B.
from (1): k can be 2^6 * 5 and be divisible by 2^6 and cannot be expressed in terms of 2^r. hence, not sufficient.

from (2): k does not have any odd number as its factor, which means k is a product of only 2's and hence, can be expressed as 2^r.

thanks arvind...

but why do u assume k is product of 2 only...6 is not an odd number but if its is factor of K, we can not express it as 2^r..any thought on this?

@ vaivish

You cannot take 6 or any even number that has an odd factor because the second st says that k is not divisible by any odd integer greater than 1. 6 is a multiple of 3 and hence any number divisible by 6 is also divisible by 3.

So we cannot choose numbers like 6, 18, 24, 36 etc which have odd factors greater than 1

Hope you get the logic

we need to look at all the prime factors of a number to see if the number is divisible by an odd number. If the number is not divisible by any odd number then, it shouldn't have any odd prime factors and must contain only even prime factor. we know that the only even prime factor is 2. Hence, it can only be multiples of 2.

thanks guys

[quote="vaivish"]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.


I dont have the ans for this one...[/quote]

Hi Vaivish..This is my soln.

1. K = (2^6) P, where P is an integer
if P= 2, then K can be written as a power of 2
if P = 6, then K can't be written as a power of 2
Therefore, Insufficient

2. K is not divisble by 3, 5, 7...or any other prime no. (as all primes > 2
are odd) or multiples of primes no's. If K is divisible by an even no. 6
then k is also divisible by 3. Therefore. K is divisible by only powers of
2.

My answer (B) i.e. statement 2 is sufficient in itself