# 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+c 0 (mod 19)
10(10a+b)+c 0 (mod 19)
10 +c 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 1.
In other words, we have to find an integer such that n such that 10n 1 mod 19.
It can be observed that the smallest n which satisfies this property is 2 as 20 1 mod 19.
Now we can multiply the original equation 10 +c 0 (mod 19)
by 2 and simplify it:
20 +2c 0 (mod 19) +2c 0 (mod 19)
We have found out that if 0 (mod 19) then, +2c 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++

 // CPP Program to validate the above logic  #include  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

 # 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

 

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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