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1 Insufficient 2- Sufficient: 1 is always a factor, so x can not be odd, for all other even they have themselves and 2 which will give even sum (Fac(4)= 4,2,1) so only two has two factors ( 2,1).

both conditions are required ....sum of two factors can be odd for lot os number but the first condition will determine that 2 is the only number which has exactly two factors ( 1&2) whose addition is odd ....

If the integer x is greater than 1, does x = 2? 1) x is evenly divisible by exactly two positive integers. 2) The sum of any two distinct positive factors of x is odd.

1 Insufficient 2- Sufficient: 1 is always a factor, so x can not be odd, for all other even they have themselves and 2 which will give even sum (Fac(4)= 4,2,1) so only two has two factors ( 2,1).

is it correct because i found an answer says C ?

You are right B should be the answer for this question and your reasoning for B is correct.

If the integer x is greater than 1, does x = 2?

Given: \(x=integer>1\). Question: does \(x=2\)?

(1) x is evenly divisible by exactly two positive integers --> \(x\) can be any prime (2, 3, 5, 7, ...). Not sufficient.

(2) The sum of ANY TWO distinct positive factors of x is odd If \(x\) is odd, then 1 (obviously one of the factors of \(x\)) plus \(x\) itself (\(x\) is also a factor of itself) will be even not odd, so \(x\) cannot be odd. Now if \(x\) is even and more than 2, then 2 (one of the factors of \(x\)) plus \(x\) itself will be even not odd, so \(x\) cannot be more than 2 --> \(x=2\) (2 has exactly two positive factors (1 and 2) and their sum is odd). Sufficient.

If the integer x is greater than 1, does x = 2? 1) x is evenly divisible by exactly two positive integers. 2) The sum of any two distinct positive factors of x is odd.

1 Insufficient 2- Sufficient: 1 is always a factor, so x can not be odd, for all other even they have themselves and 2 which will give even sum (Fac(4)= 4,2,1) so only two has two factors ( 2,1).

is it correct because i found an answer says C ?

i still have confusion with B.

it says sum of any two distinct factors of X is odd. take 10 for ex: factors - 1,2,5,10. we have 2+5=7 which is odd, thus we have numbers greater than 2 with the condition satisfied.

If the integer x is greater than 1, does x = 2? 1) x is evenly divisible by exactly two positive integers. 2) The sum of any two distinct positive factors of x is odd.

1 Insufficient 2- Sufficient: 1 is always a factor, so x can not be odd, for all other even they have themselves and 2 which will give even sum (Fac(4)= 4,2,1) so only two has two factors ( 2,1).

is it correct because i found an answer says C ?

i still have confusion with B.

it says sum of any two distinct factors of X is odd. take 10 for ex: factors - 1,2,5,10. we have 2+5=7 which is odd, thus we have numbers greater than 2 with the condition satisfied.

while considering both A&B, its sure that x=2.

am i missing something?

Stem says the sum of ANY two distinct positive factors of x is odd. It doesn't mean that the sum of some factors is odd but rather that the sum of every possible pair of distinct positive factors is odd.

In your example: 2+5=7=odd but 2+10=12=even.

Consider this: "the sum of any two distinct positive factors of x is odd" means that number must have only one odd factor and only one even factor. If odd factors, or even factors >1, then the sum of the pair of two odd factors (or even factors) will be even not odd. Only number to have only one odd and only one even factor is 2.

Re: If the integer x is greater than 1, does x = 2? 1) x is [#permalink]

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16 May 2013, 03:23

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Expert's post

nicechetan87 wrote:

If X=2 then ..... 1+2 = 3 (ODD) -------> X is 2 If X=4 then ..... 1+4 = 5 (ODD) -------> X is not 2

This part is incomplete. The problem states that sum of ANY 2 distinct positive factors of x is ODD--> This implies that any two positive factors when added will ALWAYS give an odd sum.

For x = 4, all the positive factors are = 1,2,4. If you take 1 & 4, we indeed get an odd sum, however, if we take 2 & 4, we get an even sum.Thus \(x\neq{4}\)

The point to note is that :

odd+even = odd. Any other arrangement will always give you an even sum for integers.X will have at-least one odd factor(that being 1). Now to make the sum of any of the two factors as odd, X will have to have at-least 1 even factor. If X has more than 2 factors(1 already being odd and 1 already being even) that again would be odd/even. Let us assume the no of factors for x is 3 and they are odd,even,odd.Thus, as per the problem, taking sum of any two factors must always yield an odd sum, which gets defied for (odd,odd). Similarly for 3 factors of x which are odd,even,even. This implies that x can only have 2 factors, one of them odd and one of them even. The only positive integer to have 1 odd & 1 even factor is 2. Sufficient. _________________

Re: If the integer x is greater than 1, does x = 2? 1) x is [#permalink]

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16 Jun 2013, 02:56

nicechetan87 wrote:

Hi All,

Answer for this question has to be C.

