Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

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

  • Difficulty Level : Medium
  • Last Updated : 30 Mar, 2021

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

Examples:  

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

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++




// C++ 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;
}
Java



// 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
Python3



# Python3 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.
C#



// 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.
PHP



<?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.
?>
Javascript



<script>
 
// Javascript 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)
    {
        document.write("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)
    {
        document.write("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
    let count = 0;
    let n = a - b;
    let y = Math.sqrt(a - b);
    for(let 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--;
 
    document.write(count);
}
 
// Driver Code
let a = 21;
let b = 5;
 
modularEquation(a, b);
 
// This code is contributed by _saurabh_jaiswal.
 
</script>
Output: 
2
 



My Personal Notes
arrow_drop_up
Next

Larger of a^b or b^a (a raised to power b or b raised to power a)
Recommended Articles
Page :
Find prime factors of Z such that Z is product of all even numbers till N that are product of two distinct prime numbers
28, Jun 21
Print numbers such that no two consecutive numbers are co-prime and every three consecutive numbers are co-prime
22, Jan 19
Find two numbers made up of a given digit such that their difference is divisible by N
07, May 20
Find two distinct numbers such that their LCM lies in given range
15, Aug 20
Permutation of numbers such that sum of two consecutive numbers is a perfect square
01, Mar 18
Find two Composite Numbers such that there difference is N
27, Dec 19
Find two numbers such that difference of their squares equal to N
05, Jun 20
Find two numbers with sum N such that neither of them contains digit K
30, Jun 20
Find all possible values of K such that the sum of first N numbers starting from K is G
22, Jul 21
Kth element in permutation of first N natural numbers having all even numbers placed before odd numbers in increasing order
03, Mar 21
Find two integers A and B such that A ^ N = A + N and B ^ N = B + N
30, Jul 19
Find sub-arrays from given two arrays such that they have equal sum
02, May 19
Find a permutation of 2N numbers such that the result of given expression is exactly 2K
11, Feb 19
Find two numbers with the given LCM and minimum possible difference
15, Jan 20
Find two numbers whose sum and GCD are given
21, Nov 18
Find two numbers with given sum and maximum possible LCM
18, Jul 20
Find all Ramanujan Numbers that can be formed by numbers upto L
19, Feb 21
Number of K's such that the given array can be divided into two sets satisfying the given conditions
29, Oct 19
Rearrange two given arrays such that sum of same indexed elements lies within given range
03, Nov 20
Sum of two numbers if the original ratio and new ratio obtained by adding a given number to each number is given
09, Aug 19
Probability that the sum of all numbers obtained on throwing a dice N times lies between two given integers
13, Jan 21
Print all integers that are sum of powers of two given numbers
08, Jan 19
Ways of dividing a group into two halves such that two elements are in different groups
30, Oct 18
Count triplets such that product of two numbers added with third number is N
27, Sep 21
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
Surya Priy
@Surya Priy
Vote for difficulty
Current difficulty :
Medium





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!