Rafi Rafi - 7 months ago 39
Java Question

Algorithm involving figuring out 10 digit number puzzle

To determine if a number is divisible by 7, take the last digit off the number, double it and subtract the doubled number from the remaining number. If the result is evenly divisible by 7 (e.g. 14, 7, 0, -7, etc.), then the number is divisible by seven. This may need to be repeated several times.
Example: Is 3101 evenly divisible by 7?

310 - take off the last digit of the number which was 1
-2 - double the removed digit and subtract it
308 - repeat the process by taking off the 8
-16 - and doubling it to get 16 which is subtracted
14 - the result is 14 which is a multiple of 7


The following is the code that I did to get the number:

for(int O =0; O <= 9 ; O++) {
String a = String.valueOf(number[0]);
String b = String.valueOf(number[1]);
String c = String.valueOf(number[2]);
String d = String.valueOf(number[3]);
String e = String.valueOf(number[4]);
String f = String.valueOf(number[5]);

String h = a+b+c+d+e+f;
int abcdef = Integer.valueOf(h);
if ( (abcdef -(2*O) % 7) ==0 )
number [6] = O;

}


However, it is not giving me a number of such kind.I was able to get a number up until 6 where up until each digit the number is divisible by the respective index(if I start with 1, not 0 for the index).Which means index 1 is divisible by 1, index 2 is divisible by 2, index 3 is divisible by 3,......index 7 is divisible by 7.I want to form a number of such kind.Note that I could have done it without using the algorithm by the following way:

for(int O =0; O <= 9 ; O++) {
String a = String.valueOf(number[0]);
String b = String.valueOf(number[1]);
String c = String.valueOf(number[2]);
String d = String.valueOf(number[3]);
String e = String.valueOf(number[4]);
String f = String.valueOf(number[5]);
String g = String.valueOf(O);
String h = a+b+c+d+e+f+g;
int abcdefg = Integer.valueOf(h);
if ( (abcdefg % 7) ==0 )
number [6] = O;

}


However, I really want to do it using the algorithm that I described in the beginning.

Answer

Example calling code:

int [] num = new int [7];
for (int i = 1234567; i < 9999999; i++)
{
    // put i into the array and check it
    if (checkDigitsDivisible(i, num))
    {
        System.out.println(i);
    }
}

Check if a number has the first digit evenly divisible by 1, first 2 digits evenly divisible by 2, first 3 digits divisible by 3, etc.

public static boolean checkDigitsDivisible (int num, int [] arr)
{
    String numStr = String.valueOf(num);

    int one = Integer.parseInt(numStr.substring(0, 1));
    int two = Integer.parseInt(numStr.substring(0, 2));
    int three = Integer.parseInt(numStr.substring(0, 3));
    int four = Integer.parseInt(numStr.substring(0, 4));
    int five = Integer.parseInt(numStr.substring(0, 5));
    int six = Integer.parseInt(numStr.substring(0, 6));
    int seven = Integer.parseInt(numStr.substring(0, 7));

    arr[0] = Integer.parseInt(numStr.substring(0, 1));
    arr[1] = Integer.parseInt(numStr.substring(1, 2));
    arr[2] = Integer.parseInt(numStr.substring(2, 3));
    arr[3] = Integer.parseInt(numStr.substring(3, 4));
    arr[4] = Integer.parseInt(numStr.substring(4, 5));
    arr[5] = Integer.parseInt(numStr.substring(5, 6));
    arr[6] = Integer.parseInt(numStr.substring(6, 7));

    return (one % 1 == 0) && 
            (two % 2 == 0) && 
            (three % 3 == 0) && 
            (four % 4 == 0) && 
            (five % 5 == 0) && 
            (six % 6 == 0) &&
            (isDivisibleBy7(seven));
}

A recursive solution to check if a number is divisible by 7 (algorithm as described in the question):

public static boolean isDivisibleBy7 (int n)
{
    System.out.println("Checking " + n);

    if (n == 0 || n == 7 || n == 14)
    {
        return true;
    }
    else if (n > -6 && n < 7)
    {
        return false;
    }
    else
    {
        int lastDigit = n % 10;
        int lastDigitOff = n / 10;

        return isDivisibleBy7(lastDigitOff - (lastDigit * 2));
    }
}