Python 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) =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)

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

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# 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 isnt 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

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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!




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