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15 lis 2008 · Answer. (a) The numbers in Question 2 (2, 4, 5, 10) are all divisors of 20. (b) The numbers in Question 3 (3, 7, 9) are all relatively prime to 20 (ie they have no common factors with 20). (c) The numbers in Question 4 (6, 8) all share a common factor with 20, but are not themselves divisors.
We have the following rules for modular arithmetic: Sum rule: IF a ≡ b(mod m) THEN a + c ≡ b + c(mod m). (3) m) on An inverse to. ab ≡ 1(mod m). (5) his means that ab − 1 = k · m for some integer k. As before, there are may be many solutions to this equation but we choose as a representative the smallest positiv. solution and say .
Proof If x3+ 10000 = y3then x3+ 10000 ≡ y3(mod 7) (by Theorem 2.1.3(1)). Since 10000 ≡ 4 (mod 7), x3+4 ≡ y3(mod 7). But x3≡ −1,0, or 1 (mod 7) by previous example, so x3+ 4 ≡ 3,4 or 5 (mod 7), while y3≡ −1,0, or 1 (mod 7) contradiction. This example illustrates one of the uses of modular arithmetic.
I am working through a modulo tutorial and have become stuck here: $$ 11^{32}(\operatorname{mod}13) = (11^{16})^2(\operatorname{mod}13)= 3^2(\operatorname{mod}13)= 9(\operatorname{mod}13) $$ My qu...
REDUCE User’s Manual Free Version Anthony C. Hearn and Rainer Schöpf https://reduce-algebra.sourceforge.io/ March 29, 2024
14 sty 2021 · The modulus[1] is another name for the remainder after division. For example, 17 mod 5 = 2, since if we divide 17 by 5, we get 3 with remainder 2. Modular arithmetic is sometimes called clock arithmetic, since analog clocks wrap around times past 12, meaning they work on a modulus of 12.
1 = 2 (mod 3); 2 = 1 (mod 3) and 1=2 = 2 (mod 3) remembering that 1=2 is the number you multiply by 2 to get 1. It is worth repeating the remarkable feature of mod parithmetic.
