Modular multiplicative inverse
Last Updated :
23 Jul, 2025
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 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.*/
# 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.
<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 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.
?>
Time Complexity: O(M)
Auxiliary Space: O(1)
Modular multiplicative inverse when M and A are coprime or gcd(A, M)=1:
The idea is to use Extended Euclidean algorithms that take two integers 'a' and 'b', then find their gcd, and also find 'x' and 'y' such that
ax + by = gcd(a, b)
To find the multiplicative inverse of 'A' under 'M', we put b = M in the above formula. Since we know that A and M are relatively prime, we can put the value of gcd as 1.
Ax + My = 1
If we take modulo M on both sides, we get
Ax + My ≡ 1 (mod M)
We can remove the second term on left side as 'My (mod M)' would always be 0 for an integer y.
Ax ≡ 1 (mod M)
So the 'x' that we can find using Extended Euclid Algorithm is the multiplicative inverse of 'A'
Below is the implementation of the above approach:
C++
// C++ program to find multiplicative modulo
// inverse using Extended Euclid algorithm.
#include <bits/stdc++.h>
using namespace std;
// Function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int* x, int* y);
// Function to find modulo inverse of a
void modInverse(int A, int M)
{
int x, y;
int g = gcdExtended(A, M, &x, &y);
if (g != 1)
cout << "Inverse doesn't exist";
else {
// m is added to handle negative x
int res = (x % M + M) % M;
cout << "Modular multiplicative inverse is " << res;
}
}
// Function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int* x, int* y)
{
// Base Case
if (a == 0) {
*x = 0, *y = 1;
return b;
}
// To store results of recursive call
int x1, y1;
int gcd = gcdExtended(b % a, a, &x1, &y1);
// Update x and y using results of recursive
// call
*x = y1 - (b / a) * x1;
*y = x1;
return gcd;
}
// Driver Code
int main()
{
int A = 3, M = 11;
// Function call
modInverse(A, M);
return 0;
}
// This code is contributed by khushboogoyal499
// C program to find multiplicative modulo inverse using
// Extended Euclid algorithm.
#include <stdio.h>
// C function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int* x, int* y);
// Function to find modulo inverse of a
void modInverse(int A, int M)
{
int x, y;
int g = gcdExtended(A, M, &x, &y);
if (g != 1)
printf("Inverse doesn't exist");
else {
// m is added to handle negative x
int res = (x % M + M) % M;
printf("Modular multiplicative inverse is %d", res);
}
}
// C function for extended Euclidean Algorithm
int gcdExtended(int a, int b, int* x, int* y)
{
// Base Case
if (a == 0) {
*x = 0, *y = 1;
return b;
}
int x1, y1; // To store results of recursive call
int gcd = gcdExtended(b % a, a, &x1, &y1);
// Update x and y using results of recursive
// call
*x = y1 - (b / a) * x1;
*y = x1;
return gcd;
}
// Driver Code
int main()
{
int A = 3, M = 11;
// Function call
modInverse(A, M);
return 0;
}
// java program to find multiplicative modulo
// inverse using Extended Euclid algorithm.
public class GFG {
// Global Variables
public static int x;
public static int y;
// Function for extended Euclidean Algorithm
static int gcdExtended(int a, int b)
{
// Base Case
if (a == 0) {
x = 0;
y = 1;
return b;
}
// To store results of recursive call
int gcd = gcdExtended(b % a, a);
int x1 = x;
int y1 = y;
// Update x and y using results of recursive
// call
int tmp = b / a;
x = y1 - tmp * x1;
y = x1;
return gcd;
}
static void modInverse(int A, int M)
{
int g = gcdExtended(A, M);
if (g != 1) {
System.out.println("Inverse doesn't exist");
}
else {
// m is added to handle negative x
int res = (x % M + M) % M;
System.out.println(
"Modular multiplicative inverse is " + res);
}
}
// Driver code
public static void main(String[] args)
{
int A = 3, M = 11;
// Function Call
modInverse(A, M);
}
}
// The code is contributed by Gautam goel (gautamgoel962)
