# 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 .
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// CPP program to find x such that a % x is equal ` `// to b. ` `#include ` `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

 `# 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. `

## 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

 ` ``\$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. ` `?> `

Output:

```2
```

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Improved By : vt_m, sanskar27jain