Check if any large number is divisible by 19 or not

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

Time Complexity: O(log₁₀n), time required to check if number is divisible by 19
Auxiliary Space: O(1), as no extra space is required

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.