Python Program for Number of Solutions to Modular Equations (original) (raw)

Last Updated : 4 Nov, 2025

Given two integers A and B, the task is to find how many integer values of X satisfy the modular equation: **A mod X = B. Here, X is called a solution of the modular equation. In simple terms, find all integers X such that when A is divided by X, the remainder is B.

**For Example:

**Input: A = 26, B = 2
**Output: 6
Possible X values -> {3, 4, 6, 8, 12, 24}

From the definition of modulus:

A = X * Y + B --> A - B = X * Y

So, X must be a divisor of (A − B). Also, because remainder B must be strictly less than the divisor X, only those divisors of (A − B) that are greater than B are valid solutions.

Let’s explore different methods to do this in Python.

Using square-root divisor method

This is the fastest method to count all divisors of (A - B) that are greater than B using a single loop up to √(A - B).

Python `

A = 26 B = 2

When A and B are equal → Infinite solutions

if A == B: print("For A =", A, "and B =", B, ", X can take infinitely many values greater than", A)

When A is less than B → No solution possible

elif A < B: print("For A =", A, "and B =", B, ", X cannot take any value")

When A is greater than B

else: count = 0 diff = A - B sqrt_diff = int(diff ** 0.5)

for i in range(1, sqrt_diff + 1):
    if diff % i == 0:
        if i > B:
            count += 1
        if diff // i != i and diff // i > B:
            count += 1

print("For A =", A, "and B =", B, ", X can take", count, "values")

Output

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

**Explanation:

diff = A - B -> Finds the difference to analyze divisors.
diff % i == 0 -> Checks if i divides (A - B) evenly.
i > B and (diff // i) > B -> Only counts divisors greater than B.
Counts all valid divisors and prints the result.

Using math.sqrt()

This version uses the math module for clearer square root calculation and keeps the logic simple.

Python `

import math A = 21 B = 5

if A == B: print("For A =", A, "and B =", B, ", X can take infinitely many values greater than", A)

elif A < B: print("For A =", A, "and B =", B, ", X cannot take any value")

else: count = 0 N = A - B limit = int(math.sqrt(N))

for i in range(1, limit + 1):
    if N % i == 0:
        if i > B:
            count += 1
        if N // i != i and N // i > B:
            count += 1

print("For A =", A, "and B =", B, ", X can take", count, "values")

Output

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

**Explanation:

math.sqrt() calculates the integer square root limit for iteration.
Divisors and their pairs are checked and counted if greater than B.
The final count gives all valid X values satisfying A % X == B.

Using brute force check

This method checks every possible X from B + 1 to A and counts those that satisfy the modular equation. It’s simple but not efficient for large numbers.

Python `

A = 26 B = 2

if A == B: print("For A =", A, "and B =", B, ", X can take infinitely many values greater than", A)

elif A < B: print("For A =", A, "and B =", B, ", X cannot take any value")

else: count = 0 for X in range(B + 1, A + 1): if A % X == B: count += 1

print("For A =", A, "and B =", B, ", X can take", count, "values")

Output

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

**Explanation:

Loops through all possible X values from B + 1 to A.
Uses A % X == B to verify the condition.