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# Digit fifth powers

Digit fifth powers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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.

## 11l

F fifth_power_digit_sum(n)   R sum(String(n).map(c -> Int(c) ^ 5)) print(sum((2..999999).filter(i -> i == fifth_power_digit_sum(i))))
Output:
443839


## 8080 Assembly

putch:	equ	2 		; CP/M syscall to print a characterputs:	equ	9		; CP/M syscall to print a string	org	100h	;	Find the sum of the 5-powers of the digits	;	of the current numbersum5:	mvi	b,6		; There are 6 digits	lxi 	h,dps		; Set the accumulator to zero	call 	dgzero 	lxi	d,cur		; Load the start of the current numberaddpow:	ldax	d 		; Get current digit	mov	c,a		; Multiply by 6 (width of table)	add	a	add 	c	add	a	mvi 	h,0		; HL = index of table entry	mov	l,a	push	d		; Keep pointer to current digit	lxi	d,pow5		; Add start address of pow5 table	dad	d 	xchg			; Let [DE] = n^5	lxi 	h,dps		; Get accumulator	call	dgadd		; Add the current power to it	pop	d		; Restore pointer to current digit	inx	d	dcr 	b		; If we're not done yet, do the next digit	jnz 	addpow 	lxi	d,cur 		; Is the result the same as the current number?	call	dgcmp	jnz	next		; If not, try the next number	lxi 	h,total		; But if so, it needs to be added to the total	call	dgadd	xchg			; As well as printed 	call	dgoutnext:	lxi 	h,cur		; Increment the current number	call	dginc	lxi	d,max		; Have we reached the end yet?	call	dgcmp	jnz	sum5		; If not, keep going	lxi	d,stot	mvi	c,puts	call	5	lxi	h,total	jmp	dgout	;;;;;;; Program data ;;;;;;;	;	Table of powers of 5, stored as digits in low-endian orderpow5:	db	0,0,0,0,0,0	; 0 ^ 5	db	1,0,0,0,0,0	; 1 ^ 5	db	2,3,0,0,0,0	; 2 ^ 5	db	3,4,2,0,0,0	; 3 ^ 5	db	4,2,0,1,0,0	; 4 ^ 5	db	5,2,1,3,0,0	; 5 ^ 5	db	6,7,7,7,0,0	; 6 ^ 5	db	7,0,8,6,1,0	; 7 ^ 5	db	8,6,7,2,3,0	; 8 ^ 5	db	9,4,0,9,5,0	; 9 ^ 5	; 	End of the search space (9^5 * 6)max:	db	4,9,2,4,5,3	;	Variablestotal:	db	0,0,0,0,0,0	; Total of all matching numbersdps:	db	0,0,0,0,0,0	; Current sum of 5-powers of digitscur:	db	2,0,0,0,0,0	; Current number to test (start at 2)	;	Stringsnl:	db	13,10,'$' ; Newlinestot: db 'Total:$'	;;;;;;; Math routines ;;;;;;	;	Zero out [HL]dgzero:	push	b		; Keep BC and HL	push 	h 	xra	a	mvi	b,6dgzl:	mov	m,a	inx	h	dcr	b	jnz	dgzl	pop 	h		; Restore HL and BC	pop 	b 		ret 	;	Increment [HL]dginc:	push 	h 		; Keep HLdgincl:	inr	m		; Increment current digit	mov	a,m		; Load it into the accumulator	sui	10		; Subtract 10 from it	jc	dginco		; If there is no carry, we're done	mov	m,a		; Otherewise, write it back 	inx	h		; And go increment the next digit	jmp	dgincldginco:	pop 	h		; Restore HL	ret		; 	Print the number in [HL]dgout:	push 	b		; Keep all registers	push 	d	push 	h	lxi	b,6		; Move to the last digit	dad	bdzero:	dcr	c		; Skip leading zeroes	jm	restor 		; Don't bother handling 0 case	dcx	h		; Go back	mov	a,m		; Get digit	ana	a	jz	dzero		; Keep going until we find a nonzero digitdgprn:	adi	'0'		; Write the digit	mov	e,a	push	b		; CP/M syscall destroys registers	push 	h	mvi	c,putch	call 	5 	pop 	h	pop 	b	dcx 	h	mov	a,m	dcr	c	jp	dgprn	mvi	c,puts		; Finally, print a newline	lxi	d,nl	call	5restor:	pop 	h		; And restore the registers	pop 	d	pop 	b	ret	;	Compare [DE] to [HL]dgcmp:	push 	b		; Keep the registers	push 	d	push 	h 	mvi	b,6dgcmpl:	ldax	d		; Get [DE]	cmp	m		; Compare to [HL]	jnz	restor		; If unequal, this is the result	inx	d		; Otherwise, compare next pair	inx	h	dcr 	b	jnz	dgcmpl	jmp	restor	;	Add [DE] to [HL]dgadd:	push	b	push 	d	push 	h	lxi	b,600h 		; B = counter, C = carrydgaddl:	ldax	d		; Get digit from [DE]	add	m 		; Add digit from [HL]	add	c		; Carry the one	cpi	10		; Is the result 10 or higher?	mvi 	c,0		; Assume there will be no carry	jc	dgwr		; If not, handle next digit	sui	10		; But if so, subtract 10,	inr	c 		; And set the carry flag for the next digitdgwr:	mov	m,a		; Store the resulting digit in [HL]	inx	d		; Move the pointers	inx	h	dcr	b		; Any more digits?	jnz	dgaddl	jmp	restor
Output:
4150
4151
54748
92727
93084
194979
Total: 443839

