# Check if a large number is divisible by 8 or not

• Difficulty Level : Basic
• Last Updated : 23 May, 2022

Given a number, the task is to check if a number is divisible by 8 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 = 1128
Output : Yes

Input : n = 1124
Output : No

Input  : n = 363588395960667043875487
Output : No```

Since input number may be very large, we cannot use n % 8 to check if a number is divisible by 8 or not, especially in languages like C/C++. The idea is based on following fact.

A number is divisible by 8 if number formed by last three digits of it is divisible by 8.

Illustration:

```For example, let us consider 76952
Number formed by last three digits = 952
Since 952 is divisible by 8, 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 8 is 0 if i greater
than or equal to three. Note than 10000,
1000,... etc lead to remainder 0 when divided by 8.

So remainder of "7*10000 + 6*1000 + 9*100 +
5*10 + 2" divided by 8 is equivalent to remainder
of following :
0 + 0 + 9*100 + 5*10 + 2 = 52
Therefore we can say that the whole number is
divisible by 8 if 952 is divisible by 8.```

Below is implementation of above fact :

## C++

 `// C++ program to find if a number is divisible by``// 8 or not``#include``using` `namespace` `std;` `// Function to find that number divisible by``// 8 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'``)%8 == 0);` `    ``// If there is double digit``    ``if` `(n == 2)``        ``return` `(((str[n-2]-``'0'``)*10 + (str[n-1]-``'0'``))%8 == 0);` `    ``// If number formed by last three digits is``    ``// divisible by 8.``    ``int` `last = str[n-1] - ``'0'``;``    ``int` `second_last = str[n-2] - ``'0'``;``    ``int` `third_last = str[n-3] - ``'0'``;` `    ``return` `((third_last*100 + second_last*10 + last) % 8 == 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 8 or not``class` `IsDivisible``{``    ``// Function to find that number divisible by``    ``// 8 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'``)%``8` `== ``0``);``     ` `        ``// If there is double digit``        ``if` `(n == ``2``)``            ``return` `(((str.charAt(n-``2``)-``'0'``)*``10` `+ (str.charAt(n-``1``)-``'0'``))%``8` `== ``0``);``     ` `        ``// If number formed by last three digits is``        ``// divisible by 8.``        ``int` `last = str.charAt(n-``1``) - ``'0'``;``        ``int` `second_last = str.charAt(n-``2``) - ``'0'``;``        ``int` `third_last = str.charAt(n-``3``) - ``'0'``;``     ` `        ``return` `((third_last*``100` `+ second_last*``10` `+ last) % ``8` `== ``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 8 or not`  `# Function to find that``# number divisible by 8``# or not``def` `check(st) :``    ``n ``=` `len``(st)``    ` `    ``# Empty string``    ``if` `(n ``=``=` `0``) :``        ``return` `False` `    ``# If there is single digit``    ``if` `(n ``=``=` `1``) :``        ``return` `((``int``)(st[``0``]) ``%` `8` `=``=` `0``)` `    ``# If there is double digit``    ``if` `(n ``=``=` `2``) :``        ``return` `((``int``)(st[n ``-` `2``]) ``*` `10` `+``          ``((``int``)(``str``[n ``-` `1``]) ``%` `8` `=``=` `0``))` `    ``# If number formed by last``    ``# three digits is divisible``    ``# by 8.``    ``last ``=` `(``int``)(st[n ``-` `1``])``    ``second_last ``=` `(``int``)(st[n ``-` `2``])``    ``third_last ``=` `(``int``)(st[n ``-` `3``])` `    ``return` `((third_last``*``100` `+` `second_last``*``10` `+``                               ``last) ``%` `8` `=``=` `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 8 or not``using` `System;` `class` `IsDivisible``{``    ``// Function to find that number``    ``// divisible by 8 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'``) %8  == 0);``    ` `        ``// If there is double digit``        ``if` `(n == 2)``            ``return` `(((str[n - 2] - ``'0'``) * 10 +``            ``(str[n - 1] - ``'0'``)) % 8 == 0);``    ` `        ``// If number formed by last three``        ``// digits is divisible by 8``        ``int` `last = str[n - 1] - ``'0'``;``        ``int` `second_last = str[n - 2] - ``'0'``;``        ``int` `third_last = str[n - 3] - ``'0'``;``    ` `        ``return` `((third_last * 100 + second_last ``                        ``* 10 + last) % 8 == 0);``    ``}``    ` `    ``// Driver Code``    ``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

 ``

## Javascript

 ``

Output:

`Yes`

Time Complexity: O(1), as we are not using any loops for traversing.

Auxiliary Space: O(1), as we are not using any extra space.
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