Harshad or Niven series
You are encouraged to solve this task according to the task description, using any language you may know.
The Harshad or Niven numbers are positive integers ≥ 1 that are divisible by the sum of their digits.
For example, 42 is a Harshad number as 42 is divisible by (4 + 2) without remainder.
Assume that the series is defined as the numbers in increasing order.
- Task
The task is to create a function/method/procedure to generate successive members of the Harshad sequence.
Use it to list the first twenty members of the sequence and list the first Harshad number greater than 1000.
Show your output here.
Ada
<lang Ada>with Ada.Text_IO;
procedure Harshad is
function Next(N: in out Positive) return Positive is
function Sum_Of_Digits(N: Natural) return Natural is
( if N = 0 then 0 else ((N mod 10) + Sum_Of_Digits(N / 10)) );
begin
while not (N mod Sum_Of_Digits(N) = 0) loop
N := N + 1;
end loop;
return N;
end Next;
Current: Positive := 1;
begin
for I in 1 .. 20 loop
Ada.Text_IO.Put(Integer'Image(Next(Current)));
Current := Current + 1;
end loop;
Current := 1000 + 1;
Ada.Text_IO.Put_Line(" ..." & Integer'Image(Next(Current)));
end Harshad;</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 ... 1002
ALGOL 68
<lang algol68>BEGIN
PROC digit sum = (INT i) INT :
BEGIN
INT res := i %* 10, h := i;
WHILE (h %:= 10) > 0 DO res +:= h %* 10 OD;
res
END;
INT found := 0;
FOR i WHILE found < 20 DO
(i %* digit sum (i) = 0 | found +:= 1; printf (($g(0)", "$, i)) ) OD;
FOR i FROM 1001 DO
(i %* digit sum (i) = 0 | printf (($g(0)l$, i)); stop) OD
END</lang>
- Output:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 1002
AutoHotkey
<lang AutoHotkey>H := [] n := 1
Loop n := (H[A_Index] := NextHarshad(n)) + 1 until H[H.MaxIndex()] > 1000
Loop, 20 Out .= H[A_Index] ", "
MsgBox, % Out ". . . " H[H.MaxIndex()]
NextHarshad(n) { Loop, { Loop, Parse, n sum += A_LoopField if (!Mod(n, sum)) return n n++, sum := "" } }</lang>
- Output:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, . . . 1002
AWK
<lang AWK>#!/usr/bin/awk -f BEGIN { k=0; n=0; printf("First twenty Harshad numbers are:\n "); while (k<20) { if (isharshad(++n)) { printf("%i ",n); ++k; } } n = 1000; while (!isharshad(++n)); printf("\nFirst Harshad number larger than 1000 is \n %i\n",n); }
function isharshad(n) { s = 0; for (i=0; i<length(n); ) { s+=substr(n,++i,1); } return !(n%s); }</lang>
- Output:
First twenty Harshad numbers are: 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 First Harshad number larger than 1000 is 1002
Batch File
<lang dos> @echo off setlocal enabledelayedexpansion
for /l %%i in (1,1,20) do (
call:harshad echo Harshad number %%i - !errorlevel!
)
- loop
call:harshad if %errorlevel% leq 1000 goto loop echo First Harshad number greater than 1000: %errorlevel% pause>nul exit /b
- harshad
if "%harshadnum%"=="" set harshadnum=0 set /a harshadnum+=1 call:strlength %harshadnum%
set harshadsum=0 for /l %%i in (0,1,%errorlevel%) do set /a harshadsum+=!harshadnum:~%%i,1!
set /a isharshad=%harshadnum% %% %harshadsum% if %isharshad%==0 exit /b %harshadnum% goto harshad
- strlength
setlocal enabledelayedexpansion set tempcount=1 set str=%1
- strlengthloop
set /a length=%tempcount%-1 if "!str:~%tempcount%,1!"=="" endlocal && exit /b %length% set /a tempcount+=1 goto strlengthloop </lang>
- Output:
Harshad number 1 - 1 Harshad number 2 - 2 Harshad number 3 - 3 Harshad number 4 - 4 Harshad number 5 - 5 Harshad number 6 - 6 Harshad number 7 - 7 Harshad number 8 - 8 Harshad number 9 - 9 Harshad number 10 - 10 Harshad number 11 - 12 Harshad number 12 - 18 Harshad number 13 - 20 Harshad number 14 - 21 Harshad number 15 - 24 Harshad number 16 - 27 Harshad number 17 - 30 Harshad number 18 - 36 Harshad number 19 - 40 Harshad number 20 - 42 First Harshad number greater than 1000: 1002
BBC BASIC
<lang bbcbasic> I%=1:CNT%=0
WHILE TRUE
IF FNHarshad(I%) THEN
IF CNT%<20 PRINT ;I%;" ";:CNT%+=1
IF I%>1000 PRINT ;I%:EXIT WHILE
ENDIF
I%+=1
ENDWHILE
END
DEF FNHarshad(num%)
LOCAL sum%,tmp%
tmp%=num%
sum%=0
WHILE (tmp%>0)
sum%+=tmp% MOD 10
tmp%/=10
ENDWHILE
=(num% MOD sum%)=0</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
Befunge
<lang befunge>45*1>::01-\>:55+%\vv\0< >\1+^ + <|:/<+55<` : ^_>1-\:.v@1>\:0\`#v_+\^ >^1\,+55<.^_:#%$:#<"}"v ^!:\_ ^###< !`*8<</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
C
<lang C>#include <stdio.h>
static int digsum(int n) {
int sum = 0;
do { sum += n % 10; } while (n /= 10);
return sum;
}
int main(void) {
int n, done, found;
for (n = 1, done = found = 0; !done; ++n) {
if (n % digsum(n) == 0) {
if (found++ < 20) printf("%d ", n);
if (n > 1000) done = printf("\n%d\n", n);
}
}
return 0;
}</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
C#
<lang csharp> using System; using System.Collections.Generic;
namespace Harshad {
class Program
{
public static bool IsHarshad(int n)
{
char[] inputChars = n.ToString().ToCharArray();
IList<byte> digits = new List<byte>();
foreach (char digit in inputChars)
{
digits.Add((byte)Char.GetNumericValue(digit));
}
if (n < 1)
{
return false;
}
int sum = 0;
foreach (byte digit in digits)
{
sum += digit;
}
return n % sum == 0;
}
static void Main(string[] args)
{
int i = 1;
int count = 0;
while (true)
{
if (IsHarshad(i))
{
count++;
if (count <= 20)
{
Console.Write(string.Format("{0} ", i));
}
else if (i > 1000)
{
Console.Write(string.Format("{0} ", i));
break;
}
}
i++;
}
Console.ReadKey();
}
}
} </lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
C++
<lang cpp>#include <vector>
- include <iostream>
int sumDigits ( int number ) {
int sum = 0 ;
while ( number != 0 ) {
sum += number % 10 ;
number /= 10 ;
}
return sum ;
}
bool isHarshad ( int number ) {
return number % ( sumDigits ( number ) ) == 0 ;
}
int main( ) {
std::vector<int> harshads ;
int i = 0 ;
while ( harshads.size( ) != 20 ) {
i++ ;
if ( isHarshad ( i ) )
harshads.push_back( i ) ;
}
std::cout << "The first 20 Harshad numbers:\n" ;
for ( int number : harshads )
std::cout << number << " " ;
std::cout << std::endl ;
int start = 1001 ;
while ( ! ( isHarshad ( start ) ) )
start++ ;
std::cout << "The smallest Harshad number greater than 1000 : " << start << '\n' ;
return 0 ;
}</lang>
- Output:
The first 20 Harshad numbers: 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 The smallest Harshad number greater than 1000 : 1002
Clojure
<lang Clojure>(defn digsum [n acc]
(if (zero? n) acc
(digsum (quot n 10) (+ acc (mod n 10)))))
(let [harshads (filter
#(zero? (mod % (digsum % 0)))
(iterate inc 1))]
(prn (take 20 harshads))
(prn (first (filter #(> % 1000) harshads))))</lang>
- Output:
(1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42) 1002
COBOL
<lang cobol>identification division. program-id. harshad. environment division. data division. working-storage section.
