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Task

Task desciption is taken from Project Euler (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.

Even though 1⁵ = 1, it is not expressed as a sum (a sum being the summation of a list of two or more numbers), and is therefore not included.

Ada

<lang Ada>with Ada.Text_Io;

procedure Digit_Fifth_Powers is

  subtype Number is Natural range 000_002 .. 999_999;

  function Sum_5 (N : Natural) return Natural
  is
     Pow_5 : constant array (0 .. 9) of Natural :=
       (0 => 0**5, 1 => 1**5, 2 => 2**5, 3 => 3**5, 4 => 4**5,
        5 => 5**5, 6 => 6**5, 7 => 7**5, 8 => 8**5, 9 => 9**5);
  begin
     return (if N = 0
             then 0
             else Pow_5 (N mod 10) + Sum_5 (N / 10));
  End Sum_5;

  use Ada.Text_Io;
  Sum : Natural := 0;

begin

  for N in Number loop
     if N = Sum_5 (N) then
        Sum := Sum + N;
        Put_Line (Number'Image (N));
     end if;
  end loop;
  Put ("Sum: ");
  Put_Line (Natural'Image (Sum));

end Digit_Fifth_Powers;</lang>

Output:

 4150
 4151
 54748
 92727
 93084
 194979
Sum:  443839

ALGOL 68

As noted by the Julia sample, we need only consider up to 6 digit numbers.
Also note, the digit fifth power sum is independent of the order of the digits. <lang algol68>BEGIN

   []INT fifth = []INT( 0, 1, 2^5, 3^5, 4^5, 5^5, 6^5, 7^5, 8^5, 9^5 )[ AT 0 ];
   # as observed by the Julia sample, 9^5 * 7 has only 6 digits whereas 9^5 * 6 has 6 digits #
   # so only up to 6 digit numbers need be considered #
   # also, the digit fifth power sum is independent ofg the order of the digits # 
   [ 1 : 100 ]INT sums; FOR i TO UPB sums DO sums[ i ] := 0 OD;
   [ 0 :   9 ]INT used; FOR i FROM 0 TO 9 DO used[ i ] := 0 OD;
   INT s count := 0;
   FOR d1 FROM 0 TO 9 DO
       INT s1 = fifth[ d1 ];
       used[ d1 ] +:= 1;
       FOR d2 FROM d1 TO 9 DO
           INT s2 = fifth[ d2 ] + s1;
           used[ d2 ] +:= 1;
           FOR d3 FROM d2 TO 9 DO
               INT s3 = fifth[ d3 ] + s2;
               used[ d3 ] +:= 1;
               FOR d4 FROM d3 TO 9 DO
                   INT s4 = fifth[ d4 ] + s3;
                   used[ d4 ] +:= 1;
                   FOR d5 FROM d4 TO 9 DO
                       INT s5 = fifth[ d5 ] + s4;
                       used[ d5 ] +:= 1;
                       FOR d6 FROM d5 TO 9 DO
                           INT s6 = fifth[ d6 ] + s5;
                           used[ d6 ] +:= 1;
                           # s6 is the sum of the fifth powers of the digits #
                           # check it it is composed of the digits d1 - d6   #
                           [ 0 : 9 ]INT check; FOR i FROM 0 TO 9 DO check[ i ] := 0 OD;
                           INT v := s6;
                           FOR i TO 6 DO
                               check[ v MOD 10 ] +:= 1;
                               v OVERAB 10
                           OD;
                           BOOL same := TRUE;
                           FOR i FROM 0 TO 9 WHILE ( same := used[ i ] = check[ i ] ) DO SKIP OD;
                           IF same THEN
                               # found a number that is the sum of the fifth powers of its digits #
                               sums[ s count +:= 1 ] := s6
                           FI;
                           used[ d6 ] -:= 1
                       OD # d6 # ;
                       used[ d5 ] -:= 1
                   OD # d5 # ;
                   used[ d4 ] -:= 1
               OD # d4 # ;
               used[ d3 ] -:= 1
           OD # d3 # ;
           used[ d2 ] -:= 1
       OD # d2 # ;
       used[ d1 ] -:= 1
   OD # d1 # ;
   # sum and print the sums - ignore 0 and 1 #
   INT total := 0;
   print( ( "Numbers that are the sums of the fifth powers of their digits: " ) );
   FOR i TO s count DO
       IF sums[ i ] > 1 THEN
           print( ( " ", whole( sums[ i ], 0 ) ) );
           total +:= sums[ i ]
       FI
   OD;
   print( ( newline ) );
   print( ( "Total: ", whole( total, 0 ), newline ) )

