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Number.prototype.digitSum = function() {
  return this.toString().split('').map(Number)
    .reduce(function(a, b){return a + b;});
}

let i = 1, j;
while ((j = i**2).digitSum() != 43) i++;

console.log(j < 5e5 ? `${j} = ${i}²` : 'Netrovita.');
                

Lösung von: Lisa Salander (Heidi-Klum-Gymnasium Bottrop)

print([i for i in range(1,500000) if sum([int(y) for y in str(i)])==43 and sqrt(i)%1==0][0])

                

Lösung von: rob ert (tub)

// C++ 17
#include <iostream>
#include <functional>

int main() {
    std::function<int(int)> digit_sum{ [&](int n) { return n != 0 ? n % 10 + digit_sum(n / 10) : 0; } };
    for (auto i{ 0 }, k{ 0 }; i < sqrt(5e5); k = ++i * i)
        if(digit_sum(k) == 43) std::cout << k << "\n";
}
                

Lösung von: Jens Kelm (@JKooP)

// NET 6.0 | C# 10.x | VS 2022

for (int i = 0, k = 0; i < Math.Sqrt(5e5); k = ++i * i)
    if (k.ToString().Select(x => x - '0').Sum() == 43) Console.WriteLine(k);
                

Lösung von: Jens Kelm (@JKooP)

' VBA-Funktion

Function DigitSum(n As Long) As Long
    Dim r As Long
    Do While n > 0
        r = r + n Mod 10
        n = n \ 10
    Loop
    DigitSum = r
End Function

Sub Main()
    Dim i As Long
    For i = 0 To Sqr(500000)
        If DigitSum(i * i) = 43 Then Debug.Print i * i
    Next
End Sub
                

Lösung von: Jens Kelm (@JKooP)