This was not bad! the minimum number that can be written as a sum of its 5th digits powers is
$$
\begin{align*}
2 = 0^5 + 0^5 + 0^5 + 1^5 + 1^5.
\end{align*}
$$
while the maximum number that can be written as a sum of its 5th digits powers is
$$
\begin{align*}
295245 = 9^5 + 9^5 + 9^5 + 9^5 + 9^5.
\end{align*}
$$
So let’s check all the numbers in this range
// 9^5 + 9^5 + 9^5 + 9^5 + 9^5 = 5*9^5 = 295,245
// 0^5 + 0^5 + 0^5 + 1^5 + 1^5 = 2
int total_sum = 0;
for (int n = 2; n <= 295245; n++) {
long sum = 0;
int num = n;
while (num > 0 && sum <= n) {
int d = num%10;
int p = pow(d, 5);
sum += p;
num /= 10;
}
if (sum == n) {
total_sum += sum;
}
}
printf("%d\n", total_sum);