Modular multiplicative inverse

Last Updated : 24 Jul, 2024

Given two integers A and M, find the modular multiplicative inverse of A under modulo M.
The modular multiplicative inverse is an integer X such that:

A X ≡ 1 (mod M)

Note: The value of X should be in the range {1, 2, ... M-1}, i.e., in the range of integer modulo M. ( Note that X cannot be 0 as A*0 mod M will never be 1). The multiplicative inverse of "A modulo M" exists if and only if A and M are relatively prime (i.e. if gcd(A, M) = 1)

Examples:

Input: A = 3, M = 11
Output: 4
Explanation: Since (4*3) mod 11 = 1, 4 is modulo inverse of 3(under 11).
One might think, 15 also as a valid output as "(15*3) mod 11"
is also 1, but 15 is not in range {1, 2, ... 10}, so not valid.
Input: A = 10, M = 17
Output: 12
Explamation: Since (10*12) mod 17 = 1, 12 is modulo inverse of 10(under 17).

Naive Approach: To solve the problem, follow the below idea:

A naive method is to try all numbers from 1 to m. For every number x, check if (A * X) % M is 1

Below is the implementation of the above approach:

C++

// C++ program to find modular
// inverse of A under modulo M
#include <bits/stdc++.h>
using namespace std;

// A naive method to find modular
// multiplicative inverse of 'A'
// under modulo 'M'

int modInverse(int A, int M)
{
    for (int X = 1; X < M; X++)
        if (((A % M) * (X % M)) % M == 1)
            return X;
}

// Driver code
int main()
{
    int A = 3, M = 11;

    // Function call
    cout << modInverse(A, M);
    return 0;
}

Java

// Java program to find modular inverse
// of A under modulo M
import java.io.*;

class GFG {

    // A naive method to find modulor
    // multiplicative inverse of A
    // under modulo M
    static int modInverse(int A, int M)
    {

        for (int X = 1; X < M; X++)
            if (((A % M) * (X % M)) % M == 1)
                return X;
        return 1;
    }

    // Driver Code
    public static void main(String args[])
    {
        int A = 3, M = 11;

        // Function call
        System.out.println(modInverse(A, M));
    }
}

/*This code is contributed by Nikita Tiwari.*/

Python

# Python3 program to find modular
# inverse of A under modulo M

# A naive method to find modulor
# multiplicative inverse of A
# under modulo M


def modInverse(A, M):

    for X in range(1, M):
        if (((A % M) * (X % M)) % M == 1):
            return X
    return -1


# Driver Code
if __name__ == "__main__":
    A = 3
    M = 11

    # Function call
    print(modInverse(A, M))

# This code is contributed by Nikita Tiwari.

// C# program to find modular inverse
// of A under modulo M
using System;

class GFG {

    // A naive method to find modulor
    // multiplicative inverse of A
    // under modulo M
    static int modInverse(int A, int M)
    {

        for (int X = 1; X < M; X++)
            if (((A % M) * (X % M)) % M == 1)
                return X;
        return 1;
    }

    // Driver Code
    public static void Main()
    {
        int A = 3, M = 11;

        // Function call
        Console.WriteLine(modInverse(A, M));
    }
}

// This code is contributed by anuj_67.

JavaScript

<script>

// Javascript program to find modular 
// inverse of a under modulo m

// A naive method to find modulor
// multiplicative inverse of
// 'a' under modulo 'm'
function modInverse(a, m)
{
    for(let x = 1; x < m; x++)
        if (((a % m) * (x % m)) % m == 1)
            return x;
}

// Driver Code
let a = 3; 
let m = 11;

// Function call
document.write(modInverse(a, m));

// This code is contributed by _saurabh_jaiswal.

</script>

PHP

<?php
// PHP program to find modular 
// inverse of A under modulo M

// A naive method to find modulor
// multiplicative inverse of
// A under modulo M
function modInverse( $A, $M)
{
    
    for ($X = 1; $X < $M; $X++)
        if ((($A%$M) * ($X%$M)) % $M == 1)
            return $X;
}

// Driver Code
$A = 3; 
$M = 11;

// Function call
echo modInverse($A, $M);

// This code is contributed by anuj_67.
?>

Output

Time Complexity: O(M)
Auxiliary Space: O(1)