2 votes 2 votes dm4006 asked Jan 31, 2018 dm4006 248 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Ashwin Kulkarni commented Jan 31, 2018 reply Follow Share Is it option D ?? 0 votes 0 votes dm4006 commented Jan 31, 2018 reply Follow Share Yes Please explain 0 votes 0 votes Ashwin Kulkarni commented Jan 31, 2018 reply Follow Share the subtraction might go upto one digit more because of carry bit, hence 17 bits will be required. And 2 16 bit combinations go upto $2^{32}$ hence size will be $4G * 17$ 1 votes 1 votes dm4006 commented Jan 31, 2018 reply Follow Share why it can not be 2^17*17 17 bit number can be address using 2^17 address 0 votes 0 votes Please log in or register to add a comment.