equal to 2

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If the integer n is greater than 1 is n equal to 2?

A) n has exactly two positive factors
B) the difference of any two distinct positive factors of n is odd.

[muzali]

A) any prime number has two positive factors - so insuff
B) for any prime number except 2, the difference between the two factors will always be even.
Let us consider some non-prime numbers:
4; factors are: 1,2,4 -the given condition in B is false for 4-2 (even)
9; factors are; 1,3,9 - again false.

So B gives us only one number, i.e., 2 . ---suff

Ans B

[cramya]

B)

Stmt I

n has exactly two positive factors

n is prime could be 2,3,5,7....etcc

INSUFF

Stmt II
the difference of any two distinct positive factors of n is odd

N has to be 2 since for any other prime the differecne is always even and for composites there are atleast 2 distinct factors whose difference will be even

SUFF

[schumi_gmat]

IMO C

A) is Insuff as stated by Cramya

B) the difference of any two distinct positive factors of n is odd.


IF the number has 3 factors, then difference between any 2 factors can be odd e.g 6 = 2*3*1

Here the diff between any 2 factors is odd.

Hence In suff.

A and B together

A) states that number is prime
B) the difference of any two distinct positive factors of n is odd, which is possible only if one factor is even and hence n =2

[Stuart@KaplanGMAT]

IMO C

A) is Insuff as stated by Cramya

B) the difference of any two distinct positive factors of n is odd.


IF the number has 3 factors, then difference between any 2 factors can be odd e.g 6 = 2*3*1

Here the diff between any 2 factors is odd.

Hence In suff.

In the example you chose, the difference between 3 and 1 is 2, which is even. Therefore, it's not true that if n=6 the difference of any two factors is odd.

As stated in the above posts, for all positive integers greater than 1, only 2 fits the bill if statement (2) must be true. Every other even integer has itself and 2 as factors (giving an even difference); every odd integer has itself and 1 as factors (giving an even difference).

And for the record, 6 has four factors: 1, 2, 3 and 6 (so 6-2 also gives us an even difference).

[parallel_chase]

IMO C

A) is Insuff as stated by Cramya

B) the difference of any two distinct positive factors of n is odd.


IF the number has 3 factors, then difference between any 2 factors can be odd e.g 6 = 2*3*1

you missed 6 there

6 has 4 factors 1, 2, 3, 6

6-2=4 even, cant be possible

Hence B.

Let me know if you have any other doubts.

[schumi_gmat]

Thanks Stuart and PC.

I realised the mistake I made. I was assuming that for B to be true we have to know that the number is prime and that where i went wrong.

From B, it is clear that only 2 will have 2 positive factors with difference 1. Any other number (prime or non-prime) does not matter.

Thanks again.