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;
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.

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
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
+/(⊢(/⍨)(⊢=(+/5*⍨⍎¨∘⍕))¨)1↓⍳6×9*5
Output:
443839

## AWK

 # syntax: GAWK -f DIGIT_FIFTH_POWERS.AWKBEGIN {    for (p=3; p<=6; p++) {      limit = 9^p*p      sum = 0      for (i=2; i<=limit; i++) {        if (i == main(i)) {          printf("%6d\n",i)          sum += i        }      }      printf("%6d power %d sum\n\n",sum,p)    }    exit(0)}function main(n,  i,total) {    for (i=1; i<=length(n); i++) {      total += substr(n,i,1) ^ p    }    return(total)}
Output:
   153
370
371
407
1301 power 3 sum

1634
8208
9474
19316 power 4 sum

4150
4151
54748
92727
93084
194979
443839 power 5 sum

548834
548834 power 6 sum


## BASIC

### FreeBASIC

function dig5( n as uinteger ) as uinteger    dim as string ns = str(n)    dim as uinteger ret = 0    for i as ubyte = 1 to len(ns)        ret += val(mid(ns,i,1))^5    next i    return retend function dim as uinteger i, sum = 0 for i = 2 to 999999    if i = dig5(i) then         print i        sum += i    end ifnext i print "Their sum is ", sum
Output:

