# Modular exponentiation (Recursive)

Hello Everyone,

Given three numbers a, b and c, we need to find (ab) % c
Now why do “% c” after exponentiation, because ab will be really large even for relatively small values of a, b and that is a problem because the data type of the language that we try to code the problem, will most probably not let us store such a large number.
Examples:

Input : a = 2312 b = 3434 c = 6789
Output : 6343

Input : a = -3 b = 5 c = 89
Output : 24

The idea is based on below properties.
Property 1:
(m * n) % p has a very interesting property:
(m * n) % p =((m % p) * (n % p)) % p
Property 2:
if b is even:
(a ^ b) % c = ((a ^ b/2) * (a ^ b/2))%c ? this suggests divide and conquer
if b is odd:
(a ^ b) % c = (a * (a ^( b-1))%c
Property 3:
If we have to return the mod of a negative number x whose absolute value is less than y:
then (x + y) % y will do the trick
Note:
Also as the product of (a ^ b/2) * (a ^ b/2) and a * (a ^( b-1) may cause overflow, hence we must be careful about those scenarios

`// Recursive C++ program to compute modular power`

`#include <bits/stdc++.h>`

`using` `namespace` `std;`

`int` `exponentMod(` `int` `A, ` `int` `B, ` `int` `C)`

`{`

` ` `// Base cases`

` ` `if` `(A == 0)`

` ` `return` `0;`

` ` `if` `(B == 0)`

` ` `return` `1;`

` ` `// If B is even`

` ` `long` `y;`

` ` `if` `(B % 2 == 0) {`

` ` `y = exponentMod(A, B / 2, C);`

` ` `y = (y * y) % C;`

` ` `}`

` ` `// If B is odd`

` ` `else` `{`

` ` `y = A % C;`

` ` `y = (y * exponentMod(A, B - 1, C) % C) % C;`

` ` `}`

` ` `return` `(` `int` `)((y + C) % C);`

`}`

`// Driver code`

`int` `main()`

`{`

` ` `int` `A = 2, B = 5, C = 13;`

` ` `cout << ` `"Power is "` `<< exponentMod(A, B, C);`

` ` `return` `0;`

`}`

Output:

Power is 6