1) shows that n is prime so n can be 2,3,5 , 7 and so on . Thus Insufficient.

2) we know that n>1. Now there are two possibilities n is Even or n is Odd


If n is Even --> we know that n can be 2 or an even number greater than 2. If n = 2 then both two factors 1 and 2 have a difference of 1 = Odd integer. However, if the number is greater than 2 then 2 is definitely a factor of any other Even number and the difference will always be Even as Even - 2 = Even. Thus the only possibility in this case will be n = 2

If n is Odd --> We know that 1 will be a factor of n. n is also a factor of n. Thus, n - 1 = Odd - Odd = Even. Thus, Any odd integer doesn't satisfy this condition.



Thus, the only integer that satisfies this condition is 2. Thus Sufficient.

but statement second says that difference of any distinct factor
so if we take 6,1,2,3 and 6 are factors of 6 and 6-3=3 , this satisfy the given condition , right???
hence statement second alone is not sufficient.. if i'm wrong .. please explain

suryapal wrote:but statement second says that difference of any distinct factor
so if we take 6,1,2,3 and 6 are factors of 6 and 6-3=3 , this satisfy the given condition , right???
hence statement second alone is not sufficient.. if i'm wrong .. please explain

Read the second option a bit more carefully ...

(2) the difference of any two distinct positive factors of n is odd

It means you have to consider all the pairs that you can make out of its factors.

Take for e.g 6, the # that you took. You are correct in saying that 1,2,3 and 6 are it's positive factors and difference of 6 and 3 is odd. But, what about difference of 3 and 1. Is it odd? The condition should satisfy for ANY two factors of the number.

and that is why 2 is sufficient :

* consider any prime greater than 2. It will have 1 and the number itself as the only factors. But, the difference of the two factors (only pair) will be even.

* Consider any other odd composite #. it will have atleast two odd factors and hence, difference of these two factors will be even.

*any even # greater than two. It will also have 2 and that number as its factor and hence, won't satisfy the condition.

NOW, consider 2.

its factors 1 & 2 . We can make just a single pair out of these two factors and their difference is ODD. which is what we want to know.

Therefore, second statement is sufficient to tell that the number is 2.

ohh i got it ... thanks