Question

How to find the first 3 integers equal to the sum of their digits raised to some power in Java

Answer 1

import java.util.ArrayList;

public class DigitPower {

    // compute sum of digits
    static long digitSum(long n) {
        long s = 0;

        while (n > 0) {
            s += n % 10;
            n /= 10;
        }

        return s;
    }

    public static void main(String[] args) {
        ArrayList<Long> results = new ArrayList<>();

        for (long n = 2; results.size() < 3; n++) {
            long s = digitSum(n);

            // Try powers k = 2..10 (enough for reasonable ranges)
            long p = s * s;
            for (int k = 2; k <= 10; k++) {
                if (p == n) {
                    System.out.println("Found: " + n +
                                       " = (" + s + ")^" + k);
                    results.add(n);
                    break; // stop checking powers for this n
                }
                p *= s; // next power
            }
        }

        System.out.println("\nFirst 3 numbers:");
        for (long x : results) {
            System.out.println(x);
        }
    }
}



/*
run:

Found: 81 = (9)^2
Found: 512 = (8)^3
Found: 2401 = (7)^4

First 3 numbers:
81
512
2401

*/

 

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