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)

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 

# 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.

# 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 

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)