4150
4151
54748
92727
93084
194979

Their sum is  443839

10 SUM! = 020 FOR I! = 2 TO 999999!30 GOSUB 8040 IF R! = I! THEN SUM! = SUM! + I! : PRINT I!50 NEXT I!60 PRINT "Total = ",SUM70 END80 N$= STR$(I)90 R! = 0100 FOR J = 1 TO LEN(N$)110 D = VAL(MID$(N$,J,1))120 R! = R! + D*D*D*D*D130 NEXT J140 RETURN  Output:  4150 4151 54748 92727 93084 194979 Total = 443839 ### QB64 CONST LIMIT& = 9 ^ 5 * 6 ' we don't need to search higher than this in base 10DIM AS LONG num, sum, digitSumDIM digit AS _BYTEDIM FifthPowers(9) AS _UNSIGNED INTEGER FOR i% = LBOUND(FifthPowers) TO UBOUND(FifthPowers) FifthPowers(i%) = i% ^ 5NEXT 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 IFNEXT i& PRINT "The sum is"; sum  Output:  4150 4151 54748 92727 93084 194979 The sum is 443839  ## BQN Sum5 ← { 0:0; (𝕊⌊𝕩÷10) + (10|𝕩)⋆5 } +´(⊢=Sum5)¨⊸/ 2↓↕6×9⋆5 Output: 443839 ## 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;} Output: 4150 4151 54748 92727 93084 194979 Total is 443839 ## C++ Fast version. Checks numbers up to 399,999, which is above the requirement of 6 * 95 and well below the overkill value of 999,999. #include <iostream>#include <cmath>#include <chrono> using namespace std;using namespace chrono; int main() { auto st = high_resolution_clock::now(); const uint i5 = 100000, i4 = 10000, i3 = 1000, i2 = 100, i1 = 10; uint p4[] = { 0, 1, 32, 243 }, nums[10], p5[10], t = 0, m5, m4, m3, m2, m1, m0; m5 = m4 = m3 = m2 = m1 = m0 = 0; for (uint i = 0; i < 10; i++) p5[i] = pow(nums[i] = i, 5); for (auto i : p4) { auto im = m5, ip = i; m4 = 0; for (auto j : p5) { auto jm = im + m4, jp = ip + j; m3 = 0; for (auto k : p5) { auto km = jm + m3, kp = jp + k; m2 = 0; for (auto l : p5) { auto lm = km + m2, lp = kp + l; m1 = 0; for (auto m : p5) { auto mm = lm + m1, mp = lp + m; m0 = 0; for (auto n : p5) { auto nm = mm + m0++; if (nm == mp + n && nm > 1) t += nm; } m1 += i1; } m2 += i2; } m3 += i3; } m4 += i4; } m5 += i5; } auto et = high_resolution_clock::now(); std::cout << t << " " << duration_cast<nanoseconds>(et - st).count() / 1000.0 << " μs";} Output @ Tio.run: 443839 250.514 μs ## CLU sum5 = proc (n: int) returns (int) sum: int := 0 while n > 0 do sum := sum + (n//10) ** 5 n := n/10 end return(sum)end sum5 start_up = proc () po: stream := stream$primary_output()    total: int := 0    for i: int in int$from_to(2, 6*9**5) do if sum5(i)=i then total := total + i stream$putright(po, int$unparse(i), 6) stream$putc(po, '\n')        end    end    stream$putl(po, "------ +") stream$putright(po, int$unparse(total), 6) stream$putc(po, '\n')end start_up
Output:
  4150
4151
54748
92727
93084
194979
------ +
443839

## 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.
Output:
  4150
4151
54748
92727
93084
194979
------ +
443839

## Comal

0010 FUNC sum5(n) CLOSED0020   sum:=00030   WHILE n>0 DO sum:+(n MOD 10)^5;n:=n DIV 100040   RETURN sum0050 ENDFUNC sum50060 //0070 max:=9^5*60080 total:=00090 FOR i:=2 TO max DO0100   IF i=sum5(i) THEN0110     PRINT USING "######":i0120     total:+i0130   ENDIF0140 ENDFOR i0150 PRINT "------ +"0160 PRINT USING "######":total0170 END
Output:
  4150
4151
54748
92727
93084
194979
------ +
443839

## 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();
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.

USING: kernel math math.functions math.ranges math.text.utilsmath.vectors prettyprint sequences ; 2 9 5 ^ 6 * [a,b] [ dup 1 digit-groups 5 v^n sum = ] filter sum .
Output:
443839


