The multiplicative inverse of 0x95 in AES where m(x)=x8+x4+x3+x+1 is

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answerhappygod
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The multiplicative inverse of 0x95 in AES where m(x)=x8+x4+x3+x+1 is

Post by answerhappygod »

a) 0x8F
b) 0xF8
c) 0x8A
d) 0xA8
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