# Python3 program to find multiplicative modulo
# inverse using Extended Euclid algorithm.
# Global Variables
x, y = 0, 1
# Function for extended Euclidean Algorithm
def gcdExtended(a, b):
global x, y
# Base Case
if (a == 0):
x = 0
y = 1
return b
# To store results of recursive call
gcd = gcdExtended(b % a, a)
x1 = x
y1 = y
# Update x and y using results of recursive
# call
x = y1 - (b // a) * x1
y = x1
return gcd
def modInverse(A, M):
g = gcdExtended(A, M)
if (g != 1):
print("Inverse doesn't exist")
else:
# m is added to handle negative x
res = (x % M + M) % M
print("Modular multiplicative inverse is ", res)
# Driver Code
if __name__ == "__main__":
A = 3
M = 11
# Function call
modInverse(A, M)
# This code is contributed by phasing17
// C# program to find multiplicative modulo
// inverse using Extended Euclid algorithm.
using System;
public class GFG {
public static int x, y;
// Function for extended Euclidean Algorithm
static int gcdExtended(int a, int b)
{
// Base Case
if (a == 0) {
x = 0;
y = 1;
return b;
}
// To store results of recursive call
int gcd = gcdExtended(b % a, a);
int x1 = x;
int y1 = y;
// Update x and y using results of recursive
// call
x = y1 - (b / a) * x1;
y = x1;
return gcd;
}
// Function to find modulo inverse of a
static void modInverse(int A, int M)
{
int g = gcdExtended(A, M);
if (g != 1)
Console.Write("Inverse doesn't exist");
else {
// M is added to handle negative x
int res = (x % M + M) % M;
Console.Write(
"Modular multiplicative inverse is " + res);
}
}
// Driver Code
public static void Main(string[] args)
{
int A = 3, M = 11;
// Function call
modInverse(A, M);
}
}
// this code is contributed by phasing17
<script>
// JavaScript program to find multiplicative modulo
// inverse using Extended Euclid algorithm.
// Global Variables
let x, y;
// Function for extended Euclidean Algorithm
function gcdExtended(a, b){
// Base Case
if (a == 0)
{
x = 0;
y = 1;
return b;
}
// To store results of recursive call
let gcd = gcdExtended(b % a, a);
let x1 = x;
let y1 = y;
// Update x and y using results of recursive
// call
x = y1 - Math.floor(b / a) * x1;
y = x1;
return gcd;
}
function modInverse(a, m)
{
let g = gcdExtended(a, m);
if (g != 1){
document.write("Inverse doesn't exist");
}
else{
// m is added to handle negative x
let res = (x % m + m) % m;
document.write("Modular multiplicative inverse is ", res);
}
}
// Driver Code
{
let a = 3, m = 11;
// Function call
modInverse(a, m);
}
// This code is contributed by Gautam goel (gautamgoel962)
</script>
<?php
// PHP program to find multiplicative modulo
// inverse using Extended Euclid algorithm.
// Function to find modulo inverse of a
function modInverse($A, $M)
{
$x = 0;
$y = 0;
$g = gcdExtended($A, $M, $x, $y);
if ($g != 1)
echo "Inverse doesn't exist";
else
{
// m is added to handle negative x
$res = ($x % $M + $M) % $M;
echo "Modular multiplicative " .
"inverse is " . $res;
}
}
// function for extended Euclidean Algorithm
function gcdExtended($a, $b, &$x, &$y)
{
// Base Case
if ($a == 0)
{
$x = 0;
$y = 1;
return $b;
}
$x1;
$y1; // To store results of recursive call
$gcd = gcdExtended($b%$a, $a, $x1, $y1);
// Update x and y using results of
// recursive call
$x = $y1 - (int)($b/$a) * $x1;
$y = $x1;
return $gcd;
}
// Driver Code
$A = 3;
$M = 11;
// Function call
modInverse($A, $M);
// This code is contributed by chandan_jnu
?>
OutputModular multiplicative inverse is 4
Time Complexity: O(log M)
Auxiliary Space: O(log M), because of the internal recursion stack.