## 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 );
Output:

4150
4151
54748
92727
93084
194979

The sum was  443839

## FOCAL

01.10 S M=9^5*601.20 S T=001.30 F C=2,M;D 301.40 T "TOTAL",T,!01.50 Q 02.10 S X=C02.20 S S=002.30 S Y=FITR(X/10)02.40 S S=S+(X-Y*10)^502.50 S X=Y02.60 I (-X)2.3 03.10 D 203.20 I (C-S)3.5,3.3,3.503.30 T %6,C,!03.40 S T=T+C03.50 R
Output:
=   4150
=   4151
=  54748
=  92727
=  93084
= 194979
TOTAL= 443839

## Go

Translation of: Wren
Library: Go-rcu
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)}
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

(([=[:+/10&#.^:_1^5:)"0+/@#])2}.i.6*9^5
Output:
443839

## jq

Works with: jq

Works with gojq, the Go implementation of jq

Preliminaries

# To take advantage of gojq's arbitrary-precision integer arithmetic:def power($b): . as$in | reduce range(0;$b) as$i (1; . * $in); def sum(s): reduce s as$x (0; .+$x); # Output: a stream of integersdef digits: tostring | explode[] | [.] | implode | tonumber; The Task # Output: an array of i^5 for i in 0 .. 9 inclusivedef dp5: [range(0;10) | power(5)]; def task: dp5 as$dp5  | ($dp5[9] * 6) as$limit  | sum( range(2; $limit + 1) | sum( digits |$dp5[.] ) as $s | select(. ==$s) ) ; "The sum of all numbers that can be written as the sum of the 5th powers of their digits is:", task
Output:
The sum of all numbers that can be written as the sum of the 5th powers of their digits is:
443839


## 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.

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


            NORMAL MODE IS INTEGER             INTERNAL FUNCTION(X)            ENTRY TO POW5.            FUNCTION RETURN X * X * X * X * X            END OF FUNCTION             INTERNAL FUNCTION(N)            ENTRY TO SUM5.            CUR = N            SUM = 0LOOP        WHENEVER CUR.G.0                NEXT = CUR / 10                SUM = SUM + POW5.(CUR - NEXT*10)                CUR = NEXT                TRANSFER TO LOOP            END OF CONDITIONAL            FUNCTION RETURN SUM            END OF FUNCTION             LIMIT = POW5.(9) * 6            TOTAL = 0            THROUGH TEST, FOR I = 2, 1, I.GE.LIMIT            WHENEVER SUM5.(I).E.I                TOTAL = TOTAL + I                PRINT FORMAT NUM, I            END OF CONDITIONALTEST        CONTINUE             PRINT FORMAT TOT, TOTAL            VECTOR VALUES NUM = $S7,I6*$            VECTOR VALUES TOT = $7HTOTAL: ,I6*$            END OF PROGRAM
Output:
         4150
4151
54748
92727
93084
194979
TOTAL: 443839

## PARI/GP

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);
Output:
4150
4151
54748
92727
93084
194979

Total: 443839

## Pascal

slightly modified Own_digits_power_sum checks decimals up to power 19.

