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Check if a large number is divisible by 25 or not
• Difficulty Level : Basic
• Last Updated : 15 Apr, 2021

Given a number, the task is to check if number is divisible by 25. The input number may be large and it may not be possible to store even if we use long long int.

Examples:

```Input  : n = 56945250
Output : Yes

Input  : n = 1234567589333100
Output : Yes

Input  : n = 3635883959606670431112222
Output : No```

Since input number may be very large, we cannot use n % 25 to check if a number is divisible by 25 or not, especially in languages like C/C++. The idea is based on following fact.

```A number is divisible by 25 if its digits
last two digits will be 0  or divisible by 25 .```

Illustration:

```For example, let us consider 769575
Number formed by last two digits is = 75
Since 75 is divisible by 25 , answer is YES.```
```Let us consider 5325, we can write it as
5325 = 5*1000 + 3*100 + 2*10 + 5

The proof is based on below observation:
Remainder of 10i divided by 25 is 0 if i greater
than or equal to two. Note than 100, 1000,
... etc lead to remainder 0 when divided by 25.

So remainder of " 5*1000 + 3*100 + 2*10 + 5"
divided by 25 is equivalent to remainder
of following :
0 + 0 + 20 + 5 = 25

Since 25 is divisible by 25, answer is yes.```

## C++

 `// C++ program to find if a number is``// divisible by 25 or not``#include``using` `namespace` `std;` `// Function to find that number divisible``// by 25 or not.``bool` `isDivisibleBy25(string str)``{``    ``// If length of string is single digit then``    ``// it's not divisible by 25``    ``int` `n = str.length();``    ``if` `(n == 1)``        ``return` `false``;` `    ``return` `( (str[n-1]-``'0'` `== 0  &&``              ``str[n-2]-``'0'` `== 0) ||``   ``((str[n-2]-``'0'``)*10 + (str[n-1]-``'0'``))%25 == 0 );``}` `// Driver code``int` `main()``{``    ``string str = ``"76955"``;``    ``isDivisibleBy25(str)?  cout << ``"Yes"` `:``                           ``cout << ``"No "``;``    ``return` `0;``}`

## Java

 `// Java program to find if a number is``// divisible by 25 or not``class` `IsDivisible``{``    ``// Function to find that number divisible``    ``// by 25 or not.``    ``static` `boolean` `isDivisibleBy25(String str)``    ``{``        ``// If length of string is single digit then``        ``// it's not divisible by 25``        ``int` `n = str.length();``        ``if` `(n == ``1``)``            ``return` `false``;``     ` `        ``return` `( (str.charAt(n-``1``)-``'0'` `== ``0`  `&&``                  ``str.charAt(n-``2``)-``'0'` `== ``0``) ||``        ``((str.charAt(n-``2``)-``'0'``)*``10` `+ (str.charAt(n-``1``)-``'0'``))%``25` `== ``0` `);``    ``}``    ` `    ``// main function``    ``public` `static` `void` `main (String[] args)``    ``{``        ``String str = ``"76955"``;``        ``if``(isDivisibleBy25(str))``            ``System.out.println(``"Yes"``);``        ``else``            ``System.out.println(``"No"``);``    ``}``}`

## Python3

 `# Python 3 program to find if``# a number is divisible by 25``# or not` `# Function to find that``# number divisible by 25 or not.``def` `isDivisibleBy25(st) :` `    ``# If length of string is``    ``# single digit then it's``    ``# not divisible by 25``    ``n ``=` `len``(st)``    ``if` `(n ``=``=` `1``) :``        ``return` `False` `    ``return` `((``int``)(st[n``-``1``]) ``=``=` `0` `and` `((``int``)(st[n``-``2``])``=``=` `0``) ``or``           ``((``int``)(st[n``-``2``])``*``10` `+` `(``int``)(st[n``-``1``])``%``25` `=``=` `0``))` `# Driver code``st ``=` `"76955"``if``(isDivisibleBy25(st)) :``    ``print``(``"Yes"``)``else` `:``    ``print``(``"No"``)``    `  `# This code is contributed by Nikita Tiwari.`

## C#

 `// C# program to find if a number``// is divisible by 25 or not``using` `System;` `class` `IsDivisible``{``    ``// Function to find that number``    ``// divisible by 25 or not.``    ``static` `bool` `isDivisibleBy25(String str)``    ``{``        ``// If length of string is single digit then``        ``// then it's not divisible by 25``        ``int` `n = str.Length;``        ``if` `(n == 1)``            ``return` `false``;``    ` `        ``return` `((str[n - 1] - ``'0'` `== 0 &&``                 ``str[n - 2] - ``'0'` `== 0) ||``                ``((str[n - 2] - ``'0'``) * 10 +``                 ``(str[n - 1] - ``'0'``)) % 25 == 0);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `Main ()``    ``{``        ``String str = ``"76955"``;``        ``if``(isDivisibleBy25(str))``            ``Console.Write(``"Yes"``);``        ``else``            ``Console.Write(``"No"``);``    ``}``}` `// This code is contributed by Nitin Mittal`

## PHP

 ``

## Javascript

 ``

Output:

`No`

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