Related Articles

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

• Difficulty Level : Hard
• Last Updated : 25 Mar, 2021

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

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`

## Javascript

 ``
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`

## Javascript

 `  ```
Output:
`1 16 2 8 4`

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up