integers

[cans]

If k is an integer greater than 1, does k = 2?

(1) a and b are any two positive factors of k, where a ≠ b, and a - b must be odd.

(2) k has only two positive factors.

[Frankenstein]

Hi,
From(1): a and b are any two numbers. So, lets take b=1 as 1 is factor of any integer. So a must be even(as diff. is odd). 2 is a factor of every even number greater than 2. So, the diff. between any even number and 2(factor of even number) is even(not odd). So, only possible value of k is 2.
Sufficient

From(2): k has 2 positive factors. So k is prime
Insufficient

Hence, A

[[email protected]]

IMO C
because
(1) is not sufficent as for any number k (say 12) two factors are 3,4 and a-b=1 is odd and say for k=44 two factors are 4 and 11 and a-b=7 which is odd and still value of k!=2, on the other hand if k=2 then a-b=2-1=1 is odd so k CAN be 2 but not necessary
(2) this tells us that k is a prime no. so say k=2,3,5,7 etc here too, k CAN be 2 but not necessary

now combining (1) and (2) it tells that there are only two factors for K and their difference must be odd.. there is only one such value for k ie.k=2(apart from that if u take any prime no. then the difference will be an even number eg. 7 so 7-1=6 even, 13 so 13-1=12 even) hence sufficient[spoiler][/spoiler]

[Frankenstein]

Yeah, I believe statement(1) has more to do with the wording. 'a and b are any two positive factors of k'. So, I feel it is true for any 2 factors of k. That is how I interpret it, not that I am correct.

[[email protected]]

Yeah, I believe statement(1) has more to do with the wording. 'a and b are any two positive factors of k'. So, I feel it is true for any 2 factors of k. That is how I interpret it, not that I am correct.

ya i get your point, i stand corrected. in that case IMO A

[smackmartine]

IMO C

k > 1

1) a and b are any two positive factors of k, where a ≠ b, and a - b must be odd.

one of a and b must be even and other must be odd. let a=2, b=1

for k=2, a=2 and b=1 is a factor of 2, however
for k=6, a=2 and b=1 is a factor of 6. K can be any even integer greater than equal to 2. so Insufficient.

2) k has only two positive factors

means k is prime number.
k can be 2,3,5,7... Insufficient.

Combining 1 and 2
As a and b are any two positive factors of k and because "2" is the only prime number that can divide by an even number, k=2. Sufficient.

[smackmartine]

Yeah, I believe statement(1) has more to do with the wording. 'a and b are any two positive factors of k'. So, I feel it is true for any 2 factors of k. That is how I interpret it, not that I am correct.

@Frankenstein,

According to your interpretation and explanation , should n't it be : 'a and b are ONLY two positive factors of k' ?

The word ONLY restricts us to think in terms of prime #s. However statement 1 says ANY. What do you think about this?

[phanideepak]

IMO C

k > 1

1) a and b are any two positive factors of k, where a ≠ b, and a - b must be odd.

one of a and b must be even and other must be odd. let a=2, b=1

for k=2, a=2 and b=1 is a factor of 2, however
for k=6, a=2 and b=1 is a factor of 6. K can be any even integer greater than equal to 2. so Insufficient.

2) k has only two positive factors

means k is prime number.
k can be 2,3,5,7... Insufficient.

Combining 1 and 2
As a and b are any two positive factors of k and because "2" is the only prime number that can divide by an even number, k=2. Sufficient.

you made a mistake there the point is that ANY two factors when u subtract the number should be ODD For example lets take 6 as you have taken it before

factors of 6 : 1,2,3,6 Now subtract 3 and 1 3-1 = 2 So NOT ODD

lets take another even integer greater than 2 but has no odd factors. example 8

factors of 8 are : 1,2,4,8 now subtract 4 and 2 4-2 = 2 which is NOT ODD not odd again so the only number which satisfies that condition is 2

Answer IMO is A

[smackmartine]

ANY two factors when u subtract the number should be ODD

I guess I got it.Thanks

[cans]

OA A
in a) any two factors means it should be true for all the pairs of factors and not just one.