Multiply 00100110 by 10011110 in GF(2^8) with modulus 100011011.The result is
(a) 00101111
(b) 00101100
(c) 01110011
(d) 11101111
I have been asked this question during an interview.
This interesting question is from Polynomial and Modular Arithmetic- III in section Basic Concepts in Number Theory and Finite Fields of Cryptograph & Network Security