Given a number, the task is to quickly check if the number is divisible by 19 or not. **Examples:**

Input : x = 38 Output : Yes Input : x = 47 Output : No

A solution to the problem is to extract the last digit and add 2 times of last digit to remaining number and repeat this process until a two digit number is obtained. If the obtained two digit number is divisible by 19, then the given number is divisible by 19.**Approach:**

- Extract the last digit of the number/truncated number every time
- Add 2*(last digit of the previous number) to the truncated number
- Repeat the above three steps as long as necessary.

**Illustration:**

101156-->10115+2*6 = 10127-->1012+2*7=1026-->102+2*6=114 and 114=6*19, So 101156 is divisible by 19.

Mathematical Proof :

Let be any number such that =100a+10b+c .

Now assume that is divisible by 19. Then

0 (mod 19)

100a+10b+c0 (mod 19)

10(10a+b)+c0 (mod 19)

10+c0 (mod 19)

Now that we have separated the last digit from the number, we have to find a way to use it.

Make the coefficient of 1.

In other words, we have to find an integer such that n such that 10n1 mod 19.

It can be observed that the smallest n which satisfies this property is 2 as 201 mod 19.

Now we can multiply the original equation 10+c0 (mod 19)

by 2 and simplify it:

20+2c0 (mod 19)

+2c0 (mod 19)

We have found out that if 0 (mod 19) then,

+2c0 (mod 19).

In other words, to check if a 3-digit number is divisible by 19,

we can just remove the last digit, multiply it by 2,

and then add to the rest of the two digits.

## C++

`// CPP Program to validate the above logic` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to check if the number` `// is divisible by 19 or not` `bool` `isDivisible(` `long` `long` `int` `n)` `{` ` ` `while` `(n / 100) ` `//` ` ` `{` ` ` `// Extracting the last digit` ` ` `int` `d = n % 10;` ` ` `// Truncating the number` ` ` `n /= 10;` ` ` `// Adding twice the last digit` ` ` `// to the remaining number` ` ` `n += d * 2;` ` ` `}` ` ` `// return true if number is divisible by 19` ` ` `return` `(n % 19 == 0);` `}` `// Driver code` `int` `main()` `{` ` ` `long` `long` `int` `n = 101156;` ` ` `if` `(isDivisible(n))` ` ` `cout << ` `"Yes"` `<< endl;` ` ` `else` ` ` `cout << ` `"No"` `<< endl;` ` ` `return` `0;` `}` |

## Java

`// Java Program to validate the above logic` `import` `java.io.*;` `class` `GFG {` `// Function to check if the` `// number is divisible by 19 or not` `static` `boolean` `isDivisible(` `long` `n)` `{` ` ` `while` `(n / ` `100` `>` `0` `)` ` ` `{` ` ` `// Extracting the last digit` ` ` `long` `d = n % ` `10` `;` ` ` `// Truncating the number` ` ` `n /= ` `10` `;` ` ` `// Subtracting the five times the` ` ` `// last digit from the remaining number` ` ` `n += d * ` `2` `;` ` ` `}` ` ` `// Return n is divisible by 19` ` ` `return` `(n % ` `19` `== ` `0` `);` `}` `// Driver code` ` ` `public` `static` `void` `main (String[] args) {` ` ` `long` `n = ` `101156` `;` ` ` `if` `(isDivisible(n))` ` ` `System.out.println( ` `"Yes"` `);` ` ` `else` ` ` `System.out.println( ` `"No"` `);` ` ` `}` `}` `// This code is contributed by Raj.` |

```
# Python 3 Program to check
# if the number is divisible
# by 19 or not
# Function to check if the number
# is divisible by 19 or not
def isDivisible(n) :
while (n // 100) :
# Extracting the last digit
d = n % 10
# Truncating the number
n //= 10
# Adding twice the last digit
# to the remaining number
n += d * 2
# return true if number
# is divisible by 19
return (n % 19 == 0)
# Driver Code
if __name__ == "__main__" :
n = 101156
if (isDivisible(n)) :
print("Yes" )
else :
print("No")
# This code is contributed
# by ANKITRAI1
```

## C#

`// C# Program to validate the` `// above logic` `using` `System;` `class` `GFG` `{` ` ` `// Function to check if the` `// number is divisible by 19 or not` `static` `bool` `isDivisible(` `long` `n)` `{` ` ` `while` `(n / 100 > 0)` ` ` `{` ` ` `// Extracting the last digit` ` ` `long` `d = n % 10;` ` ` `// Truncating the number` ` ` `n /= 10;` ` ` `// Subtracting the five times` ` ` `// the last digit from the` ` ` `// remaining number` ` ` `n += d * 2;` ` ` `}` ` ` `// Return n is divisible by 19` ` ` `return` `(n % 19 == 0);` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `long` `n = 101156;` ` ` ` ` `if` `(isDivisible(n))` ` ` `Console.WriteLine( ` `"Yes"` `);` ` ` `else` ` ` `Console.WriteLine( ` `"No"` `);` `}` `}` `// This code is contributed by ajit` |

## PHP

`<?php` `// PHP Program to validate` `// the above logic` `// Function to check if the number` `// is divisible by 19 or not` `function` `isDivisible( ` `$n` `)` `{` ` ` ` ` `while` `(1)` ` ` `{` ` ` `// Extracting the last digit` ` ` `$d` `= ` `$n` `% 10;` ` ` `// Truncating the number` ` ` `$n` `= ` `$n` `/ 10;` ` ` `// Adding twice the last digit` ` ` `// to the remaining number` ` ` `$n` `= ` `$n` `+ ` `$d` `* 2;` ` ` `if` `(` `$n` `< 100)` ` ` `break` `;` ` ` `}` ` ` ` ` `// return true if number is` ` ` `// divisible by 19` ` ` `return` `(` `$n` `% 19 == 0);` `}` `// Driver code` `$n` `= 38;` `if` `(isDivisible(` `$n` `))` ` ` `echo` `"Yes"` `;` `else` ` ` `echo` `"No"` `;` `// This code is contributed by ash264` `?>` |

## Javascript

`<script>` `// javascript Program to validate the above logic` `// Function to check if the` `// number is divisible by 19 or not` `function` `isDivisible(n)` `{` ` ` `while` `(parseInt(n / 100)>0)` ` ` `{` ` ` `// Extracting the last digit` ` ` `var` `d = n % 10;` ` ` `// Truncating the number` ` ` `n = parseInt(n/ 10);` ` ` `// Subtracting the five times the` ` ` `// last digit from the remaining number` ` ` `n += d * 2;` ` ` `}` ` ` `// Return n is divisible by 19` ` ` `return` `(n % 19 == 0);` `}` `// Driver code` `var` `n = 101156;` `if` `(isDivisible(n))` ` ` `document.write( ` `"Yes"` `);` `else` ` ` `document.write( ` `"No"` `);` `// This code is contributed by 29AjayKumar` `</script>` |

**Output:**

Yes

Note that the above program may not make a lot of sense as could simply do n % 19 to check for divisibility. The idea of this program is to validate the concept. Also, this might be an efficient approach if input number is large and given as string.

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