# Check if a number is divisible by 31 or not

• Last Updated : 24 Nov, 2021

Given a number N, the task is to check whether the number is divisible by 31 or not.
Examples:

Input: N = 1922
Output: Yes
Explanation:
31 * 62 = 1922
Input: N = 2722400
Output: No

Approach: The divisibility test of 31 is:

1. Extract the last digit.
2. Subtract 3 * last digit from the remaining number obtained after removing the last digit.
3. Repeat the above steps until a two-digit number, or zero, is obtained.
4. If the two-digit number is divisible by 31, or it is 0, then the original number is also divisible by 31.

For example:

```If N = 49507

Step 1:
N = 49507
Last digit = 7
Remaining number = 4950
Subtracting 3 times last digit
Resultant number = 4950 - 3*7 = 4929

Step 2:
N = 4929
Last digit = 9
Remaining number = 492
Subtracting 3 times last digit
Resultant number = 492 - 3*9 = 465

Step 3:
N = 465
Last digit = 5
Remaining number = 46
Subtracting 3 times last digit
Resultant number = 46 - 3*5 = 31

Step 4:
N = 31
Since N is a two-digit number,
and 31 is divisible by 31

Therefore N = 49507 is also divisible by 31```

Below is the implementation of the above approach:

## C++

 `// C++ program to check whether a number` `// is divisible by 31 or not` `#include` `#include`   `using` `namespace` `std;`   `// Function to check if the number is divisible by 31 or not ` `bool` `isDivisible(``int` `n) ` `{` `    ``int` `d;` `    `  `    ``// While there are at least two digits ` `    ``while` `(n / 100) ` `    ``{`   `        ``// Extracting the last ` `        ``d = n % 10;`   `        ``// Truncating the number ` `        ``n /= 10;`   `        ``// Subtracting three times the last ` `        ``// digit to the remaining number ` `        ``n = ``abs``(n-(d * 3));` `    ``}` `    `  `    ``// Finally return if the two-digit` `    ``// number is divisible by 31 or not` `    ``return` `(n % 31 == 0) ;` `}`   `// Driver Code ` `int` `main() ` `{` `    ``int` `N = 1922;`   `    ``if` `(isDivisible(N)) ` `        ``cout<<``"Yes"``<

## Java

 `// Java program to check whether a number` `// is divisible by 31 or not` `import` `java.util.*;`   `class` `GFG{` ` `  `// Function to check if the number is divisible by 31 or not ` `static` `boolean` `isDivisible(``int` `n) ` `{` `    ``int` `d;` `     `  `    ``// While there are at least two digits ` `    ``while` `((n / ``100``) > ``0``) ` `    ``{` ` `  `        ``// Extracting the last ` `        ``d = n % ``10``;` ` `  `        ``// Truncating the number ` `        ``n /= ``10``;` ` `  `        ``// Subtracting three times the last ` `        ``// digit to the remaining number ` `        ``n = Math.abs(n - (d * ``3``));` `    ``}` `     `  `    ``// Finally return if the two-digit` `    ``// number is divisible by 31 or not` `    ``return` `(n % ``31` `== ``0``) ;` `}` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{` `    ``int` `N = ``1922``;` ` `  `    ``if` `(isDivisible(N)) ` `        ``System.out.print(``"Yes"``);` `    ``else` `        ``System.out.print(``"No"``);` `}     ` `} `   `// This code is contributed by PrinciRaj1992`

## Python 3

 `# Python program to check whether a number` `# is divisible by 31 or not`   `# Function to check if the number is ` `# divisible by 31 or not ` `def` `isDivisible(n) : `   `    ``# While there are at least two digits ` `    ``while` `n ``/``/` `100` `: `   `        ``# Extracting the last ` `        ``d ``=` `n ``%` `10`   `        ``# Truncating the number ` `        ``n ``/``/``=` `10`   `        ``# Subtracting three times the last ` `        ``# digit to the remaining number ` `        ``n ``=` `abs``(n``-``(d ``*` `3``))`   `    ``# Finally return if the two-digit` `    ``# number is divisible by 31 or not` `    ``return` `(n ``%` `31` `=``=` `0``) `   `# Driver Code ` `if` `__name__ ``=``=` `"__main__"` `: `   `    ``n ``=` `1922`   `    ``if` `(isDivisible(n)) : ` `        ``print``(``"Yes"``) ` `    ``else` `: ` `        ``print``(``"No"``) `

## C#

 `// C# program to check whether a number` `// is divisible by 31 or not` `using` `System;`   `class` `GFG{` `  `  `// Function to check if the number is divisible by 31 or not ` `static` `bool` `isDivisible(``int` `n) ` `{` `    ``int` `d;` `      `  `    ``// While there are at least two digits ` `    ``while` `((n / 100) > 0) ` `    ``{` `  `  `        ``// Extracting the last ` `        ``d = n % 10;` `  `  `        ``// Truncating the number ` `        ``n /= 10;` `  `  `        ``// Subtracting three times the last ` `        ``// digit to the remaining number ` `        ``n = Math.Abs(n - (d * 3));` `    ``}` `      `  `    ``// Finally return if the two-digit` `    ``// number is divisible by 31 or not` `    ``return` `(n % 31 == 0) ;` `}` `  `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{` `    ``int` `N = 1922;` `  `  `    ``if` `(isDivisible(N)) ` `        ``Console.Write(``"Yes"``);` `    ``else` `        ``Console.Write(``"No"``);` `}     ` `}`   `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output:

`Yes`

Time Complexity: O(log10N)

Auxiliary Space: O(1)

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