Given two numbers a and b find all x such that a % x = b

Given two numbers a and b find all x such that a % x = b .

Examples:

Input : a = 21, b = 5
Output : 2
The answers of the Modular Equation are
8 and 16 since 21 % 8 = 21 % 16 = 5 .

Here 3 cases arises :

  1. If ( a < b ) then there will be no answer .
  2. If ( a = b ) then all the numbers greater than a are the answer so there will be infinite solutions possible.
  3. If ( a > b ) Suppose x is an answer to our equation. Then x divides (a – b). Also since a % x = b then b < x. These conditions are necessary and sufficient as well. So the answer is number of divisors of a – b which are strictly greater than b which can be solved in O(sqrt( a-b )). Here only one case arises which we have to deal separately when (a-b) is perfect square then we will add its square root two times so we have to subtract one times, if this case arises.

C++

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// CPP program to find x such that a % x is equal
// to b.
#include <bits/stdc++.h>
using namespace std;
  
void modularEquation(int a, int b)
{
    // if a is less than b then no solution
    if (a < b) {
        cout << "No solution possible " << endl;
        return;
    }
  
    // if a is equal to b then every number
    // greater than a will be the solution
    // so its infinity
    if (a == b) {
        cout << "Infinite Solution possible " << endl;
        return;
    }
  
    // all resultant number should be greater than
    // b and (a-b) should be divisible by resultant
    // number
  
    // count variable store the number of values
    // possible
    int count = 0;
    int n = a - b;
    int y = sqrt(a - b);
    for (int i = 1; i <= y; ++i) {
        if (n % i == 0) {
  
            // checking for both divisor and quotient
            // whether they divide ( a-b ) completely
            // and greater than b .
            if (n / i > b)
                count++;
            if (i > b)
                count++;
        }
    }
  
    // Here y is added twice in the last iteration
    // so 1 y should be decremented to get correct
    // solution
    if (y * y == n && y > b)
        count--;
  
    cout << count << endl;
}
  
// Driver code
int main()
{
    int a = 21, b = 5;
    modularEquation(a, b);
    return 0;
}

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Java

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// Java program to find x such that 
// a % x is equal to b.
import java.io.*;
class GFG {
      
static void modularEquation(int a, int b)
{
    // if a is less than b then no solution
    if (a < b) {
        System.out.println("No solution possible ");
        return;
    }
  
    // if a is equal to b then every number
    // greater than a will be the solution
    // so its infinity
    if (a == b) {
        System.out.println("Infinite Solution possible ");
        return;
    }
  
    // all resultant number should be greater
    // than b and (a-b) should be divisible 
    // by resultant number
  
    // count variable store the number of
    // values possible
    int count = 0;
    int n = a - b;
    int y = (int)Math.sqrt(a - b);
    for (int i = 1; i <= y; ++i) {
        if (n % i == 0) {
  
            // checking for both divisor and
            // quotient whether they divide
            // ( a-b ) completely and 
            // greater than b .
            if (n / i > b)
                count++;
            if (i > b)
                count++;
        }
    }
  
    // Here y is added twice in the last 
    // iteration so 1 y should be decremented
    // to get correct solution
    if (y * y == n && y > b)
        count--;
  
    System.out.println(count);
}
  
// Driver code
public static void main(String[] args)
{
    int a = 21, b = 5;
    modularEquation(a, b);
}
}
  
// This code is contributed by Prerna Saini

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Python3

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# Python 3 program to find x such 
# that a % x is equal to b.
  
import math
def modularEquation(a, b) :
      
    # if a is less than b then no solution
    if (a < b) :
        print("No solution possible "
        return 
      
    # if a is equal to b then every number
    # greater than a will be the solution
    # so its infinity
    if (a == b) :
        print("Infinite Solution possible ")
        return
      
    # all resultant number should be 
    # greater than b and (a-b) should 
    # be divisible by resultant number
   
    # count variable store the number 
    # of values possible
    count = 0
    n = a - b
    y = (int)(math.sqrt(a - b))
    for i in range(1, y+1) :
        if (n % i == 0) :
              
            # checking for both divisor 
            # and quotient whether they
            # divide ( a-b ) completely
            # and greater than b .
            if (n / i > b) :
                count = count + 1
            if (i > b) :
                count = count  + 1
          
          
   
    # Here y is added twice in the 
    # last iteration so 1 y should be
    # decremented to get correct
    # solution
    if (y * y == n and y > b) :
        count = count - 1
   
    print (count)
   
# Driver code
a = 21
b = 5
modularEquation(a, b)
  
# This code is contributed by Nikita Tiwari.

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C#

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// C# program to find x such that 
// a % x is equal to b.
using System;
  
class GFG {
      
static void modularEquation(int a, int b)
{
    // if a is less than b then no solution
    if (a < b) {
        Console.WriteLine("No solution possible ");
        return;
    }
  
    // if a is equal to b then every number
    // greater than a will be the solution
    // so its infinity
    if (a == b) {
        Console.WriteLine("Infinite Solution possible ");
        return;
    }
  
    // all resultant number should be greater
    // than b and (a-b) should be divisible 
    // by resultant number
  
    // count variable store the number of
    // values possible
    int count = 0;
    int n = a - b;
    int y = (int)Math.Sqrt(a - b);
    for (int i = 1; i <= y; ++i) {
        if (n % i == 0) {
  
            // checking for both divisor and
            // quotient whether they divide
            // ( a-b ) completely and 
            // greater than b .
            if (n / i > b)
                count++;
            if (i > b)
                count++;
        }
    }
  
    // Here y is added twice in the last 
    // iteration so 1 y should be decremented
    // to get correct solution
    if (y * y == n && y > b)
        count--;
  
    Console.WriteLine(count);
}
  
// Driver code
public static void Main()
{
    int a = 21, b = 5;
    modularEquation(a, b);
}
}
  
//This code is contributed by vt_m.

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PHP

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<?php
// PHP program to find x 
// such that a % x is equal
// to b.
  
function modularEquation($a, $b)
{
      
    // if a is less than b 
    // then no solution
    if ($a < $b
    {
        echo "No solution possible " ;
        return;
    }
  
    // if a is equal to b 
    // then every number
    // greater than a will 
    // be the solution
    // so its infinity
    if ($a == $b
    {
        echo "Infinite Solution possible ";
        return;
    }
  
    // all resultant number 
    // should be greater than
    // b and (a-b) should be 
    // divisible by resultant
    // number
  
    // count variable store 
    // the number of values
    // possible
    $count = 0;
    $n = $a - $b;
    $y = sqrt($a - $b);
    for ( $i = 1; $i <= $y; ++$i)
    {
        if ($n % $i == 0) {
  
            // checking for both 
            // divisor and quotient
            // whether they divide 
            // ( a-b ) completely
            // and greater than b .
            if ($n / $i > $b)
                $count++;
            if ($i > $b)
                $count++;
        }
    }
  
    // Here y is added twice
    // in the last iteration
    // so 1 y should be 
    // decremented to get correct
    // solution
    if ($y * $y == $n && $y > $b)
        $count--;
  
    echo $count ;
}
  
    // Driver Code
    $a = 21; 
    $b = 5;
    modularEquation($a, $b);
  
// This code is contributed by anuj_67.
?>

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Output:

2


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Improved By : vt_m, sanskar27jain