Pt. 2 Says - 2) The sum of any two distinct positive factors of x is odd.

If X=2 then ..... 1+2 = 3 (ODD) -------> X is 2 If X=4 then ..... 1+4 = 5 (ODD) -------> X is not 2

So B is not sufficient.

However if I combine Pt. 1 to it which says 1) x is evenly divisible by exactly two positive integers. ---- X is Prime

Then only we can come at a conclusion that X=2

Hence C is the answer. Hope I am Clear.

hey nope any two factors doesn't mean as u have considered, it means for 4 factors 1,2,4 of these any two , so 2 and 4 can be considered which is not correct. and hence 4 is not the valid number to check only 2 is the number avialable in NS so b is quite good

If the integer x is greater than 1, does x = 2? 1) x is evenly divisible by exactly two positive integers. 2) The sum of any two distinct positive factors of x is odd.

1 Insufficient 2- Sufficient: 1 is always a factor, so x can not be odd, for all other even they have themselves and 2 which will give even sum (Fac(4)= 4,2,1) so only two has two factors ( 2,1).

is it correct because i found an answer says C ?

You are right B should be the answer for this question and your reasoning for B is correct.

If the integer x is greater than 1, does x = 2?

Given: \(x=integer>1\). Question: does \(x=2\)?

(1) x is evenly divisible by exactly two positive integers --> \(x\) can be any prime (2, 3, 5, 7, ...). Not sufficient.

(2) The sum of ANY TWO distinct positive factors of x is odd If \(x\) is odd, then 1 (obviously one of the factors of \(x\)) plus \(x\) itself (\(x\) is also a factor of itself) will be even not odd, so \(x\) cannot be odd. Now if \(x\) is even and more than 2, then 2 (one of the factors of \(x\)) plus \(x\) itself will be even not odd, so \(x\) cannot be more than 2 --> \(x=2\) (2 has exactly two positive factors (1 and 2) and their sum is odd). Sufficient.

Hey Bunuel Sorry I still don't understand B if I choose x=14 then factors are {1,2,7,14} for which the sum of any two factors could be odd. Take 7+2=9 or 2+1=3 0r 14+1=15 -----> all these option qualify for (B) but x is not 2! where is the error? _________________

If the integer x is greater than 1, does x = 2? 1) x is evenly divisible by exactly two positive integers. 2) The sum of any two distinct positive factors of x is odd.

1 Insufficient 2- Sufficient: 1 is always a factor, so x can not be odd, for all other even they have themselves and 2 which will give even sum (Fac(4)= 4,2,1) so only two has two factors ( 2,1).

is it correct because i found an answer says C ?

You are right B should be the answer for this question and your reasoning for B is correct.

If the integer x is greater than 1, does x = 2?

Given: \(x=integer>1\). Question: does \(x=2\)?

(1) x is evenly divisible by exactly two positive integers --> \(x\) can be any prime (2, 3, 5, 7, ...). Not sufficient.

(2) The sum of ANY TWO distinct positive factors of x is odd If \(x\) is odd, then 1 (obviously one of the factors of \(x\)) plus \(x\) itself (\(x\) is also a factor of itself) will be even not odd, so \(x\) cannot be odd. Now if \(x\) is even and more than 2, then 2 (one of the factors of \(x\)) plus \(x\) itself will be even not odd, so \(x\) cannot be more than 2 --> \(x=2\) (2 has exactly two positive factors (1 and 2) and their sum is odd). Sufficient.

Hey Bunuel Sorry I still don't understand B if I choose x=14 then factors are {1,2,7,14} for which the sum of any two factors could be odd. Take 7+2=9 or 2+1=3 0r 14+1=15 -----> all these option qualify for (B) but x is not 2! where is the error?

(2) does not say that the sum could be odd it says that is IS odd: the sum of ANY TWO distinct positive factors of x is odd.

If we take x=14, then 2+14=16=even, so it's not true that the sum of ANY TWO distinct positive factors of x is odd. _________________

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