program PowerOwnDigits2;{$IFDEF FPC} {$R+,O+}  {$MODE DELPHI}{$OPTIMIZATION ON,ALL}{$COPERATORS ON}{$ELSE}  {$APPTYPE CONSOLE}{$ENDIF}uses  SysUtils,StrUtils;const  CPU_hz = 1000*1000*1000;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 GetCPU_Time: Uint64;  type    TCpu = record              HiCpu,              LoCpu : Dword;           end;  var    Cput : TCpu;  begin  {$ASMMODE INTEL} asm RDTSC; MOV Dword Ptr [CpuT.LoCpu],EAX; MOV Dword Ptr [CpuT.HiCpu],EDX end; with Cput do result := Uint64(HiCPU) shl 32 + LoCpu; end; 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,T1 : UInt64; tmp : Uint64; Pot,dgtCnt,i, j : Int32; begin T0 := GetCPU_Time; For pot := 2 to MaxDgtCount do begin Write('Exponent : ',Pot,' used '); T1 := GetCPU_Time; digits := Init(MaxDgtCount,pot); 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; writeln(rec_cnt,' recursions in ',(GetCPU_Time-T1)/CPU_hz:0:6,' GigaCyles'); 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)-1 do begin write(Numb2USA(IntToStr(Numbers[i])),' + '); tmp +=Numbers[i]; end; write(Numb2USA(IntToStr(Numbers[High(Numbers)])),' = '); tmp +=Numbers[High(Numbers)]; writeln('sum to ',Numb2USA(IntToStr(tmp))); end; writeln; setlength(Numbers,0); end; T0 := GetCPU_Time-T0; Writeln('Max Uint64 ',Numb2USA(IntToStr(High(Uint64)))); writeln('Total runtime : ',T0/CPU_hz:0:6,' GigaCyles'); {$IFDEF WINDOWS}  readln;  {$ENDIF} setlength(CombIdx,0);end. Output @ Tio.run: TIO.RUN User time: 12.905 s //Total runtime : 29.470650 GigaCyles estimated ~2,28 Ghz Exponent : 2 used 275 recursions in 0.000030 GigaCyles Exponent : 3 used 990 recursions in 0.000161 GigaCyles 153 + 370 + 371 + 407 = sum to 1,301 Exponent : 4 used 2992 recursions in 0.000367 GigaCyles 1,634 + 8,208 + 9,474 = sum to 19,316 Exponent : 5 used 7997 recursions in 0.001193 GigaCyles // /2.28 -> 523 µs 4,150 + 4,151 + 54,748 + 92,727 + 93,084 + 194,979 = sum to 443,839 Exponent : 6 used 19437 recursions in 0.003013 GigaCyles 548,834 = sum to 548,834 Exponent : 7 used 43747 recursions in 0.007827 GigaCyles 1,741,725 + 4,210,818 + 9,800,817 + 9,926,315 + 14,459,929 = sum to 40,139,604 Exponent : 8 used 92367 recursions in 0.017619 GigaCyles 24,678,050 + 24,678,051 + 88,593,477 = sum to 137,949,578 Exponent : 9 used 184745 recursions in 0.037308 GigaCyles 146,511,208 + 472,335,975 + 534,494,836 + 912,985,153 = sum to 2,066,327,172 Exponent : 10 used 352705 recursions in 0.090797 GigaCyles 4,679,307,774 = sum to 4,679,307,774 Exponent : 11 used 646635 recursions in 0.207819 GigaCyles 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 used 1144055 recursions in 0.295691 GigaCyles Exponent : 13 used 1961245 recursions in 0.532789 GigaCyles 564,240,140,138 = sum to 564,240,140,138 Exponent : 14 used 3268749 recursions in 0.937579 GigaCyles 28,116,440,335,967 = sum to 28,116,440,335,967 Exponent : 15 used 5311724 recursions