- > for storing first 20 harshad-niven numbers
01 harshads.
03 harshad pic 9(5) occurs 20 times indexed by niven.
- > numbers tested for harshad-niven-ness.
01 first-num pic 9(5). 01 second-num pic 9(5).
- > loop counter
01 i pic 9(5).
- > for calculating sum of digits
01 div pic 9(5). 01 mod pic 9(5). 01 tot pic 9(5).
- > for harshad-niven calculation and display
01 harshad-div pic 9(5). 01 harshad-mod pic 9(5).
88 evenly-divisible value 0.
01 harshad-disp pic zzzz9. 01 harshad-result pic 9(5).
- > for selecting what to do with results of harshad calculation
01 pass pic 9.
88 first-pass value 1. 88 second-pass value 2.
procedure division. 10-main section.
move 1 to pass.
set niven to 1.
perform 20-calculate-harshad with test before varying first-num from 1 by 1 until niven = 21.
move 2 to pass.
move first-num to second-num.
perform 20-calculate-harshad with test after varying first-num from second-num by 1 until harshad-result > 1000.
perform with test after varying i from 1 by 1 until i = 20
move harshad(i) to harshad-disp
display function trim(harshad-disp) space with no advancing
end-perform.
move harshad-result to harshad-disp.
display "... " function trim(harshad-disp).
stop run.
20-calculate-harshad.
move first-num to div.
move zero to harshad-result.
perform 100-calculate-sum-of-digits.
divide first-num by tot giving harshad-div remainder harshad-mod.
if evenly-divisible
if first-pass
move first-num to harshad(niven)
set niven up by 1
else
move first-num to harshad-result
end-if
end-if.
exit paragraph.
100-calculate-sum-of-digits.
move zero to tot.
perform with test after until div = 0
divide div by 10 giving div remainder mod
add mod to tot
end-perform.
*> if tot >= 10
*> move tot to div
*> go to 100-calculate-sum-of-digits
*> end-if.
exit paragraph.
</lang>
ColdFusion
<lang cfm> <Cfset harshadNum = 0> <Cfset counter = 0>
<Cfloop condition="harshadNum lte 1000">
<Cfset startnum = harshadNum + 1>
<Cfset digits = 0>
<Cfset harshad = 0>
<Cfloop condition="Harshad eq 0">
<Cfset current_i = startnum>
<Cfset digits = 0>
<cfloop condition="len(current_i) gt 1">
<Cfset digit = left(current_i, 1)>
<Cfset current_i = right(current_i, len(current_i)-1)>
<Cfset digits = digits + digit>
</cfloop>
<Cfset digits = digits + current_i>
<Cfif Startnum MOD digits eq 0>
<Cfset harshad = 1>
<Cfelse>
<cfset startnum = startnum + 1>
</Cfif>
</Cfloop>
<cfset harshadNum = startnum>
<Cfset counter = counter + 1>
<Cfif counter lte 20>
<Cfoutput>#harshadNum# </Cfoutput>
</Cfif>
</Cfloop>
<Cfoutput>... #harshadNum# </Cfoutput> </lang>
Common Lisp
<lang lisp>(defun harshadp (n)
(zerop (rem n (digit-sum n))))
(defun digit-sum (n &optional (a 0))
(cond ((zerop n) a)
(t (digit-sum (floor n 10) (+ a (rem n 10))))))
(defun list-harshad (n &optional (i 1) (lst nil))
"list the first n Harshad numbers starting from i (default 1)" (cond ((= (length lst) n) (reverse lst))
((harshadp i) (list-harshad n (+ i 1) (cons i lst)))
(t (list-harshad n (+ i 1) lst))))</lang>
- Output:
CL-USER> (list-harshad 20) (1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42) CL-USER> (list-harshad 1 1001) (1002)
D
<lang d>void main() {
import std.stdio, std.algorithm, std.range, std.conv;
enum digSum = (int n) => n.text.map!(d => d - '0').sum; enum harshads = iota(1, int.max).filter!(n => n % digSum(n) == 0);
harshads.take(20).writeln; harshads.filter!(h => h > 1000).front.writeln;
}</lang>
- Output:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42] 1002
EchoLisp
<lang scheme> (define (harsh? n)
(zero? (modulo n
(apply + (map string->number (string->list (number->string n)))))))
(harsh? 42)
→ #t
(define H (stream-filter harsh? (in-naturals 1)))
(take H 20)
→ (1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42)
(for ((n H)) #:break (> n 1000) => n)
→ 1002
</lang>
Eiffel
<lang eiffel> note description : "project application root class" date : "$October 10, 2014$" revision : "$Revision$"
class NIVEN_SERIES
create make
feature make local number : INTEGER count : INTEGER last : BOOLEAN do number := 1
from number := 1 last := false
until last = true
loop
if (number \\ sum_of_digits(number) = 0) then count := count + 1
if (count <= 20 ) then print("%N") print(number) end
if (number > 1000) then print("%N") print(number) last := true end end
number := number + 1 end end
sum_of_digits(no:INTEGER):INTEGER
local sum : INTEGER num : INTEGER do sum := 0
from num := no
until num = 0
loop sum := sum + num \\ 10 num := num // 10 end
Result := sum end end
</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
Elixir
<lang elixir>defmodule Harshad do
def series, do: Stream.iterate(1, &(&1+1)) |> Stream.filter(&(number?(&1))) def number?(n), do: rem(n, digit_sum(n, 0)) == 0 defp digit_sum(0, sum), do: sum defp digit_sum(n, sum), do: digit_sum(div(n, 10), sum + rem(n, 10))
end
IO.inspect Harshad.series |> Enum.take(20)
IO.inspect Harshad.series |> Stream.drop_while(&(&1 <= 1000)) |> Enum.take(1) |> hd</lang>
- Output:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42] 1002
Erlang
<lang Erlang> -module( harshad ).
-export( [greater_than/1, sequence/1, task/0] ).
greater_than( N ) when N >= 1 ->
greater_than( 2, N, acc(1, {0, []}) ).
sequence( Find_this_many ) when Find_this_many >= 1 ->
sequence( 2, Find_this_many, acc(1, {0, []}) ).
task() ->
io:fwrite( "~p~n", [sequence(20)] ),
io:fwrite( "~p~n", [greater_than(1000)] ).
acc( N, Acc ) -> acc( N rem lists:sum([X - $0|| X <- erlang:integer_to_list(N)]), N, Acc ).
acc( 0, N, {Found, Acc} ) -> {Found + 1, [N | Acc]}; acc( _Reminder, _N, Acc ) -> Acc.
greater_than( _N, Find, {_, [Found | _T]} ) when Found > Find -> Found; greater_than( N, Find, Acc ) -> greater_than( N + 1, Find, acc(N, Acc) ).
sequence( _N, Found, {Found, Acc} ) -> lists:reverse( Acc ); sequence( N, Find, Acc ) -> sequence( N + 1, Find, acc(N, Acc) ). </lang>
- Output:
39> harshad:task(). [1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,36,40,42] 1002
Erlang 2
A somewhat more simple approach. Somewhat more efficient since it produces the partial list 23 times for the 20 element case whereas the above does so 36 or 37 times.
<lang Erlang> -module(harshad). -export([main/0,harshad/1,seq/1]).
% We return the number R if harshad, else 0 harshad(R) ->
case R
rem lists:sum([X - $0|| X <- erlang:integer_to_list(R)]) of 0
-> R; _ -> 0 end.
% build a list of harshads retrieving input from harshad(R) % filter out the nulls and return hlist(A,B) ->
RL = [ harshad(X) || X <- lists:seq(A,B) ],
lists:filter( fun(X) -> X > 0 end, RL).
seq(Total) -> seq(Total, [], 0).
seq(Total,L,_) when length(L) == Total-> L; seq(Total,L,Acc) when length(L) < Total ->
NL = hlist(1,Total + Acc),
seq(Total,NL,Acc+1).
gt(_,L) when length(L) == 1 -> hd(L); gt(X,_) ->
NL = hlist(X+1,X+2),
gt(X+2,NL).
main() ->
io:format("seq(20): ~w~n", [ seq(20) ]),
io:format("gt(1000): ~w~n", [ gt(1000,[]) ]).