END</lang>

Output:

Numbers that are the sums of the fifth powers of their digits:  4150 4151 93084 92727 54748 194979
Total: 443839

APL

Works with: Dyalog APL

Output:

BASIC

FreeBASIC

<lang freebasic>function dig5( n as uinteger ) as uinteger

   dim as string ns = str(n)
   dim as uinteger ret = 0
   for i as ubyte = 2 to len(ns)
       ret += val(mid(ns,i,1))^5
   next i
   return ret

end function

dim as uinteger i, sum = 0

for i = 0 to 999999

   if i = dig5(i) then 
       print i
       sum += i
   end if

next i

print "Their sum is ", sum</lang>

Output:

4150 4151 54748 92727 93084 194979

Their sum is 443839

<lang gwbasic>10 SUM! = 0 20 FOR I! = 2 TO 999999! 30 GOSUB 80 40 IF R! = I! THEN SUM! = SUM! + I! : PRINT I! 50 NEXT I! 60 PRINT "Total = ",SUM 70 END 80 N$ = STR$(I) 90 R! = 0 100 FOR J = 1 TO LEN(N$) 110 D = VAL(MID$(N$,J,1)) 120 R! = R! + D*D*D*D*D 130 NEXT J 140 RETURN </lang>

Output:

4150 4151 54748 92727 93084 194979

Total = 443839

QB64

<lang qbasic>CONST LIMIT& = 9 ^ 5 * 6 ' we don't need to search higher than this in base 10 DIM AS LONG num, sum, digitSum DIM digit AS _BYTE DIM FifthPowers(9) AS _UNSIGNED INTEGER

FOR i% = LBOUND(FifthPowers) TO UBOUND(FifthPowers)

   FifthPowers(i%) = i% ^ 5

NEXT i%

FOR i& = 2 TO LIMIT&

   num& = i&
   digitSum& = 0
   WHILE num& > 0
       digit%% = num& MOD 10
       digitSum& = digitSum& + FifthPowers(digit%%)
       num& = INT(num& / 10)
   WEND
   IF digitSum& = i& THEN
       PRINT digitSum&
       sum& = sum& + digitSum&
   END IF

NEXT i&

PRINT "The sum is"; sum </lang>

Output:

 4150
 4151
 54748
 92727
 93084
 194979
The sum is 443839

C

<lang c>#include<stdio.h>

include<stdlib.h>
include<math.h>

int sum5( int n ) {

   if(n<10) return pow(n,5);
   return pow(n%10,5) + sum5(n/10);

}

int main(void) {

   int i, sum = 0;
   for(i=2;i<=999999;i++) {
       if(i==sum5(i)) {
           printf( "%d\n", i );
           sum+=i;
       }
   }
   printf( "Total is %d\n", sum );
   return 0;

}</lang>

Output:

4150

4151 54748 92727 93084 194979

Total is 443839

COBOL

<lang cobol> IDENTIFICATION DIVISION.

      PROGRAM-ID. DIGIT-FIFTH-POWER.

      DATA DIVISION.
      WORKING-STORAGE SECTION.
      01 VARIABLES.
         03 CANDIDATE          PIC 9(6).
         03 MAXIMUM            PIC 9(6).
         03 DIGITS             PIC 9 OCCURS 6 TIMES,
                               REDEFINES CANDIDATE.
         03 DIGIT              PIC 9.
         03 POWER-SUM          PIC 9(6).
         03 TOTAL              PIC 9(6).

