Question

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

Answer 1

#include <iostream>
#include <vector>

// 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() {
    std::vector<long long> results;

    for (long long n = 2; results.size() < 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) {
                std::cout << "Found: " << n
                          << " = (" << s << ")^" << k << "\n";
                results.push_back(n);
                break; // stop checking powers for this n
            }
            p *= s; // next power
        }
    }

    std::cout << "\nFirst 3 numbers:\n";
    for (auto x : results) {
        std::cout << x << "\n";
    }
}


/*
run:

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

First 3 numbers:
81
512
2401

*/

 

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