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In modular arithmetic, the modular multiplicative inverse of a is also defined: it is the number x such that ax ≡ 1 (mod n). This multiplicative inverse exists if and only if a and n are coprime. For example, the inverse of 3 modulo 11 is 4 because 4 ⋅ 3 ≡ 1 (mod 11). The extended Euclidean algorithm may be used to compute it.
The set {3,19} generates the group, which means that every element of (/) is of the form 3 a × 19 b (where a is 0, 1, 2, or 3, because the element 3 has order 4, and similarly b is 0 or 1, because the element 19 has order 2).
The number 3 is a primitive root modulo 7 [5] because = = = = = = = = = = = = (). Here we see that the period of 3 k modulo 7 is 6. The remainders in the period, which are 3, 2, 6, 4, 5, 1, form a rearrangement of all nonzero remainders modulo 7, implying that 3 is indeed a primitive root modulo 7.
The sum, the difference and the product are the remainder of the division by of the result of the corresponding integer operation. The multiplicative inverse of an element may be computed by using the extended Euclidean algorithm (see Extended Euclidean algorithm § Modular integers). Let be a finite field.
Once we have defined multiplication for formal power series, we can define multiplicative inverses as follows. The multiplicative inverse of a formal power series A is a formal power series C such that AC = 1, provided that such a formal power series exists. It turns out that if A has a multiplicative inverse, it is unique, and we denote it by ...
For example, to multiply 7 and 15 modulo 17 in Montgomery form, again with R = 100, compute the product of 3 and 4 to get 12 as above. The extended Euclidean algorithm implies that 8⋅100 − 47⋅17 = 1, so R′ = 8. Multiply 12 by 8 to get 96 and reduce modulo 17 to get 11. This is the Montgomery form of 3, as expected.
This integer a −1 is called a modular multiplicative inverse of a modulo m. If a ≡ b (mod m) and a −1 exists, then a −1 ≡ b −1 (mod m) (compatibility with multiplicative inverse, and, if a = b, uniqueness modulo m). If ax ≡ b (mod m) and a is coprime to m, then the solution to this linear congruence is given by x ≡ a −1 b (mod m).
t 2 = 6 is the modular multiplicative inverse of 5 × 11 (mod 7) and t 3 = 6 is the modular multiplicative inverse of 5 × 7 (mod 11). Thus, X = 3 × (7 × 11) × 4 + 6 × (5 × 11) × 4 + 6 × (5 × 7) × 6 = 3504. and in its unique reduced form X ≡ 3504 ≡ 39 (mod 385) since 385 is the LCM of 5,7 and 11. Also, the modular multiplicative ...