Multiplicative order
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) = 1 A^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 N1 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 <a href="#">multiplicative order</a>
#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 N1
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)
Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
We have built the accelerating growthoriented website for budding engineers and aspiring job holders of technology companies such as Google, Facebook, and Amazon
If you would like to study our free courses you can join us at
http://www.techcodebit.com. #techcodebit #google #microsoft #facebook #interview portal #jobplacements
#technicalguide