Python Program for Number of solutions to Modular Equations

Last Updated : 27 Jul, 2023

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) =Output: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)

Python Program for Number of solutions to Modular Equations

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

Python3

# Python Program to find number of possible
# values of X to satisfy A mod X = B 
import math
 
# Returns the number of divisors of (A - B)
# greater than B
def calculateDivisors (A, B):
    N = A - B
    noOfDivisors = 0
     
    a = math.sqrt(N)
    for i in range(1, int(a + 1)):
        # if N is divisible by i
        if ((N % i == 0)):
            # count only the divisors greater than B
            if (i > B):
                noOfDivisors +=1
                 
            # checking if a divisor isnot counted twice
            if ((N / i) != i and (N / i) > B):
                noOfDivisors += 1;
                 
    return noOfDivisors
     
# Utility function to calculate number of all
# possible values of X for which the modular
# equation holds true 
    
def numberOfPossibleWaysUtil (A, 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    
     
    noOfDivisors = 0
    noOfDivisors = calculateDivisors;
    return noOfDivisors
         
     
# Wrapper function for numberOfPossibleWaysUtil() 
def numberOfPossibleWays(A, B):
    noOfSolutions = numberOfPossibleWaysUtil(A, B)
     
    #if infinitely many solutions available
    if (noOfSolutions == -1):
        print ("For A = " , A , " and B = " , B
                , ", X can take Infinitely many values"
                , " greater than "  , A)
     
    else:
        print ("For A = " , A , " and B = " , B
                , ", X can take " , noOfSolutions
                , " values")
# main()
A = 26
B = 2
numberOfPossibleWays(A, B)
 
 
A = 21
B = 5
numberOfPossibleWays(A, B)
 
# Contributed by _omg

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!

Method 2:

The new approach for solving the problem of finding the number of possible values of X to satisfy A mod X = B involves:

Checking for special cases where the solution is either infinite or non-existent.
Calculating the difference between A and B, and finding all its divisors.
Counting the number of divisors that are greater than B.
Returning the count of such divisors as the number of possible values of X.

Python3

# function to calculate the number of possible values for X
def numberOfPossibleWays(A, B):
    # if A and B are equal, there are no possible values for X
    if A == B:
        return -1
    # if A is less than B, X cannot be greater than B
    elif A < B:
        return 0
    else:
        count = 0
        diff = A - B
        sqrt_diff = int(diff ** 0.5)
        # iterate over the range of square root of difference
        for i in range(1, sqrt_diff + 1):
            # check if i is a factor of the difference
            if diff % i == 0:
                # if i is greater than B, increment the count
                if i > B:
                    count += 1
                # check if the quotient is greater than B and not equal to i, then increment the count
                if diff // i != i and diff // i > B:
                    count += 1
        return count
 
# test case 1
A = 26
B = 2
print("For A =", A, "and B =", B, ", X can take", numberOfPossibleWays(A, B), "values")
 
# test case 2
A = 21
B = 5
print("For A =", A, "and B =", B, ", X can take", numberOfPossibleWays(A, B), "values")
 
# Contributed by adityasha4x71

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: O(sqrt(A-B)), because the loop iterates from 1 to the square root of A-B.

Auxiliary Space: O(1)

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Python Program for Number of solutions to Modular Equations

Python Program for Number of solutions to Modular Equations

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