in 1.623457 GigaCyles Exponent : 16 used 8436274 recursions in 2.680338 GigaCyles 4,338,281,769,391,370 + 4,338,281,769,391,371 = sum to 8,676,563,538,782,741 Exponent : 17 used 13123099 recursions in 4.432118 GigaCyles 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 used 20029999 recursions in 7.169892 GigaCyles Exponent : 19 used 30045004 recursions in 11.431518 GigaCyles 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 Total runtime : 29.470650 GigaCyles ## 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**$power: " . join(' + ', @matches) . ' = ' . sum @matches;} Output: Sum of powers of n**3: 153 + 370 + 371 + 407 = 1301 Sum of powers of n**4: 1634 + 8208 + 9474 = 19316 Sum of powers of n**5: 4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839 Sum of powers of n**6: 548834 = 548834 ## Phix with javascript_semantics function sum5(integer n) return iff(n<10?power(n,5):power(remainder(n,10),5) + sum5(floor(n/10))) end function integer total = 0 sequence res = {} for i=2 to power(9,5)*6 do if i=sum5(i) then res = append(res,sprint(i)) total += i end if end for printf(1,"%s = %d\n",{join(res," + "),total})  Output: 4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839  ## PicoLisp  (de sum5th (N) (sum '((D) (** (format D) 5)) (chop N))) (setq solutions (cdr # exclude 1 (make (for N (* 6 (** 9 5)) (when (= N (sum5th N)) (link N)))))) (prinl "The numbers that can be written as the sum of the 5th power of their digits are:" )(prin " ") (println solutions)(prinl "Their sum is " (apply + solutions))(bye)  Output: The numbers that can be written as the sum of the 5th power of their digits are: (4150 4151 54748 92727 93084 194979) Their sum is 443839  ## PILOT C :max=9*(9*(9*(9*(9*6)))) :sum=0 :n=2*numberC :dps=0 :cur=n*digitC :next=cur/10 :d=cur-(next*10) :dps=dps+d*(d*(d*(d*#d))) :cur=nextJ (cur>0):*digitT (dps=n):#nC (dps=n):sum=sum+n :n=n+1J (n<max):*numberT :Total: #sumE : Output: 4150 4151 54748 92727 93084 194979 Total: 443839 ## PL/M 100H:/* BDOS ROUTINES */BDOS: PROCEDURE (F,A); DECLARE F BYTE, A ADDRESS; GO TO 5; END BDOS;EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;PUT$CHAR: PROCEDURE (C); DECLARE C BYTE; CALL BDOS(2,C); END PUT$CHAR;PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;NEW$LINE: PROCEDURE; CALL PRINT(.(13,10,'$')); END NEW$LINE; /* THE NATIVE INTEGER TYPES ARE NOT BIG ENOUGH, SO WE NEED TO    MAKE OUR OWN */DECLARE DGT$SIZE LITERALLY '6';MAKE$DEC: PROCEDURE (N, BUF) ADDRESS;    DECLARE (N, BUF) ADDRESS, (I, D BASED BUF) BYTE;    DO I=0 TO DGT$SIZE-1; D(I) = N MOD 10; N = N/10; END; RETURN BUF;END MAKE$DEC; ADD: PROCEDURE (ACC, ADDEND) ADDRESS;    DECLARE (ACC, ADDEND) ADDRESS;    DECLARE (I, C, A BASED ACC, D BASED ADDEND) BYTE;    C = 0;    DO I=0 TO DGT$SIZE-1; A(I) = A(I) + D(I) + C; IF A(I) < 10 THEN C = 0; ELSE DO; A(I) = A(I) - 10; C = 1; END; END; RETURN ACC;END