# Modular multiplicative inverse

Given two integers ‘a’ and ‘m’, find modular multiplicative inverse of ‘a’ under modulo ‘m’.
The modular multiplicative inverse is an integer ‘x’ such that.

`a x ≅ 1 (mod m)`

The value of x should be in { 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).

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Examples:

```Input:  a = 3, m = 11
Output: 4
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 ring {1, 2, ... 10}, so not
valid.

Input:  a = 10, m = 17
Output: 12
Since (10*12) mod 17 = 1, 12 is modulo inverse of 10(under 17).```

Method 1 (Naive)
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 this method.

 `// C++ program to find modular ` `// inverse of a under modulo m` `#include ` `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` `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.`

 `<≅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
`4`

Time Complexity: O(m).

Method 2 (Works when m and a are coprime)
The idea is to use Extended Euclidean algorithms that takes two integers ‘a’ and ‘b’, finds their gcd and also find ‘x’ and ‘y’ such that

`ax + by = gcd(a, b)`

To find multiplicative inverse of ‘a’ under ‘m’, we put b = m in above formula. Since we know that a and m are relatively prime, we can put 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 algorithm.

 `// C++ program to find multiplicative modulo ` `// inverse using Extended Euclid algorithm.` `#include ` `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 `   `// 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;` `}`

 `<≅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` `≅>`

Output
`Modular multiplicative inverse is 4`

Iterative Implementation:

 `// Iterative C++ program to find modular` `// inverse using extended Euclid algorithm` `#include ` `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 `   `// 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` `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.`

 `<≅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\n"``,` `                            ``modInverse(``\$a``, ``\$m``);` `        `  `// This code is contributed by Anuj_67` `≅>`

 ``

Output
`Modular multiplicative inverse is 4`

Time Complexity: O(Log m)

Method 3 (Works when m is prime)
If we know m is prime, then we can also use Fermats’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 idea.

 `// C++ program to find modular inverse of a under modulo m` `// This program works only if m is prime.` `#include ` `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) mode 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` `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) mode 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.`

 `<≅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) mode 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.` `≅>`

 ``

Output
`Modular multiplicative inverse is 4`

Time Complexity: O(Log m)

We have discussed three methods to find multiplicative inverse modulo m.
1) Naive Method, O(m)
2) Extended Euler’s GCD algorithm, O(Log m) [Works when a and m are coprime]
3) Fermat’s Little theorem, O(Log m) [Works when ‘m’ is prime]

Applications:
Computation of the modular multiplicative inverse is an essential step in RSA public-key encryption method.

References:
https://en.wikipedia.org/wiki/Modular_multiplicative_inverse
http://e-maxx.ru/algo/reverse_element