# Check if a large number is divisible by 4 or not

Given a number, the task is to check if a number is divisible by 4 or not. The input number may be large and it may not be possible to store even if we use long long int.

Examples:

```Input : n = 1124
Output : Yes

Input  : n = 1234567589333862
Output : No

Input  : n = 363588395960667043875487
Output : No
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Since input number may be very large, we cannot use n % 4 to check if a number is divisible by 4 or not, especially in languages like C/C++. The idea is based on following fact.

A number is divisible by 4 if number formed by last two digits of it is divisible by 4.

Illustration:

```For example, let us consider 76952
Number formed by last two digits = 52
Since 52 is divisible by 4, answer is YES.
```

How does this work?

```Let us consider 76952, we can write it as
76942 = 7*10000 + 6*1000 + 9*100 + 5*10 + 2

The proof is based on below observation:
Remainder of 10i divided by 4 is 0 if i greater
than or equal to two. Note than 100, 1000,
... etc lead to remainder 0 when divided by 4.

So remainder of "7*10000 + 6*1000 + 9*100 +
5*10 + 2" divided by 4 is equivalent to remainder
of following :
0 + 0 + 0 + 5*10 + 2 = 52
Therefore we can say that the whole number is
divisible by 4 if 52 is divisible by 4.```

Below is implementation of above fact :

## C++

 `// C++ program to find if a number is divisible by ` `// 4 or not ` `#include ` `using` `namespace` `std; ` ` `  `// Function to find that number divisible by ` `// 4 or not ` `bool` `check(string str) ` `{ ` `    ``int` `n = str.length(); ` ` `  `    ``// Empty string ` `    ``if` `(n == 0) ` `       ``return` `false``; ` ` `  `    ``// If there is single digit ` `    ``if` `(n == 1) ` `       ``return` `((str-``'0'``)%4 == 0); ` ` `  `    ``// If number formed by last two digits is ` `    ``// divisible by 4. ` `    ``int` `last = str[n-1] - ``'0'``; ` `    ``int` `second_last = str[n-2] - ``'0'``; ` `    ``return` `((second_last*10 + last) % 4 == 0); ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``string str = ``"76952"``; ` `    ``check(str)?  cout << ``"Yes"` `: cout << ``"No "``; ` `    ``return` `0; ` `} `

## Java

 `// Java program to find if a number is ` `// divisible by 4 or not ` `class` `IsDivisible ` `{ ` `    ``// Function to find that number  ` `    ``// is divisible by 4 or not ` `    ``static` `boolean` `check(String str) ` `    ``{ ` `        ``int` `n = str.length(); ` `      `  `        ``// Empty string ` `        ``if` `(n == ``0``) ` `           ``return` `false``; ` `      `  `        ``// If there is single digit ` `        ``if` `(n == ``1``) ` `           ``return` `((str.charAt(``0``)-``'0'``)%``4` `== ``0``); ` `      `  `        ``// If number formed by last two digits is ` `        ``// divisible by 4. ` `        ``int` `last = str.charAt(n-``1``) - ``'0'``; ` `        ``int` `second_last = str.charAt(n-``2``) - ``'0'``; ` `        ``return` `((second_last*``10` `+ last) % ``4` `== ``0``); ` `    ``} ` ` `  `    ``// main function ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``String str = ``"76952"``; ` `        ``if``(check(str)) ` `            ``System.out.println(``"Yes"``); ` `        ``else` `            ``System.out.println(``"No"``); ` `    ``} ` `}  `

## Python3

 `# Python 3 program to find ` `# if a number is divisible ` `# by 4 or not ` ` `  `# Function to find that  ` `# number divisible by  ` `# 4 or not ` `def` `check(st) : ` `    ``n ``=` `len``(st) ` ` `  `    ``# Empty string ` `    ``if` `(n ``=``=` `0``) : ` `        ``return` `False` ` `  `    ``# If there is single  ` `    ``# digit ` `    ``if` `(n ``=``=` `1``) : ` `        ``return` `((st[``0``] ``-` `'0'``) ``%` `4` `=``=` `0``) ` ` `  `    ``# If number formed by ` `    ``# last two digits is ` `    ``# divisible by 4. ` `    ``last ``=` `(``int``)(st[n ``-` `1``]) ` `    ``second_last ``=` `(``int``)(st[n ``-` `2``]) ` `     `  `    ``return` `((second_last ``*` `10` `+` `last) ``%` `4` `=``=` `0``) ` ` `  ` `  `# Driver code ` `st ``=` `"76952"` `if``(check(st)) : ` `    ``print``(``"Yes"``) ` `else` `:  ` `    ``print``(``"No "``) ` `     `  `# This code is contributed by Nikita tiwari `

## C#

 `// C# program to find if a number is ` `// divisible by 4 or not ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Function to find that number  ` `    ``// is divisible by 4 or not ` `    ``static` `bool` `check(String str) ` `    ``{ ` `        ``int` `n = str.Length; ` `     `  `        ``// Empty string ` `        ``if` `(n == 0) ` `            ``return` `false``; ` `     `  `        ``// If there is single digit ` `        ``if` `(n == 1) ` `            ``return` `((str - ``'0'``) % 4 == 0); ` `     `  `        ``// If number formed by last two ` `        ``// digits is divisible by 4. ` `        ``int` `last = str[n-1] - ``'0'``; ` `        ``int` `second_last = str[n-2] - ``'0'``; ` `         `  `        ``return` `((second_last*10 + last) ` `                                ``% 4 == 0); ` `    ``} ` ` `  `    ``// main function ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``String str = ``"76952"``; ` `         `  `        ``if``(check(str)) ` `            ``Console.Write(``"Yes"``); ` `        ``else` `            ``Console.Write(``"No"``); ` `    ``} ` `} ` ` `  `// This code is Contributed by nitin mittal. `

## PHP

 ` `

Output:

`Yes`

This article is contributed by DANISH_RAZA . 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.

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up

Article Tags :
Practice Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.