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The multiplicative inverse of 0x95 in AES where m(x)=x8+x4+x3+x+1 is
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The multiplicative inverse of 0x95 in AES where m(x)=x8+x4+x3+x+1 is
Posted:
Thu Jul 14, 2022 8:10 am
by
answerhappygod
a) 0x8F
b) 0xF8
c) 0x8A
d) 0xA8