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But 6 is divisible by an odd integer other than 1. It's divisible by 3.

The only positive integers greater than 1 that don't have any odd factors greater than 1 are powers of 2. So statement 2 is sufficient on its own.

... And once again I just answered my own question. (I feel dumb now). I'd initially answered D, thinking that statement 1 was also sufficient in itself. But of course, 192 is a multiple of 64, and 192 is not a power of 2. So D is wrong. B it is.

But 6 is divisible by an odd integer other than 1. It's divisible by 3.

The only positive integers greater than 1 that don't have any odd factors greater than 1 are powers of 2. So statement 2 is sufficient on its own.

... And once again I just answered my own question. (I feel dumb now). I'd initially answered D, thinking that statement 1 was also sufficient in itself. But of course, 192 is a multiple of 64, and 192 is not a power of 2. So D is wrong. B it is.

You right. This is really tricky if one doesnt know these number system properties (especially if she is not good in number picking like me).

II) states that it is not divisible by any odd number > 1, which implies k is even. however, what we need to say if K is a power of 2. For example, 10 is even , but is not a power of 2. I am assuming here that the notation, 2^ r , denotes, 2 raised to r and not 2 multiplied by r.

From statement 1, we're told k is a multiple of 2^6. This could be 3(2^6), 4(2^6). Since there are so many values of k, we can't tell.

From statement 2, we're told k is not divisbile by any odd integer greater than 1. This means k is even and a power of 2. Any other even number that has a prime factor other than 2 will be divisible by any odd integer > 1. For e.g., if k=18, then 18 = 3*3*2. It's divisible by 3. However, if k = 16, then 2^4 and this is equal to 2^r if r is 4.

The answer B is correct if we're asked if k can be represented in the form 2^r where r is a positive value.

However, we're asked here if k EQUALS 2^r and we do not have information for r.