C/C++ Program for Number of solutions to Modular Equations

Given A and B, the task is to find the number of possible values that X can take such that the given modular equation (A mod X) = B holds good. Here, X is also called a solution of the modular equation. Examples:

Input : A = 26, B = 2
Output : 6
Explanation
X can be equal to any of {3, 4, 6, 8,
12, 24} as A modulus any of these values
equals 2 i. e., (26 mod 3) = (26 mod 4) 
= (26 mod 6) = (26 mod 8) = .... = 2 

Input : 21 5
Output : 2
Explanation
X can be equal to any of {8, 16} as A modulus 
any of these values equals 5 i.e. (21 mod 
8) = (21 mod 16) = 5

If we carefully analyze the equation A mod X = B its easy to note that if (A = B) then there are infinitely many values greater than A that X can take. In the Case when (A < B), there cannot be any possible value of X for which the modular equation holds. So the only case we are left to investigate is when (A > B).So now we focus on this case in depth. Now, in this case we can use a well known relation i.e.

Dividend = Divisor * Quotient + Remainder

We are looking for all possible X i.e. Divisors given A i.e Dividend and B i.e., remainder. So,

We can say,
A = X * Quotient + B

Let Quotient be represented as Y
? A = X * Y + B
A - B = X * Y

? To get integral values of Y, 
we need to take all X such that X divides (A - B)

? X is a divisor of (A - B)

So, the problem reduces to finding the divisors of (A – B) and the number of such divisors is the possible values X can take. But as we know A mod X would result in values from (0 to X – 1) we must take all such X such that X > B. Thus, we can conclude by saying that the number of divisors of (A – B) greater than B, are the all possible values X can take to satisfy A mod X = B 

CPP

/* C++ Program to find number of possible
   values of X to satisfy A mod X = B */
#include <bits/stdc++.h>
using namespace std;

/* Returns the number of divisors of (A - B)
   greater than B */
int calculateDivisors(int A, int B)
{
    int N = (A - B);
    int noOfDivisors = 0;

    for (int i = 1; i <= sqrt(N); i++) {

        // if N is divisible by i
        if ((N % i) == 0) {

            // count only the divisors greater than B
            if (i > B)
                noOfDivisors++;

            // checking if a divisor isnot counted twice
            if ((N / i) != i && (N / i) > B)
                noOfDivisors++;
        }
    }

    return noOfDivisors;
}

/* Utility function to calculate number of all
   possible values of X for which the modular
   equation holds true */
int numberOfPossibleWaysUtil(int A, int B)
{

    /* if A = B there are infinitely many solutions
       to equation  or we say X can take infinitely
       many values > A. We return -1 in this case */
    if (A == B)
        return -1;

    /* if A < B, there are no possible values of
       X satisfying the equation */
    if (A < B)
        return 0;

    /* the last case is when A > B, here we calculate
       the number of divisors of (A - B), which are
       greater than B */
    int noOfDivisors = 0;
    noOfDivisors = calculateDivisors(A, B);
    return noOfDivisors;
}

/* Wrapper function for numberOfPossibleWaysUtil() */
void numberOfPossibleWays(int A, int B)
{
    int noOfSolutions = numberOfPossibleWaysUtil(A, B);

    // if infinitely many solutions available
    if (noOfSolutions == -1) {
        cout << "For A = " << A << " and B = " << B
             << ", X can take Infinitely many values"
                " greater than "
             << A << "\n";
    }

    else {
        cout << "For A = " << A << " and B = " << B
             << ", X can take " << noOfSolutions
             << " values\n";
    }
}

// Driver code
int main()
{
    int A = 26, B = 2;
    numberOfPossibleWays(A, B);
    A = 21, B = 5;
    numberOfPossibleWays(A, B);
    return 0;
}

Output:

For A = 26 and B = 2, X can take 6 values
For A = 21 and B = 5, X can take 2 values

Time Complexity of the above approach is nothing but the time complexity of finding the number of divisors of (A – B) ie O(?(A – B)) Please refer complete article on Number of solutions to Modular Equations for more details!