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Suppose, we are given a large number and we have to check whether the number is divisible by 19.

So, if the input is like 86982, then the output will be “Divisible”.

We will solve this problem using the repeated addition method, where we extract the last digit from the number, multiply it with 2, and add the result to the remaining number until we get a two-digit number that is divisible by 19.

To solve this, we will follow these steps −

- while the number is divisible by 100, do
- last_digit := number mod 10
- number := floor value of (number divided by 10)
- number := number + last_digit * 2

- return True, if number mod 19 is same as 0.

Let us see the following implementation to get better understanding −

def solve(number) : while(number // 100) : last_digit = number % 10 number //= 10 number += last_digit * 2 return (number % 19 == 0) number = 86982 if solve(number) : print("Divisible") else : print("Not Divisible")

86982

Divisible

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