ADD; INCR: PROCEDURE (N); DECLARE N ADDRESS, (I, D BASED N) BYTE; DO I=0 TO DGT$SIZE-1;        IF (D(I) := D(I) + 1) < 10 THEN            RETURN;        ELSE            D(I) = 0;    END;END INCR; EQUAL: PROCEDURE (A, B) BYTE;    DECLARE (A, B) ADDRESS, (DA BASED A, DB BASED B, I) BYTE;     DO I=0 TO DGT$SIZE-1; IF DA(I) <> DB(I) THEN RETURN 0; END; RETURN 0FFH;END EQUAL; PRINT$NUM: PROCEDURE (N);    DECLARE N ADDRESS, (I, D BASED N) BYTE;    I = DGT$SIZE-1; DO WHILE I <> -1 AND D(I) = 0; I = I-1; END; DO WHILE I <> -1; CALL PUT$CHAR('0' + D(I));        I = I-1;    END;END PRINT$NUM; /* GENERATE A TABLE OF DIGIT POWERS BEFOREHAND */DECLARE NATIVE$POWER$5 (10) ADDRESS INITIAL (0, 1, 32, 243, 1024, 3125, 7776, 16807, 32768, 59049);DECLARE POWER$5 (10) ADDRESS;DECLARE POWER$BUF (60) BYTE;DECLARE P BYTE;DO P=0 TO 9; POWER$5(P) = MAKE$DEC(NATIVE$POWER$5(P), .POWER$BUF(DGT$SIZE * P));END; /* DIGITS OF SEARCH LIMIT (9**5 * 6) IN LOW ENDIAN ORDER */DECLARE MAX (DGT$SIZE) BYTE INITIAL (4,9,2,4,5,3); /* SUM THE 5-POWERS OF THE DIGITS OF N */SUM$5: PROCEDURE (N, BUF) ADDRESS; DECLARE (N, BUF) ADDRESS, (I, D BASED N) BYTE; BUF = MAKE$DEC(0, BUF);    DO I=0 TO DGT$SIZE-1; BUF = ADD(BUF, POWER$5(D(I)));    END;    RETURN BUF;END SUM$5; DECLARE CUR (DGT$SIZE) BYTE INITIAL (2,0,0,0,0,0);DECLARE TOTAL$BUF (DGT$SIZE) BYTE;DECLARE TOTAL ADDRESS;TOTAL = MAKE$DEC(0, .TOTAL$BUF); /* TEST EACH NUMBER */DO WHILE NOT EQUAL(.CUR, .MAX);    IF EQUAL(SUM5(.CUR, .MEMORY), .CUR) THEN DO;        TOTAL = ADD(TOTAL, .CUR);        CALL PRINT$NUM(.CUR); CALL NEWLINE; END; CALL INCR(.CUR);END; CALL PRINT(.'TOTAL:$');CALL PRINT$NUM(TOTAL);CALL NEWLINE;CALL EXIT;EOF Output: 4150 4151 54748 92727 93084 194979 TOTAL: 443839 ## Python Comparing conventional vs. faster. from time import time # conventionalst = time()print(sum([n for n in range(2, 6*9**5) if sum(int(i)**5 for i in str(n)) == n]), " ", (time() - st) * 1000, "ms") # fasterst = time()nums = list(range(10))nu = list(range(((6 * 9**5) // 100000) + 1))numbers = []p5 = []for i in nums: p5.append(i**5)for i in nu: im = i * 100000 ip = p5[i] for j in nums: jm = im + 10000 * j jp = ip + p5[j] for k in nums: km = jm + 1000 * k kp = jp + p5[k] for l in nums: lm = km + 100 * l lp = kp + p5[l] for m in nums: mm = lm + 10 * m mp = lp + p5[m] for n in nums: nm = mm + n np = mp + p5[n] if np == nm: if nm > 1: numbers.append(nm)print(sum(numbers), " ", (time() - st) * 1000, "ms", end = "") Output @ Tio.run: 443839 195.04594802856445 ms 443839 22.282838821411133 ms Around eight times faster. ## Raku print q:to/EXPANATION/;Sum of all integers (except 1 for some mysterious reason ¯\_(ツ)_/¯),for which the individual digits to the nth power sum to itself.EXPANATION sub super($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 };        $_ if %p{.comb}.sum ==$_    }}
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

