# Check whether a large number represented as array is divisible by Y

Given a large integer **X** represented as an array **arr[]** where each **arr[i]** stores a digit in **X**. The task is to check whether the number represented by the array is divisible by given integer **Y**.**Examples:**

Input:arr[] = {1, 2, 1, 5, 6}, Y = 4Output:Yes

12156 / 4 = 3039Input:arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1}, Y = 14Output:No

**Approach:** Start traversing the digits of the given number from the left and take the largest number which is smaller than or equal to **Y** and divide it with **Y**. If the remainder is something other than **0** then it will be carried to the next possible number formed from the remaining digits just like in the long division. After the complete number is processed, if the remainder is still something other than 0 then the represented number is not divisible by **Y** else it is.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <iostream>` `using` `namespace` `std;` `// Function that returns true if the number represented` `// by the given array is divisible by y` `bool` `isDivisible(` `int` `* arr, ` `int` `n, ` `int` `y)` `{` ` ` `int` `d = 0, i = 0;` ` ` `// While there are digits left` ` ` `while` `(i < n) {` ` ` `// Select the next part of the number` ` ` `// i.e. the maximum number which is <= y` ` ` `while` `(d < y && i < n)` ` ` `d = d * 10 + arr[i++];` ` ` `// Get the current remainder` ` ` `d = d % y;` ` ` `}` ` ` `// If the final remainder is 0` ` ` `if` `(d == 0)` ` ` `return` `true` `;` ` ` `return` `false` `;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 1, 2, 1, 5, 6 };` ` ` `int` `x = ` `sizeof` `(arr) / ` `sizeof` `(` `int` `);` ` ` `int` `y = 4;` ` ` `cout << (isDivisible(arr, x, y) ? ` `"Yes"` `: ` `"No"` `);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of the approach` `class` `GFG` `{` ` ` ` ` `// Function that returns true if the number represented` ` ` `// by the given array is divisible by y` ` ` `static` `boolean` `isDivisible(` `int` `[] arr, ` `int` `n, ` `int` `y)` ` ` `{` ` ` `int` `d = ` `0` `, i = ` `0` `;` ` ` ` ` `// While there are digits left` ` ` `while` `(i < n)` ` ` `{` ` ` ` ` `// Select the next part of the number` ` ` `// i.e. the maximum number which is <= y` ` ` `while` `(d < y && i < n)` ` ` `d = d * ` `10` `+ arr[i++];` ` ` ` ` `// Get the current remainder` ` ` `d = d % y;` ` ` `}` ` ` ` ` `// If the final remainder is 0` ` ` `if` `(d == ` `0` `)` ` ` `return` `true` `;` ` ` `return` `false` `;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` ` ` `int` `[] arr = { ` `1` `, ` `2` `, ` `1` `, ` `5` `, ` `6` `};` ` ` `int` `x = arr.length;` ` ` `int` `y = ` `4` `;` ` ` ` ` `System.out.println(isDivisible(arr, x, y) ? ` `"Yes"` `: ` `"No"` `);` ` ` `}` `}` `// This code is contributed by ihritik` |

## Python3

`# Python3 implementation of the approach` `# Function that returns true if the number represented` `# by the given array is divisible by y` `def` `isDivisible(arr, n, y):` ` ` `d, i ` `=` `0` `, ` `0` ` ` ` ` `# While there are digits left` ` ` `while` `i < n:` ` ` ` ` `# Select the next part of the number` ` ` `# i.e. the maximum number which is <= y` ` ` `while` `d < y ` `and` `i < n:` ` ` `d ` `=` `d ` `*` `10` `+` `arr[i]` ` ` `i ` `+` `=` `1` ` ` ` ` `# Get the current remainder` ` ` `d ` `=` `d ` `%` `y` ` ` ` ` `# If the final remainder is 0` ` ` `if` `d ` `=` `=` `0` `:` ` ` `return` `True` ` ` `return` `False` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[ ` `1` `, ` `2` `, ` `1` `, ` `5` `, ` `6` `]` ` ` `x ` `=` `len` `(arr)` ` ` `y ` `=` `4` ` ` `if` `(isDivisible(arr, x, y)):` ` ` `print` `(` `"Yes"` `)` ` ` `else` `:` ` ` `print` `(` `"No"` `)` ` ` `# This code is contributed by` `# sanjeev2552` |

## C#

`// C# implementation of the approach` `using` `System;` `class` `GFG` `{` ` ` ` ` `// Function that returns true if the number represented` ` ` `// by the given array is divisible by y` ` ` `static` `bool` `isDivisible(` `int` `[] arr, ` `int` `n, ` `int` `y)` ` ` `{` ` ` `int` `d = 0, i = 0;` ` ` ` ` `// While there are digits left` ` ` `while` `(i < n)` ` ` `{` ` ` ` ` `// Select the next part of the number` ` ` `// i.e. the maximum number which is <= y` ` ` `while` `(d < y && i < n)` ` ` `d = d * 10 + arr[i++];` ` ` ` ` `// Get the current remainder` ` ` `d = d % y;` ` ` `}` ` ` ` ` `// If the final remainder is 0` ` ` `if` `(d == 0)` ` ` `return` `true` `;` ` ` `return` `false` `;` ` ` `}` ` ` ` ` `// Driver code` ` ` `public` `static` `void` `Main ()` ` ` `{` ` ` ` ` `int` `[] arr = { 1, 2, 1, 5, 6 };` ` ` `int` `x = arr.Length;` ` ` `int` `y = 4;` ` ` ` ` `Console.WriteLine(isDivisible(arr, x, y) ? ` `"Yes"` `: ` `"No"` `);` ` ` `}` `}` `// This code is contributed by ihritik` |

## Javascript

`<script>` `// JavaScript implementation of the approach` ` ` `// Function that returns true if the number represented` `// by the given array is divisible by y` `function` `isDivisible(vararr , n , y)` `{` ` ` `var` `d = 0, i = 0;` ` ` `// While there are digits left` ` ` `while` `(i < n)` ` ` `{` ` ` `// Select the next part of the number` ` ` `// i.e. the maximum number which is <= y` ` ` `while` `(d < y && i < n)` ` ` `d = d * 10 + arr[i++];` ` ` `// Get the current remainder` ` ` `d = d % y;` ` ` `}` ` ` `// If the final remainder is 0` ` ` `if` `(d == 0)` ` ` `return` `true` `;` ` ` `return` `false` `;` `}` `// Driver code` `var` `arr = [ 1, 2, 1, 5, 6 ];` `var` `x = arr.length;` `var` `y = 4;` `document.write(isDivisible(arr, x, y) ? ` `"Yes"` `: ` `"No"` `);` `// This code is contributed by 29AjayKumar` `</script>` |

**Output:**

Yes