The multiplicative inverse of 0x95 in AES where m(x)=x^8+x^4+x^3+x+1 is
(a) 0x8F
(b) 0xF8
(c) 0x8A
(d) 0xA8
I got this question by my school principal while I was bunking the class.
My question is based upon The AES Algorithm in division Advanced Encryption Algorithm and Block Cipher Operation of Cryptograph & Network Security