## REXX

/* numbers that are equal to the sum of their digits raised to the power 5 */ maximum = 9**5 * 6total = 0out = ''do i = 2 to maximum     if sum5(i) = i then do        if out \= '' then out = out || ' + '        out = out || i        total = total + i    endendsay out || ' = ' || total exit sum5: procedure     arg num    result = 0    do i = 1 to length(num)        result = result + substr(num, i, 1) ** 5    end    return result
Output:
4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839

## Ring

### Conventional

? "working..." sumEnd = 0sumList = "" pow5 = []for i = 1 to 9    add(pow5, pow(i, 5))next limitStart = 2limitEnd = 6 * pow5[9] 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    oknext ? "The sum of all the numbers that can be written as the sum of fifth powers of their digits:"? substr(sumList, 1, len(sumList) - 2) + "= " + sumEnd? "done..."
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...


### Faster

Around six times faster than the conventional version.

st = clock()lst9 = 1:10 lst3 = 1:4p5 = [] m5 = [] m4 = [] m3 = [] m2 = [] m1 = []for i in lst9  add(p5, pow(i - 1, 5)) add(m1, (i - 1) * 10) add(m2, m1[i] * 10)  add(m3, m2[i] * 10) add(m4, m3[i] * 10) add(m5, m4[i] * 10)next s = 0 t = ""for i in lst3 ip = p5[i] im = m5[i]  for j in lst9 jp = ip + p5[j] jm = im + m4[j]    for k in lst9 kp = jp + p5[k] km = jm + m3[k]      for l in lst9 lp = kp + p5[l] lm = km + m2[l]        for m in lst9 mp = lp + p5[m] mm = lm + m1[m]          for n in lst9 np = mp + p5[n] nm = mm + n - 1            if nm = np and nm > 1              if t != "" t += " + " ok              s += nm t += nm            oknext next next next next nextet = clock()put t + " = " + s + "  " + (et - st) / clockspersecond() + " sec"
Output @ Tio.run:
4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839  4.90 sec

## Seed7

\$ include "seed7_05.s7i"; const proc: main is func  local    var integer: i is 0;    var integer: n is 0;    var integer: sum is 0;    var integer: digitsum is 0;  begin    for i range 2 to 9 ** 5 * 6 do      n := i;      while n > 0 do        digitsum +:= (n mod 10) ** 5;        n := n div 10;      end while;      if digitsum = i then        sum +:= i;      end if;      digitsum := 0;    end for;    writeln(sum);  end func;
Output:
443839


## Sidef

func digit_nth_powers(n, base=10) {     var D = @(^base)    var P = D.map {|d| d**n }    var A = []    var m = (base-1)**n     for(var (k, t) = (1, 1); k*m >= t; (++k, t*=base)) {        D.combinations_with_repetition(k, {|*c|            var v = c.sum {|d| P[d] }            A.push(v) if (v.digits(base).sort == c)        })    }     A.sort.grep { _ > 1 }} for n in (3..8) {    var a = digit_nth_powers(n)    say "Sum of #{n}-th powers of their digits: #{a.join(' + ')} = #{a.sum}"}
Output:
Sum of 3-th powers of their digits: 153 + 370 + 371 + 407 = 1301
Sum of 4-th powers of their digits: 1634 + 8208 + 9474 = 19316
Sum of 5-th powers of their digits: 4150 + 4151 + 54748 + 92727 + 93084 + 194979 = 443839
Sum of 6-th powers of their digits: 548834 = 548834
Sum of 7-th powers of their digits: 1741725 + 4210818 + 9800817 + 9926315 + 14459929 = 40139604
Sum of 8-th powers of their digits: 24678050 + 24678051 + 88593477 = 137949578


## Wren

Library: Wren-math

Using the Julia entry's logic to arrive at an upper bound:

import "./math" for Int // cache 5th powers of digitsvar 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] * 6var sum = 0for (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)")
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.

\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);]
Output:
0
4150
1
4151
93084
92727
54748
194979

443840


## Zig

const std = @import("std"); fn sum5(n: u32) u32 {    var i = n;    var r: u32 = 0;    while (i != 0) : (i /= 10)       r += std.math.pow(u32, i%10, 5);    return r;} pub fn main() !void {    const stdout = std.io.getStdOut().writer();    const max = std.math.pow(u32,9,5) * 6;     var n: u32 = 2;    var total: u32 = 0;    while (n <= max) : (n += 1) {        if (sum5(n) == n) {            try stdout.print("{d:6}\n", .{n});            total += n;        }    }     try stdout.print("Total: {d:6}\n", .{total});}
Output:
  4150
4151
54748
92727
93084
194979
Total: 443839`