      01 OUT-FORMAT.
         03 OUT-NUM            PIC Z(5)9.

      PROCEDURE DIVISION.
      BEGIN.
          MOVE ZERO TO TOTAL.
          COMPUTE MAXIMUM = 9 ** 5 * 6.
          PERFORM TEST-NUMBER
              VARYING CANDIDATE FROM 2 BY 1
              UNTIL CANDIDATE IS GREATER THAN MAXIMUM.
          DISPLAY '------ +'.
          DISPLAY TOTAL.
          STOP RUN.

      TEST-NUMBER.
          MOVE ZERO TO POWER-SUM.
          PERFORM ADD-DIGIT-POWER
              VARYING DIGIT FROM 1 BY 1
              UNTIL DIGIT IS GREATER THAN 6.
          IF POWER-SUM IS EQUAL TO CANDIDATE,
              MOVE CANDIDATE TO OUT-NUM,
              DISPLAY OUT-NUM,
              ADD CANDIDATE TO TOTAL.
      
      ADD-DIGIT-POWER.
          COMPUTE POWER-SUM = POWER-SUM + DIGITS(DIGIT) ** 5.</lang>

Output:

Cowgol

<lang cowgol>include "cowgol.coh";

sub pow5(n: uint32): (p: uint32) is

   p := n*n * n*n * n;

end sub;

sub sum5(n: uint32): (r: uint32) is

   r := 0;
   while n != 0 loop
       r := r + pow5(n % 10);
       n := n / 10;
   end loop;

end sub;

var total: uint32 := 0; var n: uint32 := 2; var max: uint32 := pow5(9) * 6;

while n <= max loop

   if n == sum5(n) then
       total := total + n;
       print_i32(n);
       print_nl();
   end if;
   n := n + 1;

end loop;

print("Total: "); print_i32(total); print_nl();</lang>

Output:

4150
4151
54748
92727
93084
194979
Total: 443839

Factor

Thanks to to the Julia entry for the tip about the upper bound of the search. <lang factor>USING: kernel math math.functions math.ranges math.text.utils math.vectors prettyprint sequences ;

2 9 5 ^ 6 * [a,b] [ dup 1 digit-groups 5 v^n sum = ] filter sum .</lang>

Output:

Fermat

<lang fermat>Func Sumfp(n) = if n<10 then Return(n^5) else Return((n|10)^5 + Sumfp(n\10)) fi.; sum:=0; for i=2 to 999999 do if i=Sumfp(i) then sum:=sum+i; !!i fi od; !!('The sum was ', sum );</lang>

Output:

4150 4151 54748 92727 93084 194979

The sum was 443839

FOCAL

<lang focal>01.10 S M=9^5*6 01.20 S T=0 01.30 F C=2,M;D 3 01.40 T "TOTAL",T,! 01.50 Q

02.10 S X=C 02.20 S S=0 02.30 S Y=FITR(X/10) 02.40 S S=S+(X-Y*10)^5 02.50 S X=Y 02.60 I (-X)2.3

03.10 D 2 03.20 I (C-S)3.5,3.3,3.5 03.30 T %6,C,! 03.40 S T=T+C 03.50 R</lang>

Output:

=   4150
=   4151
=  54748
=  92727
=  93084
= 194979
TOTAL= 443839

Go

Translation of: Wren

Library: Go-rcu

<lang go>package main

import (

   "fmt"
   "rcu"

)

func main() {

   // cache 5th powers of digits
   dp5 := [10]int{0, 1}
   for i := 2; i < 10; i++ {
       ii := i * i
       dp5[i] = ii * ii * i
   }
   fmt.Println("The sum of all numbers that can be written as the sum of the 5th powers of their digits is:")
   limit := dp5[9] * 6
   sum := 0
   for i := 2; i <= limit; i++ {
       digits := rcu.Digits(i, 10)
       totalDp := 0
       for _, d := range digits {
           totalDp += dp5[d]
       }
       if totalDp == i {
           if sum > 0 {
               fmt.Printf(" + %d", i)
           } else {
               fmt.Print(i)
           }
           sum += i
       }
   }
   fmt.Printf(" = %d\n", sum)

}</lang>

Output:

The sum of all numbers that can be written as the sum of the 5th powers of their digits is:
4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839

J

Output:

Julia

In base 10, the largest digit is 9. If n is the number of digits, as n increases, 9^5 * n < 10^n. So we do not have to look beyond 9^5 * 6 since 9^5 * 6 < 1,000,000. <lang julia>println("Numbers > 1 that can be written as the sum of fifth powers of their digits:") arr = [i for i in 2 : 9^5 * 6 if mapreduce(x -> x^5, +, digits(i)) == i] println(join(arr, " + "), " = ", sum(arr))

</lang>

Output:

Numbers > 1 that can be written as the sum of fifth powers of their digits:
4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839

PARI/GP

<lang parigp>sumfp(n)=if(n<10,n^5,(n%10)^5+sumfp(n\10)); s=0; for(i=2,999999,if(i==sumfp(i),s=s+i;print(i))); print("Total: ",s);</lang>

Output:

4150

4151 54748 92727 93084 194979

Total: 443839

Pascal

slightly modified Own_digits_power_sum <lang pascal>program PowerOwnDigits2; {$IFDEF FPC}

 {$R+,O+}
 {$MODE DELPHI}{$OPTIMIZATION ON,ALL}{$COPERATORS ON}

{$ELSE}

 {$APPTYPE CONSOLE}

{$ENDIF} uses

 SysUtils,StrUtils;

const

 MAXBASE = 10;
 MaxDgtVal = MAXBASE - 1;
 MaxDgtCount = 19;

type

 tDgtCnt = 0..MaxDgtCount;
 tValues = 0..MaxDgtVal;
 tUsedDigits = array[tValues] of Int8;
 tpUsedDigits = ^tUsedDigits;

 tPower = array[tValues] of Uint64;

var

 PowerDgt:  tPower;
 gblUD   : tUsedDigits;
 CombIdx : array of Int8;
 Numbers : array of Uint64;
 rec_cnt : NativeInt;

 function InitCombIdx(ElemCount: Byte): pbyte;
 begin
   setlength(CombIdx, ElemCount + 1);
   Fillchar(CombIdx[0], sizeOf(CombIdx[0]) * (ElemCount + 1), #0);
   Result := @CombIdx[0];
   Fillchar(gblUD[0], sizeOf(tUsedDigits), #0);
   gblUD[0]:= 1;
 end;

 function Init(ElemCount:byte;Expo:byte):pByte;
 var
   pP1 : pUint64;
   p: Uint64;
   i,j: Int32;
 begin
   pP1 := @PowerDgt[0];
   for i in tValues do
   Begin
     p := 1;
     for j := 1 to Expo do
       p *= i;
     pP1[i] := p;
   end;
   result := InitCombIdx(ElemCount);
   gblUD[0]:= 1;
 end;

 function GetPowerSum(minpot:nativeInt;digits:pbyte;var UD :tUsedDigits):NativeInt;
 var
   res,r  : Uint64;
   dgt :Int32;
 begin
   dgt := minpot;
   res := 0;
   repeat
     dgt -=1;
     res += PowerDgt[digits[dgt]];
   until dgt=0;
   result := 0;
   //convert res into digits
   repeat
     r := res DIV MAXBASE;
     result+=1;
     dgt := res-r*MAXBASE;
     //substract from used digits
     UD[dgt] -= 1;
     res := r;
   until r = 0;
 end;

 procedure calcNum(minPot:Int32;digits:pbyte);
 var
   UD :tUsedDigits;
   res: Uint64;
   i: nativeInt;
 begin
   UD := gblUD;
   i:= GetPowerSum(minpot,digits,UD);
   if i = minPot then
   Begin
     //don't check 0
     i := 1;
     repeat
       If UD[i] <> 0 then
         Break;
       i +=1;
     until i > MaxDgtVal;

     if i > MaxDgtVal then
     begin
       res := 0;
       for i := minpot-1 downto 0 do
         res += PowerDgt[digits[i]];
       setlength(Numbers, Length(Numbers) + 1);
       Numbers[high(Numbers)] := res;
     end;
   end;
 end;

