[Guest]

Determine the probability that for a base ten positive integer x chosen at random from 1 to 9999 inclusively, this relationship is satisfied:

(sod(x))² = x+2, where sod(n) denotes the sum of the digits in the base ten representation of n.

[Guest]

7 out of 9999

They are 2, 23, 62, 119, 194, 287, and 398

[unreality]

public static void sumofdigitslinear(int cap)

    {

        System.out.println("~~~CAP "+cap);

        int numSatisfy = 0;

        for (int i=1; i<=cap; i++)

        {

            if (sodsquared(i) == i+2)

            {

                numSatisfy++;

                System.out.println(i);

            }

        }

        System.out.println("~~~~~\n\n\t"+numSatisfy+" out of "+cap);

        System.out.println("\t"+((double)numSatisfy/(double)cap));

    }

    public static int sodsquared(int x)

    {

        int con = 0;

        while (x > 0)

        {

            con += (x % 10);

            x /= 10;

        }   

        return con*con;

    }

~~~CAP 9999

2

23

62

119

194

287

398

~~~~~


	7 out of 9999

	7.000700070007E-4

[Guest]

2, 62, 119, 194, 287, 398. So 6/9999 or .0006 repeating

[Guest]

forgot 23, so 7/9999