Question

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

Answer 1

#include <stdio.h>

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

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

    return s;
}

int main() {
    long long results[3] = { 0 };
    int count = 0;

    for (long long n = 2; count < 3; n++) {
        long long s = digitSum(n);

        // Try powers k = 2..10 (enough for reasonable ranges)
        long long p = s * s;
        for (int k = 2; k <= 10; k++) {
            if (p == n) {
                printf("Found: %lld = (%lld)^%d\n", n, s, k);
                results[count++] = n;
                break; // stop checking powers for this n
            }
            p *= s; // next power
        }
    }

    printf("\nFirst 3 numbers:\n");
    for (int i = 0; i < 3; i++) {
        printf("%lld\n", results[i]);
    }

    return 0;
}


/*
run:

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

First 3 numbers:
81
512
2401

*/

 

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