Hi deepakk,

This type of DS question is based on some Number Properties, but if you don't know the NPs in question, you can prove/deduce them by TESTing Values.

We're asked how many DIFFERENT prime numbers are factors of the positive integer N. The word 'different' clues us in that duplicates, while allowed, are not counted. THAT Number Property will affect this question.

Fact 1: Four different prime numbers are factors of 2N.

Notice the '2' before the N? That is a prime that will affect what N could be.

If 2N = 210, then the prime factors are 2, 3, 5 and 7
So N = 105 and its prime factors are 3,5 and 7
The answer to the question is THREE.

If 2N = 420, then the prime factors are 2, 2, 3, 5, 7 (but we don't count duplicates)
So N = 210 and its prime factors are 2, 3, 5 and 7
The answer to the question is FOUR.
Fact1 is INSUFFICIENT.

Fact 2: Four different prime numbers are factors of N^2.

N^2 = (N)(N)

So the prime factors in the "first" parentheses will be the SAME as in the "second" parentheses.

(4 different prime factors)(same 4 prime factors from the first parentheses)

This means that N MUST have FOUR different prime factors
Fact 2 is SUFFICIENT.

Final Answer: B

GMAT assassins aren't born, they're made,
Rich

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 factors of n²

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