# Find (a^b)%m where ‘b’ is very large

Given three numbers a, b and m where 1<=a, m<=10^6. Given very large 'b' containing up to 10^6 digits and m is a prime number, the task is to find (a^b)%m.

Examples:

Input: a = 2, b = 3, m = 17
Output: 8
2 ^ 3 % 17 = 8

Input: a = 3, b = 100000000000000000000000000, m = 1000000007
Output: 835987331

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: According to Fermat’s little theorem,

`a^(p-1) mod p = 1, When p is prime.`

From this, as of the problem, M is prime, express A^B mod M as follows:

`A^B mod M = ( A^(M-1) * A^(M-1) *.......* A^(M-1) * A^(x) ) mod M`

Where x is B mod M-1 and A ^ (M-1) continues B/(M-1) times

Now, from Fermat’s Little Theorem,

`A ^ (M-1) mod M = 1.`

Hence,

`A^B mod M = ( 1 * 1 * ....... * 1 * A^(x) ) mod M`

Hence mod B with M-1 to reduce the number to a smaller one and then use power() method to compute (a^b)%m.

Below is the implementation of the above approach:

## C++

 `// C++ program to find ` `// (a^b)%m for b very large. ` `#include ` `#define ll long long int ` `using` `namespace` `std; ` ` `  `// Function to find power ` `ll power(ll x, ll y, ll p) ` `{ ` `    ``ll res = 1; ``// Initialize result ` ` `  `    ``// Update x if it is more than or ` `    ``// equal to p ` `    ``x = x % p;  ` ` `  `    ``while` `(y > 0) { ` `        ``// If y is odd, multiply x with the result ` `        ``if` `(y & 1) ` `            ``res = (res * x) % p; ` ` `  `        ``// y must be even now ` `        ``y = y >> 1; ``// y = y/2 ` `        ``x = (x * x) % p; ` `    ``} ` `    ``return` `res; ` `} ` `// Driver Code ` `int` `main() ` `{ ` `    ``ll a = 3; ` ` `  `    ``// String input as b is very large ` `    ``string b = ``"100000000000000000000000000"``; ` ` `  `    ``ll remainderB = 0; ` `    ``ll MOD = 1000000007; ` ` `  `    ``// Reduce the number B to a small number ` `    ``// using Fermat Little ` `    ``for` `(``int` `i = 0; i < b.length(); i++) ` `        ``remainderB = (remainderB * 10 + b[i] - ``'0'``) % (MOD - 1); ` ` `  `    ``cout << power(a, remainderB, MOD) << endl; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find ` `// (a^b)%m for b very large. ` `import` `java.io.*; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to find power ` `static` `long` `power(``long` `x,  ` `                  ``long` `y, ``long` `p) ` `{ ` `    ``long` `res = ``1``; ``// Initialize result ` ` `  `    ``// Update x if it is more  ` `    ``// than or equal to p ` `    ``x = x % p;  ` ` `  `    ``while` `(y > ``0``)  ` `    ``{ ` `        ``// If y is odd, multiply  ` `        ``// x with the result ` `        ``if` `((y & ``1``) > ``0``) ` `            ``res = (res * x) % p; ` ` `  `        ``// y must be even now ` `        ``y = y >> ``1``; ``// y = y/2 ` `        ``x = (x * x) % p; ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main (String[] args) ` `{ ` `long` `a = ``3``; ` ` `  `// String input as ` `// b is very large ` `String b = ``"100000000000000000000000000"``; ` ` `  `long` `remainderB = ``0``; ` `long` `MOD = ``1000000007``; ` ` `  `// Reduce the number B to a small  ` `// number using Fermat Little ` `for` `(``int` `i = ``0``; i < b.length(); i++) ` `    ``remainderB = (remainderB * ``10` `+  ` `                  ``b.charAt(i) - ``'0'``) %  ` `                 ``(MOD - ``1``); ` ` `  `System.out.println(power(a, remainderB, MOD)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67. `

