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nandansingh wrote:How many different prime numbers are factors of positive integer n?
1.Four different prime numbers are factors of 2n.
2.Four different prime numbers are factors of n^2.

OA: b[/list]

Statement I
Four different prime numbers are factors of 2n

let n=3*5*7, prime factors 3
2n = 2*5*7*3, prime factors, 4

Let n= 2*3*5*7,prime factors 4
2n = 2*2*3*5*7, prime factors 4

Insufficient.

Statement II

if n^2 has four different factors, n will also have 4 different prime factors. Because when we square an integer we are actually raising the powers of its prime factors.

Sufficient.

Hence B.

Hope this helps.

No rest for the Wicked....

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Re: Prime no. question (n,n^2) from prep1

by stubbornp » Fri Oct 31, 2008 12:25 am

nandansingh wrote:How many different prime numbers are factors of positive integer n?
1.Four different prime numbers are factors of 2n.
2.Four different prime numbers are factors of n^2.

OA: b[/list]

Stmt 1:We don't know whether N is odd or even...If

If N is odd,We are getting one more prime factor 2.

If N is even,We aleready have prime factor 2...-----Insufficient

Stmt 2:In case of n^2 has four different prime numbers,Then it is must that n has same # of prime factors---sufficient

B...hope it helps

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by cramya » Fri Oct 31, 2008 4:59 am

B)

Parallel Chase's explanation is what I came up with also. With 2n's different prime factors we cant determine the different prime factors of n(2 could already be present in n or not hence stmt I is insufficient) but with n^2 we can(which will be the same as n's so sufficient)