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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 question is, how does $(11^{16})^2(\operatorname{mod}13)$ get reduced to $3^2(\operatorname{mod}13)$?
What can we do to reduce the size of terms involved and make our calculation faster? Suppose we want to calculate 2^90 mod 13, but we have a calculator that can't hold any numbers larger than 2^50. Here is a simple divide and conquer strategy: smaller parts. exponent rules. 2^ 90 = 2^ 50 * 2^ 40. mod C. each part.
1 lut 2021 · How To Do Modular Arithmetic. This means that modular arithmetic finds the remainder of a number upon division! Example #1. What is 16 mod 12? Well 16 divided by 12 equals 1 remainder 4. So the answer is 4! Example #2. What about 15 mod 2? Here, 15 divided by 2 equals 7 remainder 1, so the solution is 1! Example #3. And if you have 18 mod 9?
In modular arithmetic, when we say “reduced modulo,” we mean whatever result we obtain, we divide it by \(n\), and report only the smallest possible nonnegative residue. The next theorem is fundamental to modular arithmetic.
15 lis 2008 · Reducing Modulo Divisors. If we unravel the statement that a ≡ 11 (mod 20), what is it saying? What are the possibilities for a? We’ve specified the remainder after dividing a by 20, so a must be one of the following sequence of numbers: . . . , −29, −9, 11, 31, 51, 71, . . .
Example. Reduce 101 (mod 3) to a number in the range {0, 1, 2}. Reduce −101 (mod 3) to a number in the range {0, 1, 2}. 101 = 2 (mod 3), because 3 | 101 − 2 = 99. −101 = 1 (mod 3), because 3 | −101 − 1 = −102. Proposition. Congruence mod m is an equivalence relation: (Reflexivity) a = a (mod m) for all a.
on. Suppose f(x) ∈ Z[x] is a polynomial. Consider the reduction. (x) modulo p, and suppose deg(f) = deg(f). If f(x) is reducible. in Q[x], then f(x) is reducible in Z/p[x]. Or by the contrapositive: if f(x) is irreducible i. Z/p[x], then f(x) is irreducible in Q[x].This gives a.