Iterative Implementation of the above approach:
C++
// Iterative C++ program to find modular
// inverse using extended Euclid algorithm
#include <bits/stdc++.h>
using namespace std;
// Returns modulo inverse of a with respect
// to m using extended Euclid Algorithm
// Assumption: a and m are coprimes, i.e.,
// gcd(A, M) = 1
int modInverse(int A, int M)
{
int m0 = M;
int y = 0, x = 1;
if (M == 1)
return 0;
while (A > 1) {
// q is quotient
int q = A / M;
int t = M;
// m is remainder now, process same as
// Euclid's algo
M = A % M, A = t;
t = y;
// Update y and x
y = x - q * y;
x = t;
}
// Make x positive
if (x < 0)
x += m0;
return x;
}
// Driver Code
int main()
{
int A = 3, M = 11;
// Function call
cout << "Modular multiplicative inverse is "
<< modInverse(A, M);
return 0;
}
// this code is contributed by shivanisinghss2110
// Iterative C program to find modular
// inverse using extended Euclid algorithm
#include <stdio.h>
// Returns modulo inverse of a with respect
// to m using extended Euclid Algorithm
// Assumption: a and m are coprimes, i.e.,
// gcd(A, M) = 1
int modInverse(int A, int M)
{
int m0 = M;
int y = 0, x = 1;
if (M == 1)
return 0;
while (A > 1) {
// q is quotient
int q = A / M;
int t = M;
// m is remainder now, process same as
// Euclid's algo
M = A % M, A = t;
t = y;
// Update y and x
y = x - q * y;
x = t;
}
// Make x positive
if (x < 0)
x += m0;
return x;
}
// Driver Code
int main()
{
int A = 3, M = 11;
// Function call
printf("Modular multiplicative inverse is %d\n",
modInverse(A, M));
return 0;
}
// Iterative Java program to find modular
// inverse using extended Euclid algorithm
class GFG {
// Returns modulo inverse of a with
// respect to m using extended Euclid
// Algorithm Assumption: a and m are
// coprimes, i.e., gcd(A, M) = 1
static int modInverse(int A, int M)
{
int m0 = M;
int y = 0, x = 1;
if (M == 1)
return 0;
while (A > 1) {
// q is quotient
int q = A / M;
int t = M;
// m is remainder now, process
// same as Euclid's algo
M = A % M;
A = t;
t = y;
// Update x and y
y = x - q * y;
x = t;
}
// Make x positive
if (x < 0)
x += m0;
return x;
}
// Driver code
public static void main(String args[])
{
int A = 3, M = 11;
// Function call
System.out.println("Modular multiplicative "
+ "inverse is "
+ modInverse(A, M));
}
}
/*This code is contributed by Nikita Tiwari.*/
# Iterative Python 3 program to find
# modular inverse using extended
# Euclid algorithm
# Returns modulo inverse of a with
# respect to m using extended Euclid
# Algorithm Assumption: a and m are
# coprimes, i.e., gcd(A, M) = 1
def modInverse(A, M):
m0 = M
y = 0
x = 1
if (M == 1):
return 0
while (A > 1):
# q is quotient
q = A // M
t = M
# m is remainder now, process
# same as Euclid's algo
M = A % M
A = t
t = y
# Update x and y
y = x - q * y
x = t
# Make x positive
if (x < 0):
x = x + m0
return x
# Driver code
if __name__ == "__main__":
A = 3
M = 11
# Function call
print("Modular multiplicative inverse is",
modInverse(A, M))
# This code is contributed by Nikita tiwari.
// Iterative C# program to find modular
// inverse using extended Euclid algorithm
using System;
class GFG {
// Returns modulo inverse of a with
// respect to m using extended Euclid
// Algorithm Assumption: a and m are
// coprimes, i.e., gcd(A, M) = 1
static int modInverse(int A, int M)
{
int m0 = M;
int y = 0, x = 1;
if (M == 1)
return 0;
while (A > 1) {
// q is quotient
int q = A / M;
int t = M;
// m is remainder now, process
// same as Euclid's algo
M = A % M;
A = t;
t = y;
// Update x and y
y = x - q * y;
x = t;
}
// Make x positive
if (x < 0)
x += m0;
return x;
}
// Driver Code
public static void Main()
{
int A = 3, M = 11;
// Function call
Console.WriteLine("Modular multiplicative "
+ "inverse is "
+ modInverse(A, M));
}
}
// This code is contributed by anuj_67.
