Python Program for Number of solutions to Modular Equations
Last Updated :
23 Jul, 2025
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 + BLet 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
OutputFor 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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