# Find numbers that divide X and Y to produce the same remainder

Given two integers X and Y, the task is to find and print the numbers that divide X and Y to produce the same remainder.

Examples:

Input: X = 1, Y = 5
Output: 1, 2, 4
Explanation:
Let the number be M. It can be any value in the range [1, 5]:
If M = 1, 1 % 1 = 0 and 5 % 1 = 0
If M = 2, 1 % 2 = 1 and 5 % 2 = 1
If M = 3, 1 % 3 = 1 and 5 % 3 = 2
If M = 4, 1 % 4 = 1 and 5 % 4 = 1
If M = 5, 1 % 5 = 1 and 5 % 5 = 0
Therefore, the possible M values are 1, 2, 4

Input: X = 8, Y = 10
Output: 1, 2

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

Naive Approach: The naive approach for this problem is to check the modulo value for all the possible values of M in the range [1, max(X, Y)] and print the value of M if the condition satisfies.

Below is the implementation of the above approach:

## C++

 `// C++ program to find numbers ` `// that divide X and Y ` `// to produce the same remainder ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to find ` `// the required number as M ` `void` `printModulus(``int` `X, ``int` `Y) ` `{ ` `    ``// Finding the maximum ` `    ``// value among X and Y ` `    ``int` `n = max(X, Y); ` ` `  `    ``// Loop to iterate through ` `    ``// maximum value among X and Y. ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` ` `  `        ``// If the condition satisfies, then ` `        ``// print the value of M ` `        ``if` `(X % i == Y % i) ` `            ``cout << i << ``" "``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `X, Y; ` `    ``X = 10; ` `    ``Y = 20; ` `    ``printModulus(X, Y); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find numbers ` `// that divide X and Y ` `// to produce the same remainder ` `class` `GFG{ ` `  `  `// Function to find ` `// the required number as M ` `static` `void` `printModulus(``int` `X, ``int` `Y) ` `{ ` `    ``// Finding the maximum ` `    ``// value among X and Y ` `    ``int` `n = Math.max(X, Y); ` `  `  `    ``// Loop to iterate through ` `    ``// maximum value among X and Y. ` `    ``for` `(``int` `i = ``1``; i <= n; i++) { ` `  `  `        ``// If the condition satisfies, then ` `        ``// print the value of M ` `        ``if` `(X % i == Y % i) ` `            ``System.out.print(i + ``" "``); ` `    ``} ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `  `  `    ``int` `X, Y; ` `    ``X = ``10``; ` `    ``Y = ``20``; ` `    ``printModulus(X, Y); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## Python3

 `# Python program to find numbers ` `# that divide X and Y ` `# to produce the same remainder ` ` `  `# Function to find ` `# the required number as M ` `def` `printModulus( X, Y): ` `     `  `    ``# Finding the maximum ` `    ``# value among X and Y ` `    ``n ``=` `max``(X, Y) ` ` `  `    ``# Loop to iterate through ` `    ``# maximum value among X and Y. ` `    ``for` `i ``in` `range``(``1``, n ``+` `1``): ` ` `  `        ``# If the condition satisfies, then ` `        ``# print the value of M ` `        ``if` `(X ``%` `i ``=``=` `Y ``%` `i): ` `            ``print``(i,end``=``" "``) ` ` `  `# Driver code ` `X ``=` `10` `Y ``=` `20` `printModulus(X, Y) ` ` `  `# This code is contributed by Atul_kumar_Shrivastava `

## C#

 `// C# program to find numbers ` `// that divide X and Y ` `// to produce the same remainder ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Function to find ` `// the required number as M ` `static` `void` `printModulus(``int` `X, ``int` `Y) ` `{ ` `    ``// Finding the maximum ` `    ``// value among X and Y ` `    ``int` `n = Math.Max(X, Y); ` ` `  `    ``// Loop to iterate through ` `    ``// maximum value among X and Y. ` `    ``for` `(``int` `i = 1; i <= n; i++) { ` ` `  `        ``// If the condition satisfies, then ` `        ``// print the value of M ` `        ``if` `(X % i == Y % i) ` `            ``Console.Write(i + ``" "``); ` `    ``} ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main() ` `{ ` `    ``int` `X, Y; ` `    ``X = 10; ` `    ``Y = 20; ` `    ``printModulus(X, Y); ` `} ` `} ` ` `  `// This code is contributed by AbhiThakur `

Output:

```1 2 5 10
```

Time Complexity: O(max(X, Y))

Efficient Approach: Let’s assume that Y is greater than X by a difference of D.

• Then Y can be expressed as
```Y = X + D
and
Y % M = (X + D) % M
= (X % M) + (D % M)
```
• Now, the condition becomes whether X % M and X % M + D % M are equal or not.
• Here, since X % M is common on both the sides, the value of M is true if for some M, D % M = 0.
• Therefore, the required values of M will be the factors of D.