<script>
// Iterative Javascript program to find modular
// inverse using extended Euclid algorithm
// Returns modulo inverse of a with respect
// to m using extended Euclid Algorithm
// Assumption: a and m are coprimes, i.e.,
// gcd(a, m) = 1
function modInverse(a, m)
{
let m0 = m;
let y = 0;
let x = 1;
if (m == 1)
return 0;
while (a > 1)
{
// q is quotient
let q = parseInt(a / m);
let t = m;
// m is remainder now,
// process same as
// Euclid's algo
m = a % m;
a = t;
t = y;
// Update y and x
y = x - q * y;
x = t;
}
// Make x positive
if (x < 0)
x += m0;
return x;
}
// Driver Code
let a = 3;
let m = 11;
// Function call
document.write(`Modular multiplicative inverse is ${modInverse(a, m)}`);
// This code is contributed by _saurabh_jaiswal
</script>
<?php
// Iterative PHP program to find modular
// inverse using extended Euclid algorithm
// Returns modulo inverse of a with respect
// to m using extended Euclid Algorithm
// Assumption: a and m are coprimes, i.e.,
// gcd(a, m) = 1
function modInverse($A, $M)
{
$m0 = $M;
$y = 0;
$x = 1;
if ($m == 1)
return 0;
while ($A > 1)
{
// q is quotient
$q = (int) ($A / $M);
$t = $M;
// m is remainder now,
// process same as
// Euclid's algo
$M = $A % $M;
$A = $t;
$t = $y;
// Update y and x
$y = $x - $q * $y;
$x = $t;
}
// Make x positive
if ($x < 0)
$x += $m0;
return $x;
}
// Driver Code
$A = 3;
$M = 11;
// Function call
echo "Modular multiplicative inverse is: ",
modInverse($A, $M);
// This code is contributed by Anuj_67
?>
OutputModular multiplicative inverse is 4
Time Complexity: O(log m)
Auxiliary Space: O(1)
Modular multiplicative inverse when M is prime:
If we know M is prime, then we can also use Fermat's little theorem to find the inverse.
aM-1 ≡ 1 (mod M)
If we multiply both sides with a-1, we get
a-1 ≡ a M-2 (mod M)
Below is the implementation of the above approach:
C++
// C++ program to find modular inverse of A under modulo M
// This program works only if M is prime.
#include <bits/stdc++.h>
using namespace std;
// To find GCD of a and b
int gcd(int a, int b);
// To compute x raised to power y under modulo M
int power(int x, unsigned int y, unsigned int M);
// Function to find modular inverse of a under modulo M
// Assumption: M is prime
void modInverse(int A, int M)
{
int g = gcd(A, M);
if (g != 1)
cout << "Inverse doesn't exist";
else {
// If a and m are relatively prime, then modulo
// inverse is a^(m-2) mode m
cout << "Modular multiplicative inverse is "
<< power(A, M - 2, M);
}
}
// To compute x^y under modulo m
int power(int x, unsigned int y, unsigned int M)
{
if (y == 0)
return 1;
int p = power(x, y / 2, M) % M;
p = (p * p) % M;
return (y % 2 == 0) ? p : (x * p) % M;
}
// Function to return gcd of a and b
int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Driver code
int main()
{
int A = 3, M = 11;
// Function call
modInverse(A, M);
return 0;
}
// Java program to find modular
// inverse of A under modulo M
// This program works only if
// M is prime.
import java.io.*;
class GFG {
// Function to find modular inverse of a
// under modulo M Assumption: M is prime
static void modInverse(int A, int M)
{
int g = gcd(A, M);
if (g != 1)
System.out.println("Inverse doesn't exist");
else {
// If a and m are relatively prime, then modulo
// inverse is a^(m-2) mode m
System.out.println(
"Modular multiplicative inverse is "
+ power(A, M - 2, M));
}
}
static int power(int x, int y, int M)
{
if (y == 0)
return 1;
int p = power(x, y / 2, M) % M;
p = (int)((p * (long)p) % M);
if (y % 2 == 0)
return p;
else
return (int)((x * (long)p) % M);
}
// Function to return gcd of a and b
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Driver Code
public static void main(String args[])
{
int A = 3, M = 11;
// Function call
modInverse(A, M);
}
}
// This code is contributed by Nikita Tiwari.