## Python3

 `# Python3 program to find ` `# (a^b)%m for b very large. ` ` `  `# Function to find power ` `def` `power(x, y, p): ` `    ``res ``=` `1``; ``# Initialize result ` ` `  `    ``# Update x if it is  ` `    ``# more than or equal to p ` `    ``x ``=` `x ``%` `p;  ` ` `  `    ``while` `(y > ``0``):  ` `         `  `        ``# If y is odd, multiply ` `        ``# x with the result ` `        ``if` `(y & ``1``): ` `            ``res ``=` `(res ``*` `x) ``%` `p; ` ` `  `        ``# y must be even now ` `        ``y ``=` `y >> ``1``; ``# y = y/2 ` `        ``x ``=` `(x ``*` `x) ``%` `p; ` `         `  `    ``return` `res; ` ` `  `# Driver Code ` `a ``=` `3``; ` ` `  `# String input as b ` `# is very large ` `b ``=` `"100000000000000000000000000"``; ` ` `  `remainderB ``=` `0``; ` `MOD ``=` `1000000007``; ` ` `  `# Reduce the number B  ` `# to a small number ` `# using Fermat Little ` `for` `i ``in` `range``(``len``(b)): ` `    ``remainderB ``=` `((remainderB ``*` `10` `+`  `                   ``ord``(b[i]) ``-` `48``) ``%`  `                   ``(MOD ``-` `1``)); ` ` `  `print``(power(a, remainderB, MOD)); ` ` `  `# This code is contributed by mits `

## C#

 `// C# program to find ` `// (a^b)%m for b very large. ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `// Function to find power ` `static` `long` `power(``long` `x,  ` `                  ``long` `y, ``long` `p) ` `{ ` `    ``// Initialize result ` `    ``long` `res = 1;  ` ` `  `    ``// Update x if it is more  ` `    ``// than or equal to p ` `    ``x = x % p;  ` ` `  `    ``while` `(y > 0)  ` `    ``{ ` `        ``// If y is odd, multiply  ` `        ``// x with the result ` `        ``if` `((y & 1) > 0) ` `            ``res = (res * x) % p; ` ` `  `        ``// y must be even now ` `        ``y = y >> 1; ``// y = y/2 ` `        ``x = (x * x) % p; ` `    ``} ` `    ``return` `res; ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main () ` `{ ` `    ``long` `a = 3; ` `     `  `    ``// String input as ` `    ``// b is very large ` `    ``string` `b = ``"100000000000000000000000000"``; ` `     `  `    ``long` `remainderB = 0; ` `    ``long` `MOD = 1000000007; ` `     `  `    ``// Reduce the number B to  ` `    ``// a small number using  ` `    ``// Fermat Little ` `    ``for` `(``int` `i = 0; i < b.Length; i++) ` `        ``remainderB = (remainderB * 10 +  ` `                          ``b[i] - ``'0'``) %  ` `                             ``(MOD - 1); ` `     `  `    ``Console.WriteLine(power(a, remainderB, MOD)); ` `} ` `} ` ` `  `// This code is contributed by anuj_67. `

## PHP

 ` 0)  ` `    ``{ ` `        ``// If y is odd, multiply ` `        ``// x with the result ` `        ``if` `(``\$y` `& 1) ` `            ``\$res` `= (``\$res` `* ``\$x``) % ``\$p``; ` ` `  `        ``// y must be even now ` `        ``\$y` `= ``\$y` `>> 1; ``// y = y/2 ` `        ``\$x` `= (``\$x` `* ``\$x``) % ``\$p``; ` `    ``} ` `    ``return` `\$res``; ` `} ` ` `  `// Driver Code ` `\$a` `= 3; ` ` `  `// String input as b ` `// is very large ` `\$b` `= ``"100000000000000000000000000"``; ` ` `  `\$remainderB` `= 0; ` `\$MOD` `= 1000000007; ` ` `  `// Reduce the number B  ` `// to a small number ` `// using Fermat Little ` `for` `(``\$i` `= 0; ``\$i` `< ``strlen``(``\$b``); ``\$i``++) ` `    ``\$remainderB` `= (``\$remainderB` `* 10 +  ` `                   ``\$b``[``\$i``] - ``'0'``) % ` `                  ``(``\$MOD` `- 1); ` ` `  `echo` `power(``\$a``, ``\$remainderB``, ``\$MOD``); ` ` `  `// This code is contributed by mits ` `?> `

Output:

```835987331
``` My Personal Notes arrow_drop_up Check out this Author's contributed articles.

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Improved By : vt_m, Mithun Kumar