Here key point is that for bitwise op to be 1 we must have AND operation of LSB to be 1 here since we have bitwise AND operation performed with 1 whose LSB is 1 therefore all the odd numbers shall have their LSB 's to be 1 so directly we can say that the result of bitwise AND would be 0 for all even numbers and 1 for all odd numbers.so op is option A .