</lang>
2> harshad:main(). seq(20): [1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,36,40,42] gt(1000): 1002 ok
F#
<lang fsharp>let divides d n =
match bigint.DivRem(n, d) with | (_, rest) -> rest = 0I
let splitToInt (str:string) = List.init str.Length (fun i -> ((int str.[i]) - (int "0".[0])))
let harshads =
let rec loop n = seq {
let sum = List.fold (+) 0 (splitToInt (n.ToString()))
if divides (bigint sum) n then yield n
yield! loop (n + 1I)
}
loop 1I
[<EntryPoint>] let main argv =
for h in (Seq.take 20 harshads) do printf "%A " h printfn "" printfn "%A" (Seq.find (fun n -> n > 1000I) harshads) 0</lang>
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
Factor
<lang factor> USING: math.text.utils lists lists.lazy ;
- niven? ( n -- ? ) dup 1 digit-groups sum mod 0 = ;
- first-n-niven ( n -- seq )
1 lfrom [ niven? ] lfilter ltake list>array ;
- next-niven ( n -- m ) 1 + [ dup niven? ] [ 1 + ] until ;
20 first-n-niven . 1000 next-niven . </lang>
- Output:
{ 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 }
1002
FBSL
The INITIALIZE routine fills a dynamic array with all we need, even the ellipsis. <lang qbasic>#APPTYPE CONSOLE
CLASS harshad
PRIVATE:
memo[]
SUB INITIALIZE()
DIM i = 1, c
DO
IF isNiven(i) THEN
c = c + 1
memo[c] = i
END IF
i = i + 1
IF c = 20 THEN EXIT DO
LOOP
memo[] = "..."
i = 1000
WHILE NOT isNiven(INCR(i)): WEND
memo[] = i
END SUB
FUNCTION isNiven(n)
RETURN NOT (n MOD sumdigits(n))
END FUNCTION
FUNCTION sumdigits(n)
DIM num = n, m, sum
WHILE num
sum = sum + num MOD 10
num = num \ 10
WEND
RETURN sum
END FUNCTION
PUBLIC:
METHOD Yield()
FOREACH DIM e IN memo
PRINT e, " ";
NEXT
END METHOD
END CLASS
DIM niven AS NEW harshad niven.Yield()
PAUSE </lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 ... 1002 Press any key to continue...
Fortran
Please observe compilation on GNU/linux system and output from run are in the comments at the START of the FORTRAN 2003 source. The 1--20 loop idea was stolen from the ada solution. Thank you. <lang FORTRAN> !-*- mode: compilation; default-directory: "/tmp/" -*- !Compilation started at Tue May 21 13:15:59 ! !a=./f && make $a && $a < unixdict.txt !gfortran -std=f2003 -Wall -ffree-form f.f03 -o f ! 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002 ! !Compilation finished at Tue May 21 13:15:59
program Harshad
integer :: i, h = 0 do i=1, 20 call nextHarshad(h) write(6, '(i5)', advance='no') h enddo h = 1000 call nextHarshad(h) write(6, '(i5)') h
contains
subroutine nextHarshad(h) ! alter input integer h to be the next greater Harshad number.
integer, intent(inout) :: h
h = h+1 ! bigger
do while (.not. isHarshad(h))
h = h+1
end do
end subroutine nextHarshad
logical function isHarshad(a)
integer, intent(in) :: a
integer :: mutable, digit_sum
isHarshad = .false.
if (a .lt. 1) return ! false if a < 1
mutable = a
digit_sum = 0
do while (mutable /= 0)
digit_sum = digit_sum + mod(mutable, 10)
mutable = mutable / 10
end do
isHarshad = 0 .eq. mod(a, digit_sum)
end function isHarshad
end program Harshad </lang>
FreeBASIC
<lang freebasic>' FB 1.05.0 Win64
Function sumDigits(n As Integer) As Integer
If n < 0 Then Return 0 Dim sum As Integer While n > 0 sum += n Mod 10 n \= 10 Wend Return sum
End Function
Function isHarshad(n As Integer) As Boolean
Return n Mod sumDigits(n) = 0
End Function
Print "The first 20 Harshad or Niven numbers are :" Dim count As Integer = 0 Dim i As Integer = 1
Do
If isHarshad(i) Then Print i; " "; Count += 1 If count = 20 Then Exit Do End If i += 1
Loop
Print : Print Print "The first such number above 1000 is :" i = 1001
Do
If isHarshad(i) Then Print i; " " Exit Do End If i += 1
Loop
Print Print "Press any key to quit" Sleep</lang>
- Output:
The first 20 Harshad or Niven numbers are : 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 The first such number above 1000 is : 1002
Frink
<lang frink> isHarshad[n] := n mod sum[integerDigits[n]] == 0
c = 0 i = 1 while c<20 {
if isHarshad[i]
{
c = c + 1
println[i]
}
i = i + 1
}
println[] i = 1000
do
i = i + 1
while ! isHarshad[i]
println[i] </lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
Gambas
Click this link to run this code <lang gambas>Public Sub Main() Dim siCount, siLoop, siTotal, siCounter As Short Dim sNo, sTemp As String Dim sHold, sNiven As New String[]
For siCount = 1 To 1500
sNo = Str(siCount)
For siLoop = 1 To Len(sNo)
sHold.Add(Mid(sNo, siLoop, 1))
Next
For Each sTemp In sHold
siTotal += Val(sTemp)
Next
If siCount Mod siTotal = 0 Then
Inc siCounter
If siCounter < 21 Or siCount > 1000 Then
sNiven.Add(Str(siCount))
If siCount > 1000 Then Break
Endif
Endif
siTotal = 0
sHold.Clear
Next
Print "First twenty Harshad numbers and the first Harshad number greater than 1000" Print sNiven.Join(", ")
End</lang> Output:
First twenty Harshad numbers and the first Harshad number greater than 1000 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 1002
Go
<lang go>package main
import "fmt"
type is func() int
func newSum() is {
var ms is
ms = func() int {
ms = newSum()
return ms()
}
var msd, d int
return func() int {
if d < 9 {
d++
} else {
d = 0
msd = ms()
}
return msd + d
}
}
func newHarshard() is {
i := 0
sum := newSum()
return func() int {
for i++; i%sum() != 0; i++ {
}
return i
}
}
func main() {
h := newHarshard()
fmt.Print(h())
for i := 1; i < 20; i++ {
fmt.Print(" ", h())
}
fmt.Println()
h = newHarshard()
n := h()
for ; n <= 1000; n = h() {
}
fmt.Println(n)
}</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
Groovy
<lang Groovy> class HarshadNiven{ public static boolean find(int x)
{
int sum = 0,temp,var;
var = x;
while(x>0)
{
temp = x%10;
sum = sum + temp;
x = x/10;
}
if(var%sum==0) temp = 1;
else temp = 0;
return temp;
}
public static void main(String[] args)
{
int t,i;
t = 0;
for(i=1;t<20;i++)
{
if(find(i))
{
print(i + " ");
t++;
}
}
int x = 0;
int y = 1000;
while(x!=1)
{
if(find(y)) x = 1;
y++;
}
println();
println(y+1);
}
} </lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
Haskell
<lang haskell>import Data.Char (ord)
harshads :: [Int] harshads =
let digsum = sum . map ((48 -) . ord) . show in filter ((0 ==) . (mod <*> digsum)) [1 ..]
main :: IO () main = mapM_ print [take 20 harshads, [(head . filter (> 1000)) harshads]]</lang>
- Output:
[1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,36,40,42] 1002
Or, as an alternative to strings and imports:
<lang haskell>harshadSeries :: [Int] harshadSeries = filter ((0 ==) . (rem <*> (sum . digitList))) [1 ..]
