Question

The amount of rom needed to implement a 8 bit multiplier is

Answer 1

Try to recall the structure of ROM(2(power)n*m) here m is no. of output lines so when u multiply any 2 8-bit no. then output will require 16 bit. here n is no. of input lines to denote possible no. of address which we can form using 16-bit(as we know while multiplying using ordinary method intermediate bit written then we add those bit can have any possible combination of 16-bit) so solution would be 2(power)16*16=64K*16.