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

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 76952Number 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 10^{i}divided by 4 is 0 if i greater than or equal totwo. 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<bits/stdc++.h> ` `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]-` `'0'` `)%4 == 0); ` ` ` ` ` `// If number formed by last two digits is ` ` ` `// divisble 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; ` `} ` |

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## 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 ` ` ` `// divisble 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"` `); ` ` ` `} ` `} ` |

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## 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 ` ` ` `# divisble 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 ` |

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## 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] - ` `'0'` `) % 4 == 0); ` ` ` ` ` `// If number formed by last two ` ` ` `// digits is divisble 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. ` |

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## PHP

`<?php ` `// PHP program to find if a ` `// number is divisible by ` `// 4 or not ` ` ` `// Function to find that ` `// number divisible by ` `// 4 or not ` `function` `check(` `$str` `) ` `{ ` ` ` `$n` `= ` `strlen` `(` `$str` `); ` ` ` ` ` `// Empty string ` ` ` `if` `(` `$n` `== 0) ` ` ` `return` `false; ` ` ` ` ` `// If there is single digit ` ` ` `if` `(` `$n` `== 1) ` ` ` `return` `((` `$str` `[0] - ` `'0'` `) % 4 == 0); ` ` ` ` ` `// If number formed by ` ` ` `// last two digits is ` ` ` `// divisble by 4. ` ` ` `$last` `= ` `$str` `[` `$n` `- 1] - ` `'0'` `; ` ` ` `$second_last` `= ` `$str` `[` `$n` `- 2] - ` `'0'` `; ` ` ` `return` `((` `$second_last` `* 10 + ` `$last` `) % 4 == 0); ` `} ` ` ` `// Driver code ` `$str` `= ` `"76952"` `; ` `$x` `= check(` `$str` `)? ` `"Yes"` `: ` `"No"` `; ` `echo` `(` `$x` `); ` ` ` `// This code is contributed by Ajit. ` `?> ` |

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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.

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