digitList :: Int -> [Int] digitList 0 = [] digitList n = rem n 10 : digitList (quot n 10)
main :: IO () main = mapM_ print $ [take 20, take 1 . dropWhile (<= 1000)] <*> [harshadSeries]</lang>
- Output:
[1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,36,40,42] [1002]
Icon and Unicon
<lang unicon>procedure main(A)
limit := integer(A[1]) | 20
every writes(niven(seq())\limit," ")
writes("... ")
write(niven(seq(1001))\1)
end
procedure niven(n)
n ? {s := 0; while s +:= move(1)}
if (n%s) = 0 then return n
end</lang>
- Output:
->ns 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 ... 1002 ->
J
<lang J>Until =: 2 : 'u^:(-.@:v)^:_' isHarshad =: 0 = ] |~ [: +/ #.inv NB. BASE isHarshad N assert 1 0 -: 10 isHarshad&> 42 38 nextHarshad =: (>: Until (10&isHarshad))@:>: assert 45 -: nextHarshad 42 assert 3 4 5 -: nextHarshad&> 2 3 4 assert 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 -: (, nextHarshad@:{:)Until (20 = #) 1 assert 1002 -: nextHarshad 1000
NB. next Harshad number in base 6. Input and output are in base 6. NB. Verification left to you, gentle programmer. nextHarshad_base_6 =: (>: Until (6&isHarshad))@:>: ' '-.~":6#.inv nextHarshad_base_6 6b23235
23253 </lang>
Java
<lang java5>public class Harshad{
private static long sumDigits(long n){
long sum = 0;
for(char digit:Long.toString(n).toCharArray()){
sum += Character.digit(digit, 10);
}
return sum;
}
public static void main(String[] args){
for(int count = 0, i = 1; count < 20;i++){
if(i % sumDigits(i) == 0){
System.out.println(i);
count++;
}
}
System.out.println();
for(int i = 1001; ; i++){
if(i % sumDigits(i) == 0){
System.out.println(i);
break;
}
}
}
}</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
JavaScript
ES5
<lang javascript>function isHarshad(n) {
var s = 0;
var n_str = new String(n);
for (var i = 0; i < n_str.length; ++i) {
s += parseInt(n_str.charAt(i));
}
return n % s === 0;
}
var count = 0; var harshads = [];
for (var n = 1; count < 20; ++n) {
if (isHarshad(n)) {
count++;
harshads.push(n);
}
}
console.log(harshads.join(" "));
var h = 1000; while (!isHarshad(++h)); console.log(h); </lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
ES6
One possible approach to functional composition: <lang JavaScript>(() => {
'use strict';
// HARSHADS ---------------------------------------------------------------
// nHarshads :: Int -> [Int]
const nHarshads = n => {
// isHarshad :: Int -> Bool
const isHarshad = n => 0 === n % sum(digitList(n));
return until(
dct => dct.nth === n,
dct => {
const
next = succ(dct.i),
blnHarshad = isHarshad(next);
return {
i: next,
hs: blnHarshad ? dct.hs.concat(next) : dct.hs,
nth: dct.nth + (blnHarshad ? 1 : 0)
};
}, {
i: 0,
hs: [],
nth: 0
}
)
.hs;
};
// GENERIC FUNCTIONS ------------------------------------------------------
// digitList :: Int -> [Int]
const digitList = n =>
n > 0 ? [n % 10].concat(digitList(Math.floor(n / 10))) : [];
// dropWhile :: (a -> Bool) -> [a] -> [a]
const dropWhile = (p, xs) => {
let i = 0;
for (let lng = xs.length;
(i < lng) && p(xs[i]); i++) {}
return xs.slice(i);
};
// head :: [a] -> a const head = xs => xs.length ? xs[0] : undefined;
// a -> String const show = x => JSON.stringify(x, null, 2);
// succ :: Int -> Int const succ = x => x + 1
// sum :: (Num a) => [a] -> a const sum = xs => xs.reduce((a, x) => a + x, 0);
// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = (p, f, x) => {
const go = x => p(x) ? x : go(f(x));
return go(x);
};
// TEST -------------------------------------------------------------------
return show({
firstTwenty: nHarshads(20),
firstOver1000: head(dropWhile(x => x <= 1000, nHarshads(1000)))
});
})();</lang>
- Output:
{
"firstTwenty": [
1,
2,
3,
4,
5,
6,
7,
8,
9,
10,
12,
18,
20,
21,
24,
27,
30,
36,
40,
42
],
"firstOver1000": 1002
}
jq
<lang jq>def is_harshad:
def digits: tostring | [explode[] | ([.]| implode) | tonumber]; if . >= 1 then (. % (digits|add)) == 0 else false end ;
- produce a stream of n Harshad numbers
def harshads(n):
# [candidate, count] def _harshads: if .[0]|is_harshad then .[0], ([.[0]+1, .[1]-1]| _harshads) elif .[1] > 0 then [.[0]+1, .[1]] | _harshads else empty end; [1, n] | _harshads ;
- First Harshad greater than n where n >= 0
def harshad_greater_than(n):
# input: next candidate def _harshad: if is_harshad then . else .+1 | _harshad end; (n+1) | _harshad ;
- Task:
[ harshads(20), "...", harshad_greater_than(1000)]</lang>
- Output:
$ jq -n -c -f harshad.jq [1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,36,40,42,"...",1002]
Julia
<lang julia>isharshad(x) = x % sum(digits(x)) == 0 nextharshad(x) = begin while !isharshad(x+1) x += 1 end; return x + 1 end
function harshads(n::Integer) h = Vector{typeof(n)}(n) h[1] = 1 for j in 2:n h[j] = nextharshad(h[j-1]) end return h end
println("First 20 harshad numbers: ", join(harshads(20), ", ")) println("First harshad number after 1001: ", nextharshad(1000))</lang>
- Output:
First 20 harshad numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42 First harshad number after 1001: 1002
Kotlin
<lang scala>// version 1.1
fun sumDigits(n: Int): Int = when {
n <= 0 -> 0
else -> {
var sum = 0
var nn = n
while (nn > 0) {
sum += nn % 10
nn /= 10
}
sum
}
}
fun isHarshad(n: Int): Boolean = (n % sumDigits(n) == 0)
fun main(args: Array<String>) {
println("The first 20 Harshad numbers are:")
var count = 0
var i = 0
while (true) {
if (isHarshad(++i)) {
print("$i ")
if (++count == 20) break
}
}
println("\n\nThe first Harshad number above 1000 is:")
i = 1000
while (true) {
if (isHarshad(++i)) {
println(i)
return
}
}
}</lang>
- Output:
The first 20 Harshad numbers are: 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 The first Harshad number above 1000 is: 1002
LOLCODE
<lang LOLCODE>HAI 1.3
HOW IZ I digsummin YR num
I HAS A digsum ITZ 0
IM IN YR loop
num, O RLY?
YA RLY
digsum R SUM OF digsum AN MOD OF num AN 10
num R QUOSHUNT OF num AN 10
NO WAI, FOUND YR digsum
OIC
IM OUTTA YR loop
IF U SAY SO
I HAS A found ITZ 0
IM IN YR finder UPPIN YR n
I HAS A n ITZ SUM OF n AN 1 I HAS A digsum ITZ I IZ digsummin YR n MKAY
NOT MOD OF n AN digsum, O RLY?
YA RLY
DIFFRINT found AN BIGGR OF found AN 20, O RLY?
YA RLY
VISIBLE n " "!
found R SUM OF found AN 1
OIC
DIFFRINT n AN SMALLR OF n AN 1000, O RLY?
