Cs290g — Introduction to Modern Cryptography Task 1 — Modular Arithmetic

نویسنده

  • Stefano Tessaro
چکیده

(i) Since 32 = 9 = 7 · 1 + 2, then we clearly have 32 mod 7 = 9 mod 7 = 2. (ii) The solution is easy found by inspection: We have 4 · 4 mod 15 = 16 mod 15 = 1, and thus 4−1 mod 15 = 4. (iii) I have seen many complicated solutions here, but the solution is very simple, and uses the fact that 32 mod 8 = 9 mod 8 = 1. Then, we use the fact (discussed in class) that a · b mod n = (a mod n) · (b mod n) mod n. 3 mod 8 = 32·15668+1 mod 8 = (3) · 3 mod 8 = ((3 mod 8) mod 8) · 3 mod 8 = (1 mod 8) · 3 mod 8 = 1 · 3 = 3 .

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تاریخ انتشار 2014