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Check if a larger number divisible by 36
• Difficulty Level : Easy
• Last Updated : 14 Apr, 2021

Given a number, check whether a given number is divisible by 36 or not. The number may be very large and may not fit in any numeric(int, long int, float, etc.) data type.
Examples:

```Input : 72
Output : Yes

Input : 244
Output : No

Input : 11322134
Output : No

Input : 92567812197966231384
Output : Yes```

A number is divisible by 36 if the number is divisible by 4 and 9

Below is the implementation based on above idea.

## C++

 `// C++ implementation to check divisibility by 36``#include ``using` `namespace` `std;` `// Function to check whether a number``// is divisible by 36 or not``string divisibleBy36(string num)``{``    ``int` `l = num.length();` `    ``// null number cannot``    ``// be divisible by 36``    ``if` `(l == 0)``        ``return` `"No"``;` `    ``// single digit number other than``    ``// 0 is not divisible by 36``    ``if` `(l == 1 && num != ``'0'``)``        ``return` `"No"``;` `    ``// number formed by the last 2 digits``    ``int` `two_digit_num = (num[l-2] - ``'0'``)*10 +``                        ``(num[l-1] - ``'0'``) ;` `    ``// if number is not divisible by 4``    ``if` `(two_digit_num%4 != 0)``        ``return` `"No"``;` `    ``// number is divisible by 4 calculate``    ``// sum of digits``    ``int` `sum = 0;``    ``for` `(``int` `i=0; i

## Java

 `// Java program to find if a number is``// divisible by 36 or not``class` `IsDivisible``{``    ``// Function to check whether a number``    ``// is divisible by 36 or not``    ``static` `boolean` `divisibleBy36(String num)``    ``{``        ``int` `l = num.length();``     ` `        ``// null number cannot``        ``// be divisible by 36``        ``if` `(l == ``0``)``            ``return` `false``;``     ` `        ``// single digit number other than``        ``// 0 is not divisible by 36``        ``if` `(l == ``1` `&& num.charAt(``0``) != ``'0'``)``            ``return` `false``;``     ` `        ``// number formed by the last 2 digits``        ``int` `two_digit_num = (num.charAt(l-``2``) - ``'0'``)*``10` `+``                            ``(num.charAt(l-``1``) - ``'0'``) ;``     ` `        ``// if number is not divisible by 4``        ``if` `(two_digit_num%``4` `!= ``0``)``            ``return` `false``;``     ` `        ``// number is divisible by 4 calculate``        ``// sum of digits``        ``int` `sum = ``0``;``        ``for` `(``int` `i=``0``; i

## Python3

 `# Python 3 implementation to``# check divisibility by 36` `# Function to check whether a``# number is divisible by``# 36 or not``def` `divisibleBy36(num) :``    ``l ``=` `len``(num)` `    ``# null number cannot``    ``# be divisible by 36``    ``if` `(l ``=``=` `0``) :``        ``return` `(``"No"``)` `    ``# single digit number other``    ``# than 0 is not divisible``    ``# by 36``    ``if` `(l ``=``=` `1` `and` `num[``0``] !``=` `'0'``) :``        ``return` `(``"No"``)` `    ``# number formed by the``    ``# last 2 digits``    ``two_digit_num ``=` `(((``int``)(num[l ``-` `2``])) ``*``                    ``10` `+``(``int``)(num[l ``-` `1``]))` `    ``# if number is not``    ``# divisible by 4``    ``if` `(two_digit_num``%``4` `!``=` `0``) :``        ``return` `"No"` `    ``# number is divisible``    ``# by 4 calculate sum``    ``# of digits``    ``sm ``=` `0``    ``for` `i ``in` `range``(``0``,l) :``        ``sm ``=` `sm ``+` `(``int``)(num[i])` `    ``# sum of digits is not``    ``# divisible by 9``    ``if` `(sm``%``9` `!``=` `0``) :``        ``return` `(``"No"``)` `    ``# Number is divisible``    ``# by 4 and 9 hence,``    ``# number is divisible``    ``# by 36``    ``return` `(``"Yes"``)` `# Driver program``num ``=` `"92567812197966231384"``print``(divisibleBy36(num))` `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program to find if a number is``// divisible by 36 or not``using` `System;` `class` `GFG {``    ` `    ``// Function to check whether``    ``// a number is divisible by``    ``// 36 or not``    ``static` `bool` `divisibleBy36(String num)``    ``{``        ``int` `l = num.Length;``    ` `        ``// null number cannot``        ``// be divisible by 36``        ``if` `(l == 0)``            ``return` `false``;``    ` `        ``// single digit number other than``        ``// 0 is not divisible by 36``        ``if` `(l == 1 && num != ``'0'``)``            ``return` `false``;``    ` `        ``// number formed by the last``        ``// 2 digits``        ``int` `two_digit_num = (num[l-2] - ``'0'``) * 10``                             ``+ (num[l-1] - ``'0'``) ;``    ` `        ``// if number is not divisible by 4``        ``if` `(two_digit_num % 4 != 0)``            ``return` `false``;``    ` `        ``// number is divisible by 4 calculate``        ``// sum of digits``        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < l; i++)``            ``sum += (num[i] - ``'0'``);``    ` `        ``// sum of digits is not divisible by 9``        ``if` `(sum % 9 != 0)``            ``return` `false``;``    ` `        ``// number is divisible by 4 and 9``        ``// hence, number is divisible by 36``        ``return` `true``;``    ``}` `    ``// main function``    ``public` `static` `void` `Main ()``    ``{``        ``String num = ``"92567812197966231384"``;``        ` `        ``if``(divisibleBy36(num))``            ``Console.Write(``"Yes"``);``        ``else``            ``Console.Write(``"No"``);``    ``}``}` `// This code is contributed by parashar.`

## PHP

 ``

## Javascript

 ``

Output:

`Yes`

Time Complexity: O(n)
This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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