Number of solutions to Modular Equations

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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)
= …. = 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

C++

/* C++ Program to find number of possible 
   values of X to satisfy A mod X = B */
#include <bits/stdc++.h>
using namespace std;
 
/* Returns the number of divisors of (A - B)
   greater than B */
int calculateDivisors(int A, int B)
{
    int N = (A - B);
    int noOfDivisors = 0;
 
    for (int i = 1; i <= sqrt(N); i++) {
 
        // if N is divisible by i
        if ((N % i) == 0) {
 
            // count only the divisors greater than B
            if (i > B)
                noOfDivisors++;
 
            // checking if a divisor isnot counted twice
            if ((N / i) != i && (N / i) > B)
                noOfDivisors++;
        }
    }
 
    return noOfDivisors;
}
 
/* Utility function to calculate number of all 
   possible values of X for which the modular 
   equation holds true */
int numberOfPossibleWaysUtil(int A, int 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 */
    int noOfDivisors = 0;
    noOfDivisors = calculateDivisors(A, B);
    return noOfDivisors;
}
 
/* Wrapper function for numberOfPossibleWaysUtil() */
void numberOfPossibleWays(int A, int B)
{
    int noOfSolutions = numberOfPossibleWaysUtil(A, B);
 
    // if infinitely many solutions available
    if (noOfSolutions == -1) {
        cout << "For A = " << A << " and B = " << B
             << ", X can take Infinitely many values"
             " greater than "  << A << "\n";
    }
 
    else {
        cout << "For A = " << A << " and B = " << B
             << ", X can take " << noOfSolutions
              << " values\n";
    }
}
 
// Driver code
int main()
{
    int A = 26, B = 2;
    numberOfPossibleWays(A, B);
    A = 21, B = 5;
    numberOfPossibleWays(A, B);
    return 0;
}

Java

/* Java Program to find number of possible
   values of X to satisfy A mod X = B */
import java.lang.*;
 
class GFG
{
    /* Returns the number of divisors of (A - B)
       greater than B */
    public static int calculateDivisors(int A, int B)
    {
        int N = (A - B);
        int noOfDivisors = 0;
 
        for (int i = 1; i <= Math.sqrt(N); i++)
        {
 
            // if N is divisible by i
            if ((N % i) == 0)
            {
 
                // count only the divisors greater than B
                if (i > B)
                    noOfDivisors++;
 
                // checking if a divisor isnot counted twice
                if ((N / i) != i && (N / i) > B)
                    noOfDivisors++;
            }
        }
        return noOfDivisors;
    }
 
    /* Utility function to calculate number of all
       possible values of X for which the modular
       equation holds true */
    public static int numberOfPossibleWaysUtil(int A, int 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 */
        int noOfDivisors = 0;
        noOfDivisors = calculateDivisors(A, B);
        return noOfDivisors;
    }
 
    /* Wrapper function for numberOfPossibleWaysUtil() */
    public static void numberOfPossibleWays(int A, int B)
    {
        int noOfSolutions = numberOfPossibleWaysUtil(A, B);
 
        // if infinitely many solutions available
        if (noOfSolutions == -1)
        {
            System.out.print("For A = " + A + " and B = " + B
                             + ", X can take Infinitely many values"
                             + " greater than "  + A + "\n");
        }
 
        else
        {
            System.out.print("For A = " + A + " and B = " + B
                             + ", X can take " + noOfSolutions
                             + " values\n");
        }
    }
    // Driver program
    public static void main(String[] args)
    {
        int A = 26, B = 2;
        numberOfPossibleWays(A, B);
        A = 21;
        B = 5;
        numberOfPossibleWays(A, B);
    }
}
// Contributed by _omg

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

C#

/* C# Program to find number of possible
   values of X to satisfy A mod X = B */
using System;
 
class GFG
{
    /* Returns the number of divisors of (A - B)
       greater than B */
    static int calculateDivisors(int A, int B)
    {
        int N = (A - B);
        int noOfDivisors = 0;
 
        double a = Math.Sqrt(N);
        for (int i = 1; i <= (int)(a); i++)
        {
 
            // if N is divisible by i
            if ((N % i) == 0)
            {
 
                // count only the divisors greater than B
                if (i > B)
                    noOfDivisors++;
 
                // checking if a divisor isnot counted twice
                if ((N / i) != i && (N / i) > B)
                    noOfDivisors++;
            }
        }
        return noOfDivisors;
    }
 
