hobbit wrote:

answer is D in my opinion....

st1: as AK noted, if there are even number of integers in S, and since they are consecutive, there will be equal number of even and odd numbers. hence sufficient (the answer is no).

st2: a bit more subtle, but the answer is no as well, and hence sufficient: the only way to get more even than odd in S it must start and end in even number. if it ends in odd number, either there are equal number of even an odd numbers or there is more odd than even.... in any case there are not more even than odd..... so sufficient.

its my own question, so OA is D and OE is as explained above...