# Check if any large number is divisible by 19 or not

• Last Updated : 11 Jul, 2022

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 using namespace std; // Function to check if the number// is divisible by 19 or notbool 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 codeint 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 logicimport java.io.*; class GFG { // Function to check if the// number is divisible by 19 or notstatic 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 logicusing System; class GFG{     // Function to check if the// number is divisible by 19 or notstatic 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 codepublic static void Main(){    long n = 101156;         if (isDivisible(n))        Console.WriteLine( "Yes");    else        Console.WriteLine( "No");}} // This code is contributed by ajit

## PHP

 

## Javascript

 

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.

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