    /* Utility function to calculate number of all
       possible values of X for which the modular
       equation holds true */
    static int numberOfPossibleWaysUtil(int A, int 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 */
        int noOfDivisors = 0;
        noOfDivisors = calculateDivisors(A, B);
        return noOfDivisors;
    }
 
    /* Wrapper function for numberOfPossibleWaysUtil() */
    public static void numberOfPossibleWays(int A, int B)
    {
        int noOfSolutions = numberOfPossibleWaysUtil(A, B);
 
        // if infinitely many solutions available
        if (noOfSolutions == -1)
        {
            Console.Write ("For A = " + A + " and B = " + B
                           + ", X can take Infinitely many values"
                           + " greater than "  + A + "\n");
        }
 
        else
        {
            Console.Write ("For A = " + A + " and B = " + B
                           + ", X can take " + noOfSolutions
                           + " values\n");
        }
    }
 
    public static void Main()
    {
        int A = 26, B = 2;
        numberOfPossibleWays(A, B);
        A = 21;
        B = 5;
        numberOfPossibleWays(A, B);
    }
}
// Contributed by _omg

PHP

<?php
/* PHP Program to find number of possible 
values of X to satisfy A mod X = B */
 
/* Returns the number of divisors of (A - B)
greater than B */
 
function calculateDivisors($A, $B)
{
    $N = ($A - $B);
    $noOfDivisors = 0;
 
    for ($i = 1; $i <= sqrt($N); $i++) {
 
        // if N is divisible by i
        if (($N % $i) == 0) {
 
            // count only the divisors greater than B
            if ($i > $B)
                $noOfDivisors++;
 
            // checking if a divisor isnot counted twice
            if (($N / $i) != $i && ($N / $i) > $B)
                $noOfDivisors++;
        }
    }
 
    return $noOfDivisors;
}
 
/* Utility function to calculate number of all 
possible values of X for which the modular 
equation holds true */
function 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;
    $noOfDivisors = calculateDivisors($A, $B);
    return $noOfDivisors;
}
 
/* Wrapper function for numberOfPossibleWaysUtil() */
function numberOfPossibleWays($A, $B)
{
    $noOfSolutions = numberOfPossibleWaysUtil($A, $B);
 
    // if infinitely many solutions available
    if ($noOfSolutions == -1) {
        echo "For A = " , $A, " and B = " ,$B,
            "X can take Infinitely many values
            greater than " , $A , "\n";
    }
 
    else {
        echo "For A = ", $A , " and B = " ,$B,
            " X can take ",$noOfSolutions,
            " values\n";
    }
}
 
// Driver code
 
    $A = 26; $B = 2;
    numberOfPossibleWays($A, $B);
    $A = 21; $B = 5;
    numberOfPossibleWays($A, $B);
     
// This code is contributed ajit.
?>

Javascript

<script>
 
// JavaScript Program to find number of possible
// values of X to satisfy A mod X = B 
 
// Returns the number of divisors of (A - B)
// greater than B 
function calculateDivisors(A, B)
{
    let N = (A - B);
    let noOfDivisors = 0;
   
    for(let i = 1; i <= Math.sqrt(N); i++)
    {
         
        // If N is divisible by i
        if ((N % i) == 0) 
        {
             
            // Count only the divisors
            // greater than B
            if (i > B)
                noOfDivisors++;
   
            // Checking if a divisor 
            // isnot counted twice
            if ((N / i) != i && (N / i) > B)
                noOfDivisors++;
        }
    }
   
    return noOfDivisors;
}
   
// Utility function to calculate number of all 
// possible values of X for which the modular 
// equation holds true 
function 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 
    let noOfDivisors = 0;
    noOfDivisors = calculateDivisors(A, B);
    return noOfDivisors;
}
   
// Wrapper function for numberOfPossibleWaysUtil() 
function numberOfPossibleWays(A, B)
{
    let noOfSolutions = numberOfPossibleWaysUtil(A, B);
   
    // If infinitely many solutions available
    if (noOfSolutions == -1)
    {
        document.write("For A = " + A + " and B = " +
                       B + ", X can take Infinitely " + 
                       "many values" + " greater than " +
                       A + "<br/>");
    }
    else
    {
        document.write("For A = " + A + " and B = " +
                       B + ", X can take " + 
                       noOfSolutions + " values "  +
                       "<br/>");
    }
}
 
// Driver code
let A = 26, B = 2;
numberOfPossibleWays(A, B);
 
A = 21, B = 5;
numberOfPossibleWays(A, B);
 
// This code is contributed by splevel62
 
</script>

The 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))
Auxiliary Space: O(1)

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