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 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+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 <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

`// 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

`# 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#

`// 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

`<?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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