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FinalExamSols

# FinalExamSols - CIS 3362 Final Exam Solutions Date 1(10 pts...

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CIS 3362 Final Exam Solutions Date: 12/12/08 1) (10 pts) Determine 112 -1 mod 163. 163 = 1x112 + 51 112 = 2x51 + 10 51 = 5x10 + 1 (3 pts) 51 – 5x10 = 1 51 – 5(112 – 2x51) = 1 51 – 5x112 + 10x51 = 1 11x51 – 5x112 = 1 11x(163 – 112) – 5x112 = 1 11x163 – 11x112 – 5x112 = 1 11x163 – 16x112 = 1 (5 pts) Take this equation mod 163 to yield: -16x112 = 1 (mod 163) (1 pt) Thus, 112 -1 = -16 = 147 (mod 163) (1 pt for 147 from -16)

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2) (10 pts) If the round 7 key for AES is 99 88 77 66 13 57 9A CE EC A9 75 31 2F 3A 9C DA, what will the first four bytes of the round 8 key be? i tmp rotword subword Rcon(8) XOR w[i-4] w[i] 32 2F3A9CDA 3A9CDA2F 80DE5715 80000000 00DE5715 99887766 99562073 To perform each of the XORs, just write out the binary version of the hex characters and complete the XOR. Here is the last XOR: 00DE5715: 0000 0000 1101 1110 0101 0111 0001 0101 99887766: 1001 1001 1000 1000 0111 0111 0110 0110 XOR 1001 1001 0101 0110 0010 0000 0111 0011 9 9 5 6 2 0 7 3 Grading: 2 pts for rotword, 2 pts for subword, 2 pts for Rcon(10), 2 pts after XOR, 2 pts for final answer.
3) (10 pts) Given the following state matrix right before the shift columns operation in AES, determine the entry in row 1, column 1 right after the shift columns operation is done.

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