 function NextCombWithRep(pComb: pByte;pUD :tpUsedDigits;MaxVal, ElemCount: UInt32): boolean;
 var
   i,dgt: NativeInt;
 begin
   i := -1;
   repeat
     i += 1;
     dgt := pComb[i];
     if dgt < MaxVal then
       break;
     dec(pUD^[dgt]);
   until i >= ElemCount;
   Result := i >= ElemCount;

   if i = 0 then
   begin
     dec(pUD^[dgt]);
     dgt +=1;
     pComb[i] := dgt;
     inc(pUD^[dgt]);
   end
   else
   begin
     dec(pUD^[dgt]);
     dgt +=1;
     pUD^[dgt]:=i+1;
     repeat
       pComb[i] := dgt;
       i -= 1;
     until i < 0;
   end;
 end;

var

 digits : pByte;
 T0 : Int64;
 tmp : Uint64;
 Pot,dgtCnt,i, j : Int32;

begin

 For pot := 2 to MaxDgtCount do
 begin
   Writeln('Exponent : ',Pot);
   digits := Init(MaxDgtCount,pot);
   T0 := GetTickCount64;
   rec_cnt := 0;
   // i > 0
   For dgtCnt := 2 to pot+1 do
   Begin
     digits := InitCombIdx(Pot);
     repeat
       calcnum(dgtCnt,digits);
       inc(rec_cnt);
     until NextCombWithRep(digits,@gblUD,MaxDgtVal,dgtCnt);
   end;
   T0 := GetTickCount64-T0;
   writeln(rec_cnt,' recursions');
   If length(numbers) > 0 then
   Begin
     //sort
     for i := 0 to High(Numbers) - 1 do
       for j := i + 1 to High(Numbers) do
         if Numbers[j] < Numbers[i] then
         begin
           tmp := Numbers[i];
           Numbers[i] := Numbers[j];
           Numbers[j] := tmp;
         end;

     tmp := 0;
     for i := 0 to High(Numbers) do
     begin
       writeln(Numb2USA(IntToStr(Numbers[i])));
       tmp +=Numbers[i];
     end;
     writeln('sum to ',Numb2USA(IntToStr(tmp)));
   end;
   writeln;
   setlength(Numbers,0);
 end;
 Writeln('Max Uint64 ',Numb2USA(IntToStr(High(Uint64))));
 {$IFDEF WINDOWS}
 readln;
 {$ENDIF}
 setlength(CombIdx,0);

end.</lang>

Output:

Up to 19 Digits:
TIO.RUN Real time: 10.699 s  CPU share: 98.66 %
Exponent : 2
275 recursions

Exponent : 3
990 recursions
153
370
371
407
sum to 1,301

Exponent : 4
2992 recursions
1,634
8,208
9,474
sum to 19,316

Exponent : 5
7997 recursions
4,150
4,151
54,748
92,727
93,084
194,979
sum to 443,839

Exponent : 6
19437 recursions
548,834
sum to 548,834

Exponent : 7
43747 recursions
1,741,725
4,210,818
9,800,817
9,926,315
14,459,929
sum to 40,139,604

Exponent : 8
92367 recursions
24,678,050
24,678,051
88,593,477
sum to 137,949,578

Exponent : 9
184745 recursions
146,511,208
472,335,975
534,494,836
912,985,153
sum to 2,066,327,172

Exponent : 10
352705 recursions
4,679,307,774
sum to 4,679,307,774

Exponent : 11
646635 recursions
32,164,049,650
32,164,049,651
40,028,394,225
42,678,290,603
44,708,635,679
49,388,550,606
82,693,916,578
94,204,591,914
sum to 418,030,478,906

Exponent : 12
1144055 recursions

Exponent : 13
1961245 recursions
564,240,140,138
sum to 564,240,140,138

Exponent : 14
3268749 recursions
28,116,440,335,967
sum to 28,116,440,335,967

Exponent : 15
5311724 recursions

Exponent : 16
8436274 recursions
4,338,281,769,391,370
4,338,281,769,391,371
sum to 8,676,563,538,782,741

