Digit fifth powers

Task

Task desciption is taken from Project Eulet(https://projecteuler.net/problem=30)
Find the sum of all the numbers that can be written as the sum of fifth powers of their digits.

Ring

<lang ring> see "working..." + nl

sumEnd = 0 sumList = [] limitStart = 10000 limitEnd = 99999

for n = limitStart to limitEnd

   sum = 0
   nStr = string(n)
   for m = 1 to len(nStr)
       sum = sum + pow(number(nStr[m]),5)
   next
   if sum = n
      add(sumList,n) 
      sumEnd += n
   ok

   see "" + sumList[n] + " + "

next see "" + sumList[n] + " = " + sumEnd + nl see "done..." + nl </lang>

Output:

working...
The sum of all the numbers that can be written as the sum of fifth powers of their digits:
54748 + 92727 + 93084 = 240559
done...

XPL0

<lang XPL0>\upper bound: 6*9^5 = 354294 \7*9^5 is still only a 6-digit number, so 6 digits are sufficient

int A, B, C, D, E, F, \digits, A=LSD

       A5, B5, C5, D5, E5, F5, \digits to 5th power
       A0, B0, C0, D0, E0, F0, \digits multiplied by their decimal place
       N,              \number that can be written as the sum of its 5th pwrs
       S;              \sum of all numbers

[S:= 0;

for A:= 0, 9 do \for all digits

 [A5:= A*A*A*A*A;
 A0:= A;
 for B:= 0, 9 do
   [B5:= B*B*B*B*B;
   B0:= B*10;
   for C:= 0, 9 do
     [C5:= C*C*C*C*C;
     C0:= C*100;
     for D:= 0, 9 do
       [D5:= D*D*D*D*D;
       D0:= D*1000;
       for E:= 0, 9 do
         [E5:= E*E*E*E*E;
         E0:= E*10000;
         for F:= 0, 3 do
           [F5:= F*F*F*F*F;
           F0:= F*100000;
               [N:= F0 + E0 + D0 + C0 + B0 + A0;
               if N = A5 + B5 + C5 + D5 + E5 + F5 then
                       [S:= S + N;
                       IntOut(0, N);
                       CrLf(0);
                       ];
               ];
           ];
         ];
       ];
     ];
   ];
 ];

CrLf(0); IntOut(0, S); CrLf(0); ]</lang>

Output: