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Last edited by nandansingh on Fri Oct 31, 2008 9:57 am, edited 1 time in total.
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Statement Inandansingh 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]
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.
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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]
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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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)
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)