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 \overline{a b c} be any number such that \overline{a b c}=100a+10b+c .
Now assume that \overline{a b c} is divisible by 19. Then
\overline{a b c}\equiv 0 (mod 19)
100a+10b+c\equiv 0 (mod 19)
10(10a+b)+c\equiv 0 (mod 19)
10\overline{a b}+c\equiv 0 (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 \overline{a b} 1.
In other words, we have to find an integer such that n such that 10n\equiv1 mod 19.
It can be observed that the smallest n which satisfies this property is 2 as 20\equiv1 mod 19.
Now we can multiply the original equation 10\overline{a b}+c\equiv 0 (mod 19)
by 2 and simplify it:
20\overline{a b}+2c\equiv 0 (mod 19)
\overline{a b}+2c\equiv 0 (mod 19)
We have found out that if \overline{a b c}\equiv 0 (mod 19) then,
\overline{a b}+2c\equiv 0 (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++

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// 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;
}

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Java

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// 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.

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Python 3

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# 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

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C#

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// 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

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PHP

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<?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
?> 

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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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Improved By : Ryuga, ash264, R_Raj, Sach_Code



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