The challenge

“A divisibility rule is a shorthand way of determining whether a given integer is divisible by a fixed divisor without performing the division, usually by examining its digits.”

wikipedia.com

When you divide the successive powers of 10 by 13 you get the following remainders of the integer divisions:

1, 10, 9, 12, 3, 4 because:

(For “mod” you can see: https://en.wikipedia.org/wiki/Modulo_operation)

Then the whole pattern repeats. Hence the following method:

Multiply the right most digit of the number with the left most number in the sequence shown above, the second right most digit to the second left most digit of the number in the sequence. The cycle goes on and you sum all these products. Repeat this process until the sequence of sums is stationary.

Example:

What is the remainder when 1234567 is divided by 13?

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7      6     5      4     3     2     1  (digits of 1234567 from the right)
×      ×     ×      ×     ×     ×     ×  (multiplication)
1     10     9     12     3     4     1  (the repeating sequence)

Therefore following the method we get:

7×1 + 6×10 + 5×9 + 4×12 + 3×3 + 2×4 + 1×1 = 178

We repeat the process with the number 178:

8x1 + 7x10 + 1x9 = 87

and again with 87:

7x1 + 8x10 = 87

From now on the sequence is stationary (we always get 87) and the remainder of 1234567 by 13 is the same as the remainder of 87 by 13 (i.e 9).

Task:

Call thirt the function which processes this sequence of operations on an integer n (>=0). thirt will return the stationary number.

thirt(1234567) calculates 178, then 87, then 87 and returns 87.

thirt(321) calculates 48, 48 and returns 48

The solution in Java code

Option 1 (using dual while loops):

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class Thirteen {
    
    public static long thirt(long n) {
       int[] numbers = {1, 10, 9, 12, 3, 4};
        while (n > 99){
            long nr = 0;
            int i = 0;
            while (n > 0){
                nr += n%10 * numbers[i++%6];
                n /= 10;
            }
            n = nr;
        }
        return n;
    }
}

Option 2 (using StringBuilder):

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class Thirteen {

    private static final int[] remainders = {1, 10, 9, 12, 3, 4};
    
    public static long thirt(long n) {
        long count = 0;
        int i = 0;
        for (String number: new StringBuilder().append(n).reverse().toString().split("")) {
          count += Integer.parseInt(number) * remainders[i%6];
          i++;
        }
        return count == n ? count : thirt(count);
    }
}

Option 3 (using a while case):

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class Thirteen {
    
    public static long thirt(long n) {

      long previous = n;
      long sum = 0;

      while (n != 0){
        sum += (n % 10);
        n = n / 10;
        sum += ((n % 10) * 10);
        n = n / 10;
        sum += ((n % 10) * 9);
        n = n / 10;
        sum += ((n % 10) * 12);
        n = n / 10;
        sum += ((n % 10) * 3);
        n = n / 10;
        sum += ((n % 10) * 4);
        n = n / 10;
      }

      if (sum == previous) return sum;

      return thirt(sum);
    }
}

Option 4 (using a char array):

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class Thirteen {
    
    public static long thirt(long n) {        
        long sum = 0;
        int len = (""+n).length(), i = 0;
        for (char c : (""+n).toCharArray()) sum += (c-'0') * new int[]{1,10,9,12,3,4}[(len-i++-1) % 6];        
        return sum == n ? n : thirt(sum);
    }
    
}

Option 5 (using a stack):

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import java.util.Arrays;
import java.util.Stack;
class Thirteen {
    public static long calculation(long input) {
        Stack<Long> stack = new Stack<>();
        Arrays
                .asList(String.valueOf(input)
                        .split(""))
                .forEach(x ->
                        stack.push(Long.parseLong(x))
                );

        int size = stack.size();
        long[] numbers = new long[]{1, 10, 9, 12, 3, 4};

        long sum = 0;
        for (int i = 0; i < size; i++) {
            sum = sum + stack.pop() * numbers[i % 6];
        }

        return sum;
    }

    public static long thirt(long input) {
        long lastCalc = calculation(input);
        while (true) {
            if(calculation(lastCalc)==lastCalc){
                return lastCalc;
            }else{
                lastCalc= calculation(lastCalc);
            }
        }
    }
}

Test cases to validate our Java solution

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import static org.junit.Assert.*;
import org.junit.Test;

public class ThirteenTest {

    private static int randInt(int min, int max) {
        return min + (int)(Math.random() * ((max - min) + 1));
    }
    private static void testing(long actual, long expected) {
        assertEquals(expected, actual);
    }
    @Test
    public void test1() {
        System.out.println("Fixed Tests: thirt"); 
        testing(Thirteen.thirt(8529), 79);
        testing(Thirteen.thirt(85299258), 31);
        testing(Thirteen.thirt(5634), 57);
        testing(Thirteen.thirt(1111111111), 71);
        testing(Thirteen.thirt(987654321), 30);
    }
    public static long thirtSol(long n) {
        int[] w = new int[] {1, 10, 9, 12, 3, 4};
        while (true) {
            long r = n; long q = -1; long c = 0; int j = 0;
            while (q != 0) {
                q = (long)(r / 10);
                c += (r % 10) * w[j % 6];
                r = q;
                j++;
            }
            if (c == n) 
                return c;
            n = c;
        }
    }
    @Test
    public void test4() {
        System.out.println("Random Tests");        
        for (int i = 0; i < 200; i++) { 
            int p0 = randInt(150, 1500000000);
            testing(Thirteen.thirt(p0), thirtSol(p0)); 
        }
    }
}