Below is the implementation of the above approach:

## CPP

 `// C++ program to find numbers ` `// that divide X and Y to ` `// produce the same remainder ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to print all the possible values ` `// of M such that X % M = Y % M ` `void` `printModulus(``int` `X, ``int` `Y) ` `{ ` `    ``// Finding the absolute difference ` `    ``// of X and Y ` `    ``int` `d = ``abs``(X - Y); ` ` `  `    ``// Iterating from 1 ` `    ``int` `i = 1; ` ` `  `    ``// Loop to print all the factors of D ` `    ``while` `(i * i <= d) { ` ` `  `        ``// If i is a factor of d, then print i ` `        ``if` `(d % i == 0) { ` `            ``cout << i << ``" "``; ` ` `  `            ``// If d / i is a factor of d, ` `            ``// then print d / i ` `            ``if` `(d / i != i) ` `                ``cout << d / i << ``" "``; ` `        ``} ` `        ``i++; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``int` `X = 10; ` `    ``int` `Y = 26; ` ` `  `    ``printModulus(X, Y); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find numbers ` `// that divide X and Y to ` `// produce the same remainder ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function to print all the possible values ` `// of M such that X % M = Y % M ` `static` `void` `printModulus(``int` `X, ``int` `Y) ` `{ ` `    ``// Finding the absolute difference ` `    ``// of X and Y ` `    ``int` `d = Math.abs(X - Y); ` `  `  `    ``// Iterating from 1 ` `    ``int` `i = ``1``; ` `  `  `    ``// Loop to print all the factors of D ` `    ``while` `(i * i <= d) { ` `  `  `        ``// If i is a factor of d, then print i ` `        ``if` `(d % i == ``0``) { ` `            ``System.out.print(i+ ``" "``); ` `  `  `            ``// If d / i is a factor of d, ` `            ``// then print d / i ` `            ``if` `(d / i != i) ` `                ``System.out.print(d / i+ ``" "``); ` `        ``} ` `        ``i++; ` `    ``} ` `} ` `  `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `  `  `    ``int` `X = ``10``; ` `    ``int` `Y = ``26``; ` `  `  `    ``printModulus(X, Y); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

## Python3

 `# Python program to find numbers ` `# that divide X and Y to ` `# produce the same remainder ` ` `  `# Function to prall the possible values ` `# of M such that X % M = Y % M ` `def` `printModulus(X, Y): ` `    ``# Finding the absolute difference ` `    ``# of X and Y ` `    ``d ``=` `abs``(X ``-` `Y); ` ` `  `    ``# Iterating from 1 ` `    ``i ``=` `1``; ` ` `  `    ``# Loop to prall the factors of D ` `    ``while` `(i ``*` `i <``=` `d): ` ` `  `        ``# If i is a factor of d, then pri ` `        ``if` `(d ``%` `i ``=``=` `0``): ` `            ``print``(i, end``=``""); ` ` `  `            ``# If d / i is a factor of d, ` `            ``# then prd / i ` `            ``if` `(d ``/``/` `i !``=` `i): ` `                ``print``(d ``/``/` `i, end``=``" "``); ` `         `  `        ``i``+``=``1``; ` `     `  ` `  ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` ` `  `    ``X ``=` `10``; ` `    ``Y ``=` `26``; ` ` `  `    ``printModulus(X, Y); ` ` `  `# This code contributed by Princi Singh `

## C#

 `// C# program to find numbers ` `// that divide X and Y to ` `// produce the same remainder ` `using` `System; ` ` `  `public` `class` `GFG{ ` `   `  `// Function to print all the possible values ` `// of M such that X % M = Y % M ` `static` `void` `printModulus(``int` `X, ``int` `Y) ` `{ ` `    ``// Finding the absolute difference ` `    ``// of X and Y ` `    ``int` `d = Math.Abs(X - Y); ` `   `  `    ``// Iterating from 1 ` `    ``int` `i = 1; ` `   `  `    ``// Loop to print all the factors of D ` `    ``while` `(i * i <= d) { ` `   `  `        ``// If i is a factor of d, then print i ` `        ``if` `(d % i == 0) { ` `            ``Console.Write(i+ ``" "``); ` `   `  `            ``// If d / i is a factor of d, ` `            ``// then print d / i ` `            ``if` `(d / i != i) ` `                ``Console.Write(d / i+ ``" "``); ` `        ``} ` `        ``i++; ` `    ``} ` `} ` `   `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `   `  `    ``int` `X = 10; ` `    ``int` `Y = 26; ` `   `  `    ``printModulus(X, Y); ` `} ` `} ` `  `  `// This code contributed by Princi Singh `

Output:

```1 16 2 8 4
```

Time Complexity Analysis O(sqrt(D)), where D is the difference between the values X and Y.

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