In number theory, given an integer A and a positive integer N with gcd( A , N) = 1, the multiplicative order of a modulo N is the smallest positive integer k with A^k( mod N ) = 1. ( 0 < K < N )

Examples:

Input : A = 4 , N = 7 Output : 3 explanation : GCD(4, 7) = 1A^k( mod N )= 1 ( smallest positive integer K ) 4^1 = 4(mod 7) = 4 4^2 = 16(mod 7) = 2 4^3 = 64(mod 7) = 1 4^4 = 256(mod 7) = 4 4^5 = 1024(mod 7) = 2 4^6 = 4096(mod 7) = 1 smallest positive integer K = 3 Input : A = 3 , N = 1000 Output : 100 (3^100 (mod 1000) == 1) Input : A = 4 , N = 11 Output : 5

IF we take a close look then we observe that we do not need to calculate power every time. we can be obtaining next power by multiplying ‘A’ with the previous result of a module .

Explanation : A = 4 , N = 11 initially result = 1 with normal with modular arithmetic (A * result) 4^1 = 4 (mod 11 ) = 4 || 4 * 1 = 4 (mod 11 ) = 4 [ result = 4] 4^2 = 16(mod 11 ) = 5 || 4 * 4 = 16(mod 11 ) = 5 [ result = 5] 4^3 = 64(mod 11 ) = 9 || 4 * 5 = 20(mod 11 ) = 9 [ result = 9] 4^4 = 256(mod 11 )= 3 || 4 * 9 = 36(mod 11 ) = 3 [ result = 3] 4^5 = 1024(mod 5 ) = 1 || 4 * 3 = 12(mod 11 ) = 1 [ result = 1] smallest positive integer 5

Run a loop from 1 to N-1 and Return the smallest +ve power of A under modulo n which is equal to 1.

Below C++ implementation of above idea.

// C++ program to implement multiplicative order #include<bits/stdc++.h> using namespace std; // fuction for GCD int GCD ( int a , int b ) { if (b == 0 ) return a; return GCD( b , a%b ) ; } // Fucnction return smallest +ve integer that // holds condition A^k(mod N ) = 1 int multiplicativeOrder(int A, int N) { if (GCD(A, N ) != 1) return -1; // result store power of A that rised to // the power N-1 unsigned int result = 1; int K = 1 ; while (K < N) { // modular arithmetic result = (result * A) % N ; // return samllest +ve integer if (result == 1) return K; // increment power K++; } return -1 ; } //driver program to test above function int main() { int A = 4 , N = 7; cout << multiplicativeOrder(A, N); return 0; }

Output:

3

**Time Complexity:** O(N)

Reference : https://en.wikipedia.org/wiki/Multiplicative_order

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