YA RLY, VISIBLE ":)" n, GTFO
OIC
OIC
IM OUTTA YR finder
KTHXBYE</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
Lua
<lang lua>function isHarshad(n)
local s=0
local n_str=tostring(n)
for i=1,#n_str do
s=s+tonumber(n_str:sub(i,i))
end
return n%s==0
end
local count=0 local harshads={} local n=1
while count<20 do
if isHarshad(n) then
count=count+1
table.insert(harshads, n)
end
n=n+1
end
print(table.concat(harshads, " "))
local h=1001 while not isHarshad(h) do
h=h+1
end print(h) </lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
Mathematica / Wolfram Language
<lang mathematica>nextHarshad =
NestWhile[# + 1 &, # + 1, ! Divisible[#, Total@IntegerDigits@#] &] &;
Print@Rest@NestList[nextHarshad, 0, 20]; Print@nextHarshad@1000;</lang>
- Output:
{1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,36,40,42}
1002
MATLAB / Octave
Define a testing function whether n is harshad or not <lang MATLAB>function v = isharshad(n) v = isinteger(n) && ~mod(n,sum(num2str(n)-'0')); end; </lang> Check numbers <lang MATLAB>k=1; n=1; while (k<=20) if isharshad(n) printf('%i ',n); k=k+1; end; n=n+1; end n = 1001; while ~isharshad(n) n=n+1; end; printf('\nFirst harshad number larger than 1000 is %i\n',n);</lang>
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 First harshad number larger than 1000 is 1002
MLite
<lang ocaml>fun sumdigits
(0, n) = n
| (m, n) = sumdigits (m div 10, m rem 10) + n
| n = sumdigits (n div 10, n rem 10)
fun is_harshad n = (n rem (sumdigits n) = 0)
fun next_harshad_after (n, ~1) = if is_harshad n then n else next_harshad_after (n + 1, ~1) | n = next_harshad_after (n + 1, ~1)
fun harshad
(max, _, count > max, accum) = rev accum
| (max, here, count, accum) =
if is_harshad here then
harshad (max, here + 1, count + 1, here :: accum)
else
harshad (max, here + 1, count, accum)
| max = harshad (max, 1, 1, [])
print "first twenty harshad numbers = "; println ` harshad 20; print "first harshad number after 1000 = "; println ` next_harshad_after 1000;</lang>
NetRexx
<lang netrexx>/* NetRexx ------------------------------------------------------------
- 21.01.2014 Walter Pachl translated from ooRexx (from REXX version 1)
- --------------------------------------------------------------------*/
options replace format comments java crossref symbols nobinary
Parse Arg x y . /* get optional arguments: X Y */ If x= Then x=20 /* Not specified? Use default */ If y= Then y=1000 /* " " " " */ n=0 /* Niven count */ nl= /* Niven list. */
Loop j=1 By 1 Until n=x /* let's go Niven number hunting.*/
If j//sumdigs(j)=0 Then Do /* j is a Niven number */
n=n+1 /* bump Niven count */
nl=nl j /* add to list. */
End
End
Say 'first' n 'Niven numbers:'nl
Loop j=y+1 By 1 /* start with first candidate */
If j//sumdigs(j)=0 Then /* j is a Niven number */
Leave
End
Say 'first Niven number >' y 'is:' j Exit
method sumdigs(n) public static returns Rexx
sum=n.left(1) Loop k=2 To n.length() sum=sum+n.substr(k,1) End Return sum</lang>
output same as ooRexx's
Nim
<lang nim>import strutils
proc slice[T](iter: iterator(): T {.closure.}, sl): seq[T] =
var result {.gensym.}: seq[int64] = @[]
var i = 0
for n in iter():
if i > sl.b:
break
if i >= sl.a:
result.add(n)
inc i
result
iterator harshad(): int64 {.closure.} =
for n in 1 .. < int64.high:
var sum = 0
for ch in string($n):
sum += parseInt("" & ch)
if n mod sum == 0:
yield n
echo harshad.slice 0 .. <20
for n in harshad():
if n > 1000: echo n break</lang>
- Output:
@[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42] 1002
Objeck
<lang objeck> class Harshad {
function : Main(args : String[]) ~ Nil {
count := 0;
for(i := 1; count < 20; i += 1;) {
if(i % SumDigits(i) = 0){
"{$i} "->Print();
count += 1;
};
};
for(i := 1001; true; i += 1;) {
if(i % SumDigits(i) = 0){
"... {$i}"->PrintLine();
break;
};
};
}
function : SumDigits(n : Int) ~ Int {
sum := 0;
do {
sum += n % 10;
n /= 10;
} while(n <> 0);
return sum; }
} </lang>
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 ... 1002
Oforth
<lang Oforth>: sumDigits(n) 0 while(n) [ n 10 /mod ->n + ] ;
- isHarshad dup sumDigits mod 0 == ;
1100 seq filter(#isHarshad) dup left(20) println dup filter(#[ 1000 > ]) first println</lang>
- Output:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42] 1002
ooRexx
<lang oorexx>/* REXX ---------------------------------------------------------------
- 21.01.2014 Walter Pachl modi-(simpli-)fied from REXX version 1
- --------------------------------------------------------------------*/
Parse Arg x y . /* get optional arguments: X Y */ If x= Then x=20 /* Not specified? Use default */ If y= Then y=1000 /* " " " " */ n=0 /* Niven count */ nl= /* Niven list. */
Do j=1 Until n=x /* let's go Niven number hunting.*/
If j//sumdigs(j)=0 Then Do /* j is a Niven number */
n=n+1 /* bump Niven count */
nl=nl j /* add to list. */
End
End
Say 'first' n 'Niven numbers:'nl
Do j=y+1 /* start with first candidate */
If j//sumdigs(j)=0 Then /* j is a Niven number */
Leave
End
Say 'first Niven number >' y 'is:' j Exit
sumdigs: Procedure /* compute sum of n's digits */
Parse Arg n sum=left(n,1) Do k=2 To length(n) sum=sum+substr(n,k,1) End Return sum</lang>
- Output:
first 20 Niven numbers: 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 first Niven number > 1000 is: 1002
PARI/GP
<lang parigp>isHarshad(n)=n%sumdigits(n)==0 n=0;k=20;while(k,if(isHarshad(n++),k--;print1(n", "))); n=1000;while(!isHarshad(n++),);print("\n"n)</lang>
- Output:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 1002
Pascal
Optimized for speed, by using the state before in IncSumDigit. <lang pascal>program Niven; const
base = 10;
type
tNum = longword;{Uint64}
const
cntbasedigits = trunc(ln(High(tNum))/ln(base))+1;
type
tSumDigit = record
sdNumber : tNum;
sdDigits : array[0..cntbasedigits-1] of byte;
sdSumDig : byte;
sdIsNiven : boolean;
end;
function InitSumDigit( n : tNum):tSumDigit; var
sd : tSumDigit; qt : tNum; i : integer;
begin
with sd do
begin
sdNumber:= n;
fillchar(sdDigits,SizeOf(sdDigits),#0);
sdSumDig :=0;
sdIsNiven := false;
i := 0;
// calculate Digits und sum them up
while n > 0 do
begin
qt := n div base;
{n mod base}
sdDigits[i] := n-qt*base;
inc(sdSumDig,sdDigits[i]);
n:= qt;
inc(i);
end;
IF sdSumDig >0 then
sdIsNiven := (sdNumber MOD sdSumDig = 0);
end;
InitSumDigit:=sd;
end;
procedure IncSumDigit(var sd:tSumDigit); var
i,d: integer;
begin
i := 0;
with sd do
begin
inc(sdNumber);
repeat
d := sdDigits[i];
inc(d);
inc(sdSumDig);
//base-1 times the repeat is left here
if d < base then
begin
sdDigits[i] := d;
BREAK;
end
else
begin
sdDigits[i] := 0;
dec(sdSumDig,base);
inc(i);
end;
until i > high( sdDigits);
sdIsNiven := (sdNumber MOD sdSumDig) = 0;
end;
end;
var
MySumDig : tSumDigit; ln : tNum; cnt: integer;
begin
MySumDig:=InitSumDigit(0);
cnt := 0;
repeat
IncSumDigit(MySumDig);
IF MySumDig.sdIsNiven then
begin
write(MySumDig.sdNumber,'.');
inc(cnt);
end;
until cnt >= 20;
write('....');
MySumDig:=InitSumDigit(1000);
repeat
IncSumDigit(MySumDig);
until MySumDig.sdIsNiven;
writeln(MySumDig.sdNumber,'.');
// searching for big gaps between two niven-numbers // MySumDig:=InitSumDigit(18879989100-276);
MySumDig:=InitSumDigit(1);
cnt := 0;
ln:= MySumDig.sdNumber;
repeat
IncSumDigit(MySumDig);
if MySumDig.sdIsNiven then
begin
IF cnt < (MySumDig.sdNumber-ln) then
begin
cnt :=(MySumDig.sdNumber-ln);
writeln(ln,' --> ',MySumDig.sdNumber,' d=',cnt);
end;
ln:= MySumDig.sdNumber;
end;
until MySumDig.sdNumber= High(tNum);
{ 689988915 --> 689989050 d=135 879987906 --> 879988050 d=144 989888823 --> 989888973 d=150 2998895823 --> 2998895976 d=153 ~ 24 Cpu-cycles per test i3- 4330 1..2^32-1} end.</lang> output:
1.2.3.4.5.6.7.8.9.10.12.18.20.21.24.27.30.36.40.42.....1002.
