From GMATprep (test 2)
If k is an integer greater than 1, is k equal to 2^r for some positive integer r?
(1) k is divisible by 2^6
(2) k is not divisible by any odd integer greater than 1
OA is B
Please state your reasoning.

Hi,

1.) K is divisible by 2^6 i.e. 64. then, K value can be equal to 64,128,192, etc... Here,
if k=128 can be written as 2^7, which satisfies the given equation.
But, if k=192, it cannot be written in the form of 2^r. So, 1.) is insufficient to answer.

Consider 2.) now. If K is not divisible by any integer greater than 1, that means K cannot itself be an odd integer (in that case it will be divisible by itself i.e odd integer). The only values K can have will be of the form 2^r i.e 2, 2^2, 4^2, 8^2 etc (not even 6^2 as it will be divisble by 3 greater than 1).. which satisfies the criteria.

So, 2.) is sufficient to answer the question... :)

The first step I took was to rephrase the question. does k only have 2's in its prime box? That is the only way k could be equal to 2^r.

1) although k is divisible by 2^6, it could also be divisible by 3,5 etc.

2) if you do not allow any odd integers greater than 1 than the only even integers that work are multiples of 2.

as usual, consider the prime factorization of the number in question. in fact, the actual problem is just a disguised question about prime factorizations:
is k = 2^r for some positive integer r? --> same question as --> does the prime factorization of k contain only 2's?
note that "2^r" is actually a prime factorization in and of itself. it may not necessarily look like one - mostly because it doesn't contain any primes other than 2 - but it is one. if you notice this, it's easier to realize that the question is just asking about such prime factorizations (i.e., those that don't contain any primes other than 2).

statement (1)
* it's possible that k could be a pure power of 2. for instance, 2^8, or even 2^6 itself, is divisible by 2. therefore, the answer to the prompt question could be "yes".
* it's also possible that k could include primes other than 2, as long as it has the requisite six 2's. for instance, k could be something like (2^6)(3^3), or (2^8)(5)(7). therefore, the answer to the prompt question could be "no".
insufficient.

statement (2)
remember that all primes other than 2 are odd.
therefore, this statement implies that k is not divisible by any prime numbers other than 2.
therefore, the answer is "yes".
sufficient.

answer = (b).

Thank you.
I feel much more clear about this problem when viewing it as a prime factorization.
You the man, Ron.