How many different prime numbers are factors of the positive integer n?
1. 4 different prime number ae factors of 2n
2. 4 different prime numbers are facors of n^2
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teejaycrown
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manngmat2013
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1. 2*n , can have 2 as a different prime factor. so it is insufficient.
example: n can be 2*3*5*7 , then 2n will have the same different prime factors.
but if n is 3*5*7, then 2n will be 2*3*5*7, the number can have either 3 or 4 differnt factors.
2. n*n , will have only those prime factor with doubled power.
example: n =2*3*5*7, then n*n = 2*2*3*3*5*5*7*7, the different factors will remain same.
hence Statement 2 is suffiecient.
example: n can be 2*3*5*7 , then 2n will have the same different prime factors.
but if n is 3*5*7, then 2n will be 2*3*5*7, the number can have either 3 or 4 differnt factors.
2. n*n , will have only those prime factor with doubled power.
example: n =2*3*5*7, then n*n = 2*2*3*3*5*5*7*7, the different factors will remain same.
hence Statement 2 is suffiecient.
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adaftprig
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If 4 different prime numbers are factors of 2n then one of the 4 numbers is definitely 2. That would mean there are 3 numbers which are factors of n.
If 4 numbers a b c and d are prime factors of n then the prime factors of n^2 are a^2, b^2, c^2 and d^2.
If 4 numbers a b c and d are prime factors of n then the prime factors of n^2 are a^2, b^2, c^2 and d^2.
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Brent@GMATPrepNow
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Target question: How many different prime numbers are factors of the positive integer nteejaycrown wrote:How many different prime numbers are factors of the positive integer n?
1. 4 different prime numbers are factors of 2n
2. 4 different prime numbers are facors of n^2
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 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 n^2
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 n^2 = (2)(3)(5)(2)(3)(5), so n^2 still has 3 different prime numbers hiding in its prime factorization.
So, if n^2 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
Brent Hanneson - Creator of GMATPrepNow.com
