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How many different prime factors does N have? (1) 2N has

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How many different prime factors does N have? (1) 2N has [#permalink]

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New post 24 Mar 2005, 13:23
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How many different prime factors does N have?

(1) 2N has 4 different prime factors.
(2) N ^2 has 4 different prime factors.

OPEN DISCUSSION OF THIS QUESTION IS HERE: how-many-different-prime-numbers-are-factors-of-the-positive-126744.html
[Reveal] Spoiler: OA
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New post 24 Mar 2005, 13:46
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How many different prime factors does N have?

(1) 2N has 4 different prime factors.
(2) N ^2 has 4 different prime factors.


Lets look at (1).
2N has 4 different prime factors
=> If N is odd, then N has 3 prime factors(Example N = 3x5x7)
If N is even, then N has 4 prime factors.(Example N = 2x3x5x7)
Therefor (1) is not sufficient by itself.

Lets look at (2)
N ^2 has 4 different prime factors.
Any number N, can be written as product of prime factors(not all distinct).
N = p1 x p2 x ..... x pm

=> N^2 = p1^2 x p2^2 x p3^2 x ..........x pm ^2

You can see that N^2 will have the same set of prime factors, only repeated twice. Therefore if N^2 has 4 distinct prime factors, N will also have 4 distinct prime factors.
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New post 24 Mar 2005, 14:36
"B"

state 1.....2N....N can have 4 distinct factor one involving 2, or it can have 3 distinct factors not involving 2.....insuff


state 2...

If N^2....has 4 prime factors, so does N.....e.g. N = 2, # prime factor is 1...N^2...2^2....still 1 prime factor....suff
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Re: How many different prime factors does N have? (1) 2N has [#permalink]

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New post 22 Oct 2013, 08:50
Answer is B.

post the OA.

2N -- > introduces a multiplication, which introduces a prime number.
so it cannot be said if N had 2 or not.

where as N square -- > clearly does not introduce new primes, it just changes their order.


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Re: How many different prime factors does N have? (1) 2N has [#permalink]

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New post 22 Oct 2013, 08:53
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: how-many-different-prime-numbers-are-factors-of-the-positive-126744.html
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Re: How many different prime factors does N have? (1) 2N has   [#permalink] 22 Oct 2013, 08:53
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