Perl
<lang perl>#!/usr/bin/perl use strict ; use warnings ; use List::Util qw ( sum ) ;
sub createHarshads {
my @harshads ;
my $number = 1 ;
do {
if ( $number % sum ( split ( // , $number ) ) == 0 ) {
push @harshads , $number ;
}
$number++ ;
} until ( $harshads[ -1 ] > 1000 ) ;
return @harshads ;
} my @harshadnumbers = createHarshads ; for my $i ( 0..19 ) {
print "$harshadnumbers[ $i ]\n" ;
} print "The first Harshad number greater than 1000 is $harshadnumbers[ -1 ]!\n" ;</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 The first Harshad number greater than 1000 is 1002!
Perl 6
<lang perl6>constant @harshad = grep { $_ %% .comb.sum }, 1 .. *;
say @harshad[^20]; say @harshad.first: * > 1000;</lang>
- Output:
(1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42) 1002
Phix
<lang Phix>integer n = 0 sequence digits={0}
procedure nNiven()
while 1 do
n += 1
for i=length(digits) to 0 by -1 do
if i=0 then
digits = prepend(digits,1)
exit
end if
if digits[i]<9 then
digits[i] += 1
exit
end if
digits[i] = 0
end for
if remainder(n,sum(digits))=0 then exit end if
end while
end procedure
sequence s = {} for i=1 to 20 do
nNiven() s &= n
end for ?s while n<=1000 do
nNiven()
end while ?n</lang>
- Output:
{1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,36,40,42}
1002
Alternative version <lang Phix>function isHarshad(integer n)
return remainder(n,sum(sq_sub(sprint(n),'0')))=0
end function
sequence s = {} integer n = 0 while length(s)<20 do
n += 1
if isHarshad(n) then
s &= n
end if
end while n = 1001 while not isHarshad(n) do n += 1 end while ?s&n</lang>
- Output:
{1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,36,40,42,1002}
PL/I
<lang pli>*process source or(!) xref attributes;
niven: Proc Options(main); /********************************************************************* * 08-06.2013 Walter Pachl translated from Rexx * with a slight improvement: Do j=y+1 By 1; *********************************************************************/ Dcl (ADDR,HBOUND,MOD,SUBSTR,VERIFY) Builtin; Dcl SYSPRINT Print;
Dcl (x,y) dec fixed(8);
x=20;
y=1000;
Begin;
Dcl (n(x),j) Dec Fixed(8);
Dcl ni Bin Fixed(31) Init(0);
Dcl result Char(100) Var Init();
loop:
Do j=1 By 1;
If mod(j,sumdigs(j))=0 Then Do;
ni+=1;
n(ni)=j;
result=result!!' '!!d2c(j);
If ni=x Then Leave loop;
End;
End;
Put Edit('first 20 Niven numbers: ',result)(Skip,a,a);
Do j=y+1 By 1;
If mod(j,sumdigs(j))=0 Then
Leave;
End;
Put Edit('first Niven number > ',d2c(y),' is: ',d2c(j))(Skip,4(a));
End;
sumDigs: proc(z) Returns(Dec Fixed(3)); Dcl z Pic'(8)9'; Dcl d(8) Pic'9' Based(addr(z)); Dcl i Bin Fixed(31); Dcl sd Dec Fixed(3) Init(0); Do i=1 To hbound(d); sd+=d(i); End; Return(sd); End;
d2c: Proc(z) Returns(char(8) Var); Dcl z Pic'(8)z'; Dcl p Bin Fixed(31); p=verify(z,' '); Return(substr(z,p)); End;
End;</lang>
- Output:
first 20 Niven numbers: 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 first Niven number > 1000 is: 1002
PowerShell
In PowerShell, we generally don't wrap every little thing in a function. If you have something simple to do, you just do it. <lang PowerShell>
1..1000 | Where { $_ % ( [int[]][string[]][char[]][string]$_ | Measure -Sum ).Sum -eq 0 } | Select -First 20
1001..2000 | Where { $_ % ( [int[]][string[]][char[]][string]$_ | Measure -Sum ).Sum -eq 0 } | Select -First 1 </lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
But if we do have a need for the code to be reusable, we can do that. <lang PowerShell> function Get-HarshadNumbers
{
<#
.SYNOPSIS
Returns numbers in the Harshad or Niven series.
.DESCRIPTION
Returns all integers in the given range that are evenly divisible by the sum of their digits
in ascending order.
.PARAMETER Minimum
Lower bound of the range to search for Harshad numbers. Defaults to 1.
.PARAMETER Maximum
Upper bound of the range to search for Harshad numbers. Defaults to 2,147,483,647
.PARAMETER Count
Maximum number of Harshad numbers to return.
#>
[cmdletbinding()]
Param (
[int]$Minimum = 1,
[int]$Maximum = [int]::MaxValue,
[int]$Count )
# Skip any non-positive numbers in the specified range
$Minimum = [math]::Max( 1, $Minimum )
# If the adjusted range has any numbers in it...
If ( $Maximum -ge $Minimum )
{
# If a count was specified, build a parameter for the Select statement to kill the pipeline when the count is achieved.
If ( $Count ) { $SelectParam = @{ First = $Count } }
Else { $SelectParam = @{} }
# For each number in the range, test the remainder of it divided it by iteself (converted to a string,
# then a character array, then a string array, then an integer array, then summed).
$Minimum..$Maximum | Where { $_ % ( [int[]][string[]][char[]][string]$_ | Measure -Sum ).Sum -eq 0 } | Select @SelectParam
}
}
</lang> <lang PowerShell> Get-HarshadNumbers -Count 20 Get-HarshadNumbers -Minimum 1001 -Count 1 </lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
Prolog
Works with SWI-Prolog and module lambda.pl written by Ulrich Neumerkel, it can be found there : http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/lambda.pl. <lang Prolog>:- use_module(library(lambda)).
niven :- nb_setval(go, 1),
L = [1 | _], print_niven(L, 1), gen_niven(1, L).
print_niven([X|T], N) :-
when(ground(X),
( ( nb_getval(go, 1)
-> ( N < 20
-> writeln(X),
N1 is N+1,
print_niven(T, N1)
; ( X > 1000
-> writeln(X),
nb_setval(go, 0)
; N1 is N+1,
print_niven(T, N1)))
; true))).
gen_niven(X, [N | T]) :- ( nb_getval(go, 1) -> X1 is X+1, sum_of_digit(X, S), ( X mod S =:= 0 -> N = X, gen_niven(X1, T) ; gen_niven(X1, [N | T])) ; true).
sum_of_digit(N, S) :-
number_chars(N, LC),
maplist(\X^Y^number_chars(Y, [X]), LC, LN),
sum_list(LN, S).
</lang>
- Output:
?- niven. 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 1002 true.
Python
<lang python>>>> import itertools >>> def harshad(): for n in itertools.count(1): if n % sum(int(ch) for ch in str(n)) == 0: yield n
>>> list(itertools.islice(harshad(), 0, 20))
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42]
>>> for n in harshad():
if n > 1000:
print(n)
break
1002
>>> </lang>
Python: Functional
The for loop above could be changed to the following to find the number > 1000; in fact the harshad generator function could become a generator expression creating this more functional version: <lang python>>>> from itertools import count, islice >>> harshad = (n for n in count(1) if n % sum(int(ch) for ch in str(n)) == 0) >>> list(islice(harshad, 0, 20)) [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42] >>> next(x for x in harshad if x > 1000) 1002 >>> </lang>
Racket
<lang scheme>#lang racket
(define (digsum n)
(for/sum ([c (number->string n)]) (string->number [string c])))
(define harshads
(stream-filter (λ (n) (= (modulo n (digsum n)) 0)) (in-naturals 1)))
- First 20 harshad numbers
(displayln (for/list ([i 20]) (stream-ref harshads i)))
- First harshad greater than 1000
(displayln (for/first ([h harshads] #:when(> h 1000)) h))</lang>
- Output:
(1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42) 1002
Different to the Scheme implementation in that it illustrates Racket's native iterators, and let-values with quotient/remainder: <lang racket>#lang racket (require math/number-theory) (define (digital-sum n)
(let inner
((n n) (s 0))
(if (zero? n) s
(let-values ([(q r) (quotient/remainder n 10)])
(inner q (+ s r))))))
(define (harshad-number? n)
(and (>= n 1)
(divides? (digital-sum n) n)))
- find 1st 20 Harshad numbers
(for ((i (in-range 1 (add1 20)))
(h (sequence-filter harshad-number? (in-naturals 1)))) (printf "#~a ~a~%" i h))
- find 1st Harshad number > 1000
(displayln (for/first ((h (sequence-filter harshad-number? (in-naturals 1001)))) h))</lang>
- Output:
#1 1 #2 2 #3 3 #4 4 #5 5 #6 6 #7 7 #8 8 #9 9 #10 10 #11 12 #12 18 #13 20 #14 21 #15 24 #16 27 #17 30 #18 36 #19 40 #20 42 1002
REXX
These REXX examples allow the user to specify how many Niven numbers to list,
as well as find the first Niven number greater than a specified positive integer.
