St1: K is a multiple of 2^6. So can't say that its 2^r. It may be (2^7) or may not be (3*2^6): INSUFF
St2: If K is not divisible by any odd integer then it is divisible only by 2 and numbers which are raised to the powers of 2. SUFF.
K is not divisible by 6,10,14 etc. Its divisible by only 2,4,8,16 because factors of these numbers are all even. So K can be represented in 2^r.
SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008