Modular Exponentiation (Power in Modular Arithmetic)

Last Updated : 19 Jun, 2025

Given three integers x, n, and M, compute (x^n) % M (remainder when x raised to the power n is divided by M).

Examples :

Input: x = 3, n = 2, M = 4
Output: 1
Explanation: 3² % 4 = 9 % 4 = 1.
Input: x = 2, n = 6, M = 10
Output: 4
Explanation: 2⁶ % 10 = 64 % 10 = 4.

Table of Content

[Naive Approach] Repeated Multiplication Method - O(n) Time and O(1) Space

We initialize the result as 1 and iterate from 1 to n, updating the result by multiplying it with x and taking the modulo by M in each step to keep the number within integer bounds.

C++

#include<iostream> using namespace std;  int powMod(int x, int n, int M) {          // Initialize result as 1 (since anything power 0 is 1)     int res = 1;      // n times to multiply x with itself     for(int i = 1; i <= n; i++) {                          // Multiplying res with x          // and taking modulo to avoid overflow         res = (res * x) % M;     }      return res; }  int main() {          int x = 3, n = 2, M = 4;     cout << powMod(x, n, M) << endl;     return 0; }

Java

public class GfG {      public static int powMod(int x, int n, int M) {         int res = 1;          for (int i = 1; i <= n; i++) {                                  // Multiplying res with x and              // taking modulo to avoid overflow             res = (res * x) % M;         }          return res;     }      public static void main(String[] args) {         int x = 3, n = 2, M = 4;         System.out.println(powMod(x, n, M));     } }

Python

def powMod(x, n, M):     res = 1      # loop from 1 to n     for _ in range(n):                          # Multiplying res with x         # and taking modulo to avoid overflow         res = (res * x) % M     return res  if __name__ == "__main__":     x, n, M = 3, 2, 4     print(powMod(x, n, M))

using System;  class GfG {     static int powMod(int x, int n, int M) {         int res = 1;         for (int i = 1; i <= n; i++) {                  // Multiplying res with x and              // taking modulo to avoid overflow             res = (res * x) % M;         }          return res;     }      static void Main() {         int x = 3, n = 2, M = 4;         Console.WriteLine(powMod(x, n, M));     } }

JavaScript

function powMod(x, n, M) {     let res = 1;          // Loop n times, multiplying x and taking modulo at each step     for (let i = 1; i <= n; i++) {                  // Multiplying res with x and         // taking modulo to avoid overflow         res = (res * x) % M;     }      return res; }  // Driver Code let x = 3, n = 2, M = 4; console.log(powMod(x, n, M));

Output

[Expected Approach] Modular Exponentiation Method - O(log(n)) Time and O(1) Space

The idea of binary exponentiation is to reduce the exponent by half at each step, using squaring, which lowers the time complexity from O(n) to O(log n).
-> xⁿ= (x^n/2)²if n is even.
-> xⁿ= x*x^n-1 if n is odd.

Step by step approach:

Start with the result as 1.
Use a loop that runs while the exponent n is greater than 0.
If the current exponent is odd, multiply the result by the current base and apply the modulo.
Square the base and take the modulo to keep the value within bounds.
Divide the exponent by 2 (ignore the remainder).
Repeat the process until the exponent becomes 0.

C++

#include<iostream> using namespace std;  int powMod(int x, int n, int M) {     int res = 1;      // Loop until exponent becomes 0     while(n >= 1) {                  // n is odd, multiply result by current x and take modulo         if(n & 1) {             res = (res * x) % M;                          // Reduce exponent by 1 to make it even             n--;           }                  // n is even, square the base and halve the exponent         else {             x = (x * x) % M;             n /= 2;         }     }     return res; }  int main() {     int x = 3, n = 2, M = 4;     cout << powMod(x, n, M) << endl; }

Java

class GfG {     public int powMod(int x, int n, int M) {         int res = 1;          // Loop until exponent becomes 0         while (n >= 1) {                          // n is odd, multiply result by current x and take modulo             if ((n & 1) == 1) {                 res = (res * x) % M;                                  // Decrease n to make it even                 n--;              } else {                              // n is even, square the base and halve the exponent                 x = (x * x) % M;                 n /= 2;             }         }          return res;     }      public static void main(String[] args) {         int x = 3, n = 2, M = 4;         GfG obj = new GfG();         System.out.println(obj.powMod(x, n, M));     } }

Python

def powMod(x, n, M):     res = 1      # Loop until exponent becomes 0     while n >= 1:                  # n is odd, multiply result by current x and take modulo         if n % 2 == 1:             res = (res * x) % M                          # Make n even             n -= 1          else:                          # n is even, square the base and halve the exponent             x = (x * x) % M             n //= 2      return res  if __name__ == "__main__":     x, n, M = 3, 2, 4     print(powMod(x, n, M))

using System;  class GfG {     public int powMod(int x, int n, int M) {         int res = 1;          // Loop until exponent becomes 0         while (n >= 1) {                          // n is odd, multiply result by current x and take modulo             if ((n & 1) == 1) {                 res = (int)((1L * res * x) % M);                                  // Reduce exponent by 1                 n--;               } else {                              // n is even, square the base and halve the exponent                 x = (int)((1L * x * x) % M);                 n /= 2;             }         }          return res;     }      public static void Main() {         int x = 3, n = 2, M = 4;         GfG obj = new GfG();         Console.WriteLine(obj.powMod(x, n, M));     } }

JavaScript

function powMod(x, n, M) {     let res = 1;      // Loop until exponent becomes 0     while (n >= 1) {                  // If n is odd, multiply result by current x and take modulo         if (n % 2 === 1) {             res = (res * x) % M;             n -= 1;         } else {                          // If n is even, square the base and halve the exponent             x = (x * x) % M;             n /= 2;         }     }      return res; }  // Driver Code let x = 3, n = 2, M = 4; console.log(powMod(x, n, M));

Output

Modular multiplicative inverse

Shivam Agrawal

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Practice Tags :

Modular Exponentiation (Power in Modular Arithmetic)

[Naive Approach] Repeated Multiplication Method - O(n) Time and O(1) Space

[Expected Approach] Modular Exponentiation Method - O(log(n)) Time and O(1) Space

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