Also, gihugeic integers are supported (essentially no limit).
generic
<lang rexx>/*REXX program finds the first A Niven numbers; it also finds first Niven number > B.*/ parse arg A B . /*obtain optional arguments from the CL*/ if A== | A==',' then A= 20 /*Not specified? Then use the default.*/ if B== | B==',' then B=1000 /* " " " " " " */ numeric digits 1+max(8, length(A), length(B)) /*enable the use of any sized numbers. */
- =0; $= /*set Niven numbers count; Niven list.*/
do j=1 until #==A /*◄───── let's go Niven number hunting.*/
if j//sumDigs(j)==0 then do; #=#+1; $=$ j; end
end /*j*/ /* [↑] bump count; append J ──► list.*/
say 'first' A 'Niven numbers:' $
do t=B+1 until t//sumDigs(t)==0; end /*hunt for a Niven (or Harshad) number.*/
say 'first Niven number >' B " is: " t exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ sumDigs: procedure; parse arg x; s=0; do k=1 for length(x); s=s+substr(x,k,1); end /*k*/</lang> output when using the default inputs:
first 20 Niven numbers: 1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 first Niven number > 1000 is: 1002
idomatic
This REXX version idiomatically uses a isNiven function. <lang rexx>/*REXX program finds the first A Niven numbers; it also finds first Niven number > B.*/ parse arg A B . /*obtain optional arguments from the CL*/ if A== | A==',' then A= 20 /*Not specified? Then use the default.*/ if B== | B==',' then B=1000 /* " " " " " " */ numeric digits 1+max(8, length(A), length(B)) /*enable the use of any sized numbers. */
- =0; $= /*set Niven numbers count; Niven list.*/
do j=1 until #==A /*◄───── let's go Niven number hunting.*/
if isNiven(j) then do; #=#+1; $=$ j; end
end /*j*/ /* [↑] bump count; append J ──► list.*/
say 'first' A 'Niven numbers:' $
do t=B+1 until isNiven(t); end /*hunt for a Niven (or Harshad) number.*/
say 'first Niven number >' B " is: " t exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ isNiven: procedure; parse arg x; s=0; do k=1 for length(x); s=s+substr(x,k,1); end /*k*/
return x//s==0</lang>
output is identical to the 1st REXX version.
esoteric
This REXX version optimizes the isNiven function by using parse statements instead of the substr BIF,
yielding a faster algorithm.
<lang rexx>/*REXX program finds the first A Niven numbers; it also finds first Niven number > B.*/
parse arg A B . /*obtain optional arguments from the CL*/
if A== | A==',' then A= 20 /*Not specified? Then use the default.*/
if B== | B==',' then B=1000 /* " " " " " " */
numeric digits 1+max(8, length(A), length(B)) /*enable the use of any sized numbers. */
- =0; $= /*set Niven numbers count; Niven list.*/
do j=1 until #==A /*◄───── let's go Niven number hunting.*/
if isNiven(j) then do; #=#+1; $=$ j; end
end /*j*/ /* [↑] bump count; append J ──► list.*/
say 'first' A 'Niven numbers:' $
do t=B+1 until isNiven(t); end /*hunt for a Niven (or Harshad) number.*/
say 'first Niven number >' B " is: " t exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ isNiven: procedure; parse arg x 1 sum 2 q /*use the first decimal digit for SUM.*/
do while q\==; parse var q _ 2 q; sum=sum+_; end /*k*/
/* ↑ */
return x//sum==0 /* └───◄ is destructively parsed. */</lang>
output is identical to the 1st REXX version.
array of numbers
This REXX version builds an array of numbers instead of a list (building an array is much faster than building a list, especially if the list is very long).
In addition, if the A number is negative, the numbers in the array aren't displayed, but the last number in the array is displayed. <lang rexx>/*REXX program finds the first A Niven numbers; it also finds first Niven number > B.*/ parse arg A B . /*obtain optional arguments from the CL*/ if A== | A==',' then A= 20 /*Not specified? Then use the default.*/ if B== | B==',' then B=1000 /* " " " " " " */ tell= A>0; A=abs(A) /*flag for showing a Niven numbers list*/ A=abs(a) numeric digits 1+max(8, length(A), length(B)) /*enable the use of any sized numbers. */
- =0; $= /*set Niven numbers count; Niven list.*/
do j=1 until #==A /*◄───── let's go Niven number hunting.*/
if isNiven(j) then do; #=#+1; !.#=j; end
end /*j*/ /* [↑] bump count; append J ──► list.*/
w=length(!.w) /*W: is the width of largest Niven #.*/ if tell then do
say 'first' A 'Niven numbers:'; do k=1 for #; say right(!.k, w); end /*k*/
end
else say 'last of the' A 'Niven numbers: ' !.#
say
do t=B+1 until isNiven(t); end /*hunt for a Niven (or Harshad) number.*/
say 'first Niven number >' B " is: " t exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ isNiven: procedure; parse arg x 1 sum 2 q /*use the first decimal digit for SUM.*/
do while q\==; parse var q _ 2 q; sum=sum+_; end /*k*/
/* ↑ */
return x//sum==0 /* └───◄ is destructively parsed. */</lang>
output when the input used is: -1000000 66777888
last of the 1000000 Niven numbers: 12150510 first Niven number > 66777888 is: 66777900
Ring
<lang ring> i = 1 count = 0 while true
sum = 0
if niven(i) = 1
if count < 20 see "" + i + " is a Niven number" + nl count +=1 ok
if i > 1000 see "" + i + " is a Niven number" exit ok ok
i + =1
end
func niven nr
nrString = string(nr)
for j = 1 to len(nrString)
sum = sum + number(nrString[j])
next
niv = ((nr % sum) = 0)
return niv
</lang> Output:
1 is a Niven number 2 is a Niven number 3 is a Niven number 4 is a Niven number 5 is a Niven number 6 is a Niven number 7 is a Niven number 8 is a Niven number 9 is a Niven number 10 is a Niven number 12 is a Niven number 18 is a Niven number 20 is a Niven number 21 is a Niven number 24 is a Niven number 27 is a Niven number 30 is a Niven number 36 is a Niven number 40 is a Niven number 42 is a Niven number 1002 is a Niven number
Ruby
Ruby 2.4 gave Integers a digits method, and Arrays a sum method. <lang Ruby>harshad = 1.step.lazy.select { |n| n % n.digits.sum == 0 }
p harshad.first(20) p harshad.find { |n| n > 1000 }</lang>
- Output:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42] 1002
Run BASIC
<lang runbasic>while count < 20
h = h + 1 if neven(h) = 0 then count = count + 1 print count;": ";h end if
wend
h = 1000 while 1 = 1
h = h + 1 if neven(h) = 0 then print h exit while end if
wend
function neven(h) h$ = str$(h) for i = 1 to len(h$)
d = d + val(mid$(h$,i,1))
next i neven = h mod d end function</lang>
- Output:
1: 1 2: 2 3: 3 4: 4 5: 5 6: 6 7: 7 8: 8 9: 9 10: 10 11: 12 12: 18 13: 20 14: 21 15: 24 16: 27 17: 30 18: 36 19: 40 20: 42 1002
Scala
<lang Scala>object Harshad extends App {
val harshads = Stream from 1 filter (i => i % i.toString.map(_.asDigit).sum == 0)
println(harshads.take(20).toList) println(harshads.filter(_ > 1000).head)
}</lang>
- Output:
List(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42) 1002
Scheme
<lang scheme>#!/usr/local/bin/gosh
- Show the first 20 niven numbers and the
- first one greater than 1000.
