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

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.

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

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

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.

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.

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.

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.

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.

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.

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!

Japinder

Dear hari003,
I'm happy to help.

We want to know the number of different prime number factors of n. In other words, when we take the prime factorization of n, how many primes do we see?

Statement #1: 4 different prime numbers are factors of 2n

This is a tricky one. If n = (3)(5)(7) = 105, then 2n = 210 would have four different prime number factors, {2, 3, 5, 7}. With this example, n started with just three different prime number factors.

By contrast, if n = (2)(3)(5)(7) = 210, then 2n = 420 would have four different prime number factors, {2, 3, 5, 7}. With this example, n started with four different prime number factors.

With this statement, it's possible for n to start with either three different prime number factors (not including 2) or with four different prime number factors (including 2). Because we can give two different answers to the prompt question, this statement, alone and by itself, is not sufficient.


Statement #2: 4 different prime numbers are factors of n^2

When we square a number, we change the total number of factors, but no new prime number factors are created in the process of squaring. It is impossible for (n^2) to have a prime factor that is not also a factor of n. Thus, however many different prime number factors n has, (n^2) has to have the exact same number. If (n^2) has four different prime number factors, then it absolutely must be true that n has four different prime number factors. This statement allows us to give a definitive numerical answer to the prompt question. This statement, alone and by itself, is sufficient.

First statement is not sufficient, second statement is. OA = (B).

Does all this make sense?
Mike

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

Hi Bunuel，very detailed explanation really appreciate it. This is exactly my thinking process when I was approaching this question.
But at the end I chose E, it's a weird choice I know. Here's my logic:

1. is not sufficient for sure.
2. I'm using things I learnt from sentence correction: 4 different prime numbers are factors of n^2
that doesn't necessarily mean there isn't a 5th, a 6th and so on.
Please correct me am I getting too nerdy on this or did I misunderstand the sentence.
It's been such a long time your reply may save my life in Quan if I encountered a similar situation. LOT!

4 different prime numbers are factors of n^2 means that EXACTLY 4 different prime numbers are factors of n^2.

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

(1) Four different prime numbers are factors of 2n.

Let's list down all possible cases of n.

Case 1:
\(n = p^a* q^b* r^c\) where \(p, q, r\) are distinct prime numbers and \(a, b, c \)are positive integers
\(2n = 2*p^a* q^b* r^c \)

In Case 1, no of different prime factors of 2n is 4 and of n is 3.
The different prime factors of 2n are \(2, p, q, r\) .
The different prime factors of n are \(p, q, r\)

Case 2:
\(n = 2*p^a* q^b* r^c \)
\(2n = 2^2*p^a* q^b* r^c \)
Here also no of different prime factors of 2n is 4 i.e. \(2,p,q,r\) but there are 4 different prime factors of n i.e \(2,p,q,r\)

Since no of prime factors of n could be 3 or 4, Statement 1 alone is insufficient.

(2) Four different prime numbers are factors of \(n^2\).

The no of different prime factors of \( n\) and \(n^2\) will be the same.

Example: n= 2*3, The different prime factors of n are 2 and 3
\(n^2\) = \((2*3)^2 \)= \(2^2 * 3^2 \). The prime factors of \(n^2\) are the same i.e 2 and 3.

So we can conclude that no of different prime factors of n is 4.
Statement 2 alone is sufficient.

Option B is the answer.

Thanks,
Clifin J Francis,
GMAT SME

The minimum value of a product of four different prime integers = 2 * 3 * 5 * 7 = 210
(1)
if 2n = 210 (4 prime factors), n = 105 (3 prime factors)
if 2n = 420 (4 prime factors), n = 210 (4 prime factors) … so we are not sure … n may have 3 or
4 prime factors. NS
(2)
If n is a positive integer, n and n2 will have an exact number of factors.
Imagine … n2 has four factors 2, 3, 5, 7 … then the minimum value of n2 will be 22 * 32 * 52 * 72 …
when we take the square root, we must have even powers of each of the factors.
So n will have the same number of prime factors as n2.
Ans. B