Find power of power under mod of a prime


Given four numbers A, B, C and M, where M is prime number. Our task is to find ABC (mod M).

Input  : A = 2, B = 4, C = 3, M = 23
Output : 6
243(mod 23) = 6

A Naive Approach is to calculate res = BC and then calculate Ares % M by modular exponential. The problem of this approach is that we can’t apply directly mod M on BC, so we have to calculate this value without mod M. But if we solve it directly then we will come up with the large value of exponent of A which will definitely overflow in final answer.

An Efficient approach is to reduce the BC to a smaller value by using the Fermat’s Little Theorem, and then apply Modular exponential.

According the Fermat's little
a(M - 1) = 1 (mod M) if M is a prime.

So if we rewrite BC as x*(M-1) + y, then the
task of computing ABC becomes Ax*(M-1) + y
which can be written as Ax*(M-1)*Ayx*(M-1) = 1.
So task of computing ABC reduces to computing Ay

What is the value of y?
From BC = x * (M - 1) + y,
y can be written as BC % (M-1)

We can easily use the above theorem such that we can get
A ^ (B ^ C) % M = (A ^ y ) %  M

Now we only need to find two things as:-
1. y = (B ^ C) % (M - 1)
2. Ans = (A ^ y) % M


// C++ program to find (a^b) mod m for a large 'a'
using namespace std;

// Iterative Function to calculate (x^y)%p in O(log y)
unsigned int power(unsigned int x, unsigned int y, 
                                   unsigned int p)
    unsigned int res = 1;      // Initialize result

    x = x % p;  // Update x if it is more than or
                // equal to p

    while (y > 0)
        // If y is odd, multiply x with 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;

unsigned int Calculate(unsigned int A, unsigned int B,
                       unsigned int C, unsigned int M)
    unsigned int res, ans;

    // Calculate B ^ C (mod M - 1)
    res = power(B, C, M-1);

    // Calculate A ^ res ( mod M )
    ans = power(A, res, M);

    return ans;

// Driver program to run the case
int main()
{   // M must be be a Prime Number
    unsigned int A = 3, B = 9, C = 4, M = 19;

    cout << Calculate(A, B, C, M);

    return 0;


# Python program to calculate the ans
def calculate(A, B, C, M):

    # Calculate B ^ C (mod M - 1)
    res = pow(B, C, M-1)

    # Calculate A ^ res ( mod M )
    ans = pow(A, res, M)

    return ans

# Driver program to run the case
A = 3
B = 9
C = 4

# M must be Prime Number
M = 19
print( calculate(A, B, C, M) )

Time Complexity: O(log(B) + log(C))
Auxiliary space: O(1)

This article is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using or mail your article to See your article appearing on the GeeksforGeeks main page and help other Geeks.

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