(define (main args)
(display (iota-filtered 20 1 niven?))(newline) (display (iota-filtered 1 1001 niven?))(newline))
- Return a list of length n
- for numbers starting at start
- that satisfy the predicate fn.
(define (iota-filtered n start fn)
(let loop ((num start)(lst (list)))
(if (= (length lst) n)
lst
(loop (+ 1 num) (if (fn num) (append lst (list num)) lst)))))
- Is a number a niven number?
(define (niven? n)
(and (> n 0) (= 0 (remainder n (sum-of-digits n)))))
- Get the sum of the digits of a number.
(define (sum-of-digits n)
(apply + (map string->number (map string (string->list (number->string n))))))
</lang>
- Output:
(1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42) (1002)
Seed7
<lang seed7>$ include "seed7_05.s7i";
const func integer: sumOfDigits (in var integer: num) is func
result
var integer: sum is 0;
begin
repeat
sum +:= num rem 10;
num := num div 10;
until num = 0;
end func;
const func integer: nextHarshadNum (inout integer: num) is func
result
var integer: harshadNumber is 0;
begin
while num mod sumOfDigits(num) <> 0 do
incr(num);
end while;
harshadNumber := num;
end func;
const proc: main is func
local
var integer: current is 1;
var integer: count is 0;
begin
for count range 1 to 20 do
write(nextHarshadNum(current) <& " ");
incr(current);
end for;
current := 1001;
writeln(" ... " <& nextHarshadNum(current));
end func;</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 ... 1002
Sidef
<lang ruby>func harshad() {
var n = 0;
{
++n while !(n %% n.digits.sum);
n;
}
}
var iter = harshad(); say 20.of { iter.run };
var n; do {
n = iter.run
} while (n <= 1000);
say n;</lang>
- Output:
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42] 1002
Sinclair ZX81 BASIC
Works with 1k of RAM. FAST isn't all that fast.
<lang basic> 10 FAST
20 LET N=0 30 LET H=0 40 LET N=N+1 50 LET N$=STR$ N 60 LET SD=0 70 FOR I=1 TO LEN N$ 80 LET SD=SD+VAL N$(I) 90 NEXT I
100 IF N/SD<>INT (N/SD) THEN GOTO 40 110 LET H=H+1 120 IF H<=20 OR N>1000 THEN PRINT N 130 IF N>1000 THEN GOTO 150 140 GOTO 40 150 SLOW</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
Tcl
<lang tcl># Determine if the given number is a member of the class of Harshad numbers proc isHarshad {n} {
if {$n < 1} {return false}
set sum [tcl::mathop::+ {*}[split $n ""]]
return [expr {$n%$sum == 0}]
}
- Get the first 20 numbers that satisfy the condition
for {set n 1; set harshads {}} {[llength $harshads] < 20} {incr n} {
if {[isHarshad $n]} {
lappend harshads $n
}
} puts [format "First twenty Harshads: %s" [join $harshads ", "]]
- Get the first value greater than 1000 that satisfies the condition
for {set n 1000} {![isHarshad [incr n]]} {} {} puts "First Harshad > 1000 = $n"</lang>
- Output:
First twenty Harshads: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42 First Harshad > 1000 = 1002
uBasic/4tH
<lang>C=0
For I = 1 Step 1 Until C = 20 ' First 20 Harshad numbers
If FUNC(_FNHarshad(I)) Then Print I;" "; : C = C + 1
Next
For I = 1001 Step 1 ' First Harshad greater than 1000
If FUNC(_FNHarshad(I)) Then Print I;" " : Break
Next
End
_FNHarshad Param(1)
Local(2)
c@ = a@
b@ = 0
Do While (c@ > 0)
b@ = b@ + (c@ % 10)
c@ = c@ / 10
Loop
Return ((a@ % b@) = 0)</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002 0 OK, 0:185
VBScript
<lang vb>n = 0 m = 1 first20 = "" after1k = ""
Do If IsHarshad(m) And n <= 20 Then first20 = first20 & m & ", " n = n + 1 m = m + 1 ElseIf IsHarshad(m) And m > 1000 Then after1k = m Exit Do Else m = m + 1 End If Loop
WScript.StdOut.Write "First twenty Harshad numbers are: " WScript.StdOut.WriteLine WScript.StdOut.Write first20 WScript.StdOut.WriteLine WScript.StdOut.Write "The first Harshad number after 1000 is: " WScript.StdOut.WriteLine WScript.StdOut.Write after1k
Function IsHarshad(s) IsHarshad = False sum = 0 For i = 1 To Len(s) sum = sum + CInt(Mid(s,i,1)) Next If s Mod sum = 0 Then IsHarshad = True End If End Function</lang>
- Output:
First twenty Harshad numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, The first Harshad number after 1000 is: 1002
Visual FoxPro
<lang vfp> LOCAL lnCount As Integer, k As Integer CLEAR lnCount = 0 k = 0
- !* First 20 numbers
? "First 20 numbers:" DO WHILE lnCount < 20
k = k + 1 IF Harshad(k)
lnCount = lnCount + 1 ? lnCount, k
ENDIF
ENDDO
- !* First such number > 1000
k = 1001 DO WHILE NOT Harshad(k)
k = k + 1
ENDDO ? "First such number > 1000", k
FUNCTION Harshad(n As Integer) As Boolean LOCAL cn As String, d As Integer, i As Integer cn = TRANSFORM(n) d = 0 FOR i = 1 TO LEN(cn)
d = d + VAL(SUBSTR(cn, i, 1))
ENDFOR RETURN n % d = 0 ENDFUNC </lang>
- Output:
First 20 numbers:
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 12
12 18
13 20
14 21
15 24
16 27
17 30
18 36
19 40
20 42
First such number > 1000: 1002
Whitespace
<lang Whitespace> </lang>
</lang>
This solution was generated from the pseudo-Assembly below.
A live run is available for the inquiring skeptic.
<lang asm>push 0 ; Harshad numbers found
push 0 ; counter
0: ; Increment the counter, call "digsum", branch on the modulus.
push 1 add dup dup
push 0 call 1 mod
jz 2
jump 0
1: ; [n 0] => [digsum(n)]
copy 1
push 10 mod add swap
push 10 div swap
push 0 copy 2 sub
jn 1
slide 1 ret
2: ; Should we print this Harshad number?
push 1000 copy 1 sub jn 3 ; We're done if it's greater than 1000. swap push 1 add swap ; Increment how many we've found so far. push 20 copy 2 sub jn 0 ; If we've already got 20, go back to the top. dup onum push 32 ochr ; Otherwise, print it and a space. jump 0 ; And /then/ go back to the top.
3: ; Print the > 1000 Harshad number on its own line and exit clean.
push 10 ochr onum pop push 10 ochr exit</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
XPL0
<lang XPL0>include c:\cxpl\codes; \intrinsic 'code' declarations int H, C, N, S; \Harshad number, Counter, Number, Sum [H:= 1; C:= 0; loop [N:= H; S:= 0; \sum digits
repeat N:= N/10;
S:= S + rem(0);
until N = 0;
if rem(H/S) = 0 then \Harshad no.is evenly divisible by sum of digits
[if C < 20 then [IntOut(0, H); ChOut(0, ^ ); C:= C+1];
if H > 1000 then [IntOut(0, H); CrLf(0); quit];
];
H:= H+1;
];
]</lang>
- Output:
1 2 3 4 5 6 7 8 9 10 12 18 20 21 24 27 30 36 40 42 1002
zkl
<lang zkl>fcn harshad(n){ 0==n%(n.split().sum(0)) } [1..].tweak(fcn(n){ if(not harshad(n)) return