# Python3 program to find modular
# inverse of A under modulo M
# This program works only if M is prime.
# Function to find modular
# inverse of A under modulo M
# Assumption: M is prime
def modInverse(A, M):
g = gcd(A, M)
if (g != 1):
print("Inverse doesn't exist")
else:
# If A and M are relatively prime,
# then modulo inverse is A^(M-2) mod M
print("Modular multiplicative inverse is ",
power(A, M - 2, M))
# To compute x^y under modulo M
def power(x, y, M):
if (y == 0):
return 1
p = power(x, y // 2, M) % M
p = (p * p) % M
if(y % 2 == 0):
return p
else:
return ((x * p) % M)
# Function to return gcd of a and b
def gcd(a, b):
if (a == 0):
return b
return gcd(b % a, a)
# Driver Code
if __name__ == "__main__":
A = 3
M = 11
# Function call
modInverse(A, M)
# This code is contributed by Nikita Tiwari.
// C# program to find modular
// inverse of a under modulo M
// This program works only if
// M is prime.
using System;
class GFG {
// Function to find modular
// inverse of A under modulo
// M Assumption: M is prime
static void modInverse(int A, int M)
{
int g = gcd(A, M);
if (g != 1)
Console.Write("Inverse doesn't exist");
else {
// If A and M are relatively
// prime, then modulo inverse
// is A^(M-2) mod M
Console.Write(
"Modular multiplicative inverse is "
+ power(A, M - 2, M));
}
}
// To compute x^y under
// modulo M
static int power(int x, int y, int M)
{
if (y == 0)
return 1;
int p = power(x, y / 2, M) % M;
p = (p * p) % M;
if (y % 2 == 0)
return p;
else
return (x * p) % M;
}
// Function to return
// gcd of a and b
static int gcd(int a, int b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Driver Code
public static void Main()
{
int A = 3, M = 11;
// Function call
modInverse(A, M);
}
}
// This code is contributed by nitin mittal.
<script>
// Javascript program to find modular inverse of a under modulo m
// This program works only if m is prime.
// Function to find modular inverse of a under modulo m
// Assumption: m is prime
function modInverse(a, m)
{
let g = gcd(a, m);
if (g != 1)
document.write("Inverse doesn't exist");
else
{
// If a and m are relatively prime, then modulo
// inverse is a^(m-2) mode m
document.write("Modular multiplicative inverse is "
+ power(a, m - 2, m));
}
}
// To compute x^y under modulo m
function power(x, y, m)
{
if (y == 0)
return 1;
let p = power(x, parseInt(y / 2), m) % m;
p = (p * p) % m;
return (y % 2 == 0) ? p : (x * p) % m;
}
// Function to return gcd of a and b
function gcd(a, b)
{
if (a == 0)
return b;
return gcd(b % a, a);
}
// Driver code
let a = 3, m = 11;
// Function call
modInverse(a, m);
// This code is contributed by subham348.
</script>
<?php
// PHP program to find modular
// inverse of A under modulo M
// This program works only if M
// is prime.
// Function to find modular
// inverse of A under modulo
// M Assumption: M is prime
function modInverse( $A, $M)
{
$g = gcd($A, $M);
if ($g != 1)
echo "Inverse doesn't exist";
else
{
// If A and M are relatively
// prime, then modulo inverse
// is A^(M-2) mod M
echo "Modular multiplicative inverse is "
, power($A, $M - 2, $M);
}
}
// To compute x^y under modulo m
function power( $x, $y, $M)
{
if ($y == 0)
return 1;
$p = power($x, $y / 2, $M) % $M;
$p = ($p * $p) % $M;
return ($y % 2 == 0)? $p : ($x * $p) % $M;
}
// Function to return gcd of a and b
function gcd($a, $b)
{
if ($a == 0)
return $b;
return gcd($b % $a, $a);
}
// Driver Code
$A = 3;
$M = 11;
// Function call
modInverse($A, $M);
// This code is contributed by anuj_67.
?>
OutputModular multiplicative inverse is 4
Time Complexity: O(log M)
Auxiliary Space: O(log M), because of the internal recursion stack.
Applications:
Computation of the modular multiplicative inverse is an essential step in RSA public-key encryption method.
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