Exponent : 17
13123099 recursions
233,411,150,132,317
21,897,142,587,612,075
35,641,594,208,964,132
35,875,699,062,250,035
sum to 93,647,847,008,958,559

Exponent : 18
20029999 recursions

Exponent : 19
30045004 recursions
1,517,841,543,307,505,039
3,289,582,984,443,187,032
4,498,128,791,164,624,869
4,929,273,885,928,088,826
sum to 14,234,827,204,843,405,766

Max Uint64 18,446,744,073,709,551,615

Perl

<lang perl>use strict; use warnings; use feature 'say'; use List::Util 'sum';

for my $power (3..6) {

   my @matches;
   for my $n (2 .. 9**$power * $power) {
       push @matches, $n if $n == sum map { $_**$power } split , $n;
   }
   say "\nSum of powers of n**$power: " . join(' + ', @matches) . ' = ' . sum @matches;

}</lang>

Output:

Sum of powers of n**3: 153 + 370 + 371 + 407 = 1301
Sum of powers of n**4: 1634 + 8208 + 9474 = 19316
Sum of powers of n**5: 4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839
Sum of powers of n**6: 548834 = 548834

Python

<lang>print(sum([n for n in range(2, 6*9**5) if sum(int(i)**5 for i in str(n)) == n]))</lang>

Output:

Raku

<lang perl6>print q:to/EXPANATION/; Sum of all integers (except 1 for some mysterious reason ¯\_(ツ)_/¯), for which the individual digits to the nth power sum to itself. EXPANATION

sub super($i) { $i.trans('0123456789' => '⁰¹²³⁴⁵⁶⁷⁸⁹') }

for 3..8 -> $power {

   print "\nSum of powers of n{super $power}: ";
   my $threshold = 9**$power * $power;
   put .join(' + '), ' = ', .sum with cache
   (2..$threshold).race.map: {
       state %p = ^10 .map: { $_ => $_ ** $power };
       $_ if %p{.comb}.sum == $_
   }

}</lang>

Output:

Sum of all integers (except 1 for some mysterious reason ¯\_(ツ)_/¯),
for which the individual digits to the nth power sum to itself.

Sum of powers of n³: 153 + 370 + 371 + 407 = 1301

Sum of powers of n⁴: 1634 + 8208 + 9474 = 19316

Sum of powers of n⁵: 4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839

Sum of powers of n⁶: 548834 = 548834

Sum of powers of n⁷: 1741725 + 4210818 + 9800817 + 9926315 + 14459929 = 40139604

Sum of powers of n⁸: 24678050 + 24678051 + 88593477 = 137949578

Ring

<lang ring>? "working..."

sumEnd = 0 sumList = ""

pow5 = [] for i = 1 to 9

   add(pow5, pow(i, 5))

for n = limitStart to limitEnd

   sum = 0
   m = n
   while m > 0
       d = m % 10
       if d > 0 sum += pow5[d] ok
       m = unsigned(m, 10, "/")
   end
   if sum = n
      sumList += "" + n + " + "
      sumEnd += n
   ok

Output:

working...
The sum of all the numbers that can be written as the sum of fifth powers of their digits:
4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839
done...

Wren

Library: Wren-math

Using the Julia entry's logic to arrive at an upper bound: <lang ecmascript>import "./math" for Int

// cache 5th powers of digits var dp5 = (0..9).map { |d| d.pow(5) }.toList

System.print("The sum of all numbers that can be written as the sum of the 5th powers of their digits is:") var limit = dp5[9] * 6 var sum = 0 for (i in 2..limit) {

   var digits = Int.digits(i)
   var totalDp = digits.reduce(0) { |acc, d| acc + dp5[d] }
   if (totalDp == i) {
       System.write((sum > 0) ? " + %(i)" : i)
       sum = sum + i
   }

} System.print(" = %(sum)")</lang>

Output:

The sum of all numbers that can be written as the sum of the 5th powers of their digits is:
4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839

XPL0

Since 1 is not actually a sum, it should not be included. Thus the answer should be 443839. <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: