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Number property question - Puzzled by OG explanation,

Expert replies
by ssy » Tue Oct 16, 2007 5:52 am
I am slightly stumped by this answer explanation given in OG11 (DS, Q132, Pg 331). Would appreciate it if someone could explain, thanks!

Q: If the integer n is greater than 1, is n equal 2?

1) n has exactly two positive factors
2) The difference of any two distinct positive factors of n is odd.

[spoiler Answer=B, statement 2 is sufficient]

OG's explanation:

1) If n has exactly two positive factors, then n is a prime number. However, n could be any prime number so statement 1 is not sufficient. (I agree)

2) [Copying their explanation word for word here.] "Note that if n>2 and n is odd, then 1 and n are factors of n, and their difference is even. Also, if n >2 and n is even,then 2 and n are factors of n, and their difference is even. Thus, no integer greater than 2 satisfies this statement. However, n=2 does satisfy this statement since 1 and 2 are the only positive factors of 2, and their difference is odd; Statement 2 alone is sufficient."

My question: Statement 2 only states that the difference of "ANY two distinct positive factors" of n is odd. I literally took this to mean ANY and not ONLY..I.e. 3 and 2 are distinct factors of 6 and their difference (=1) is odd. Therefore n can be 2 or n can be 6 and we cannot establish a unique value for n using statement 2.

However, using statements 1 and 2 combine, we can eliminate all non-prime numbers and the only prime number where its two distinct positive factors are odd is 2. Therefore, statements 1 and 2 are sufficient, in my opinion, but not statement 2 alone.

Where have I gone wrong? [/spoiler]
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Source: — Data Sufficiency |

ssy wrote:I literally took this to mean ANY and not ONLY..I.e. 3 and 2 are distinct factors of 6 and their difference (=1) is odd. Therefore n can be 2 or n can be 6 and we cannot establish a unique value for n using statement 2. [/spoiler]
This is where you went wrong:

Factors of 6: 1, 2, 3, 6
All combinations of two distinct positive factors of 6 (does not need to be prime):

1, 6
1, 2
1, 3
2, 3
2, 6
3, 6

(2) notes the difference between any two distinct factors, which means any and all combinations of two factors. (1,3) and (2,6) have an even difference so 6 cannot be n.

For any even greater than 2, the number itself (e.g. 4, 6, 8) and 2 will both be factors and since an even minus an even is even, no even number greater than 2 can possibly be n.

The constraints state that n is greater than 1 so you only need to check 2 and 3 now. 3 is prime so the only factors are 1 and 3, but the difference is also even so n cannot be 3. Thus 2 is the only choice for n.
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