect on the number of different prime numbers hiding in its prime factorization.
For example, if n=(2)(3)(5), then n has 3 different prime numbers hiding in its prime factorization.
Also n² = (2)(3)(5)(2)(3)(5), so n² still has 3 different prime numbers hiding in its prime factorization.
So, if n² has 4 different prime numbers hiding in its prime factorization, then we can be certain that n has 4 different prime numbers hiding in its prime factorization
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent
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Re: How many different prime numbers are factors of the positive integer n   [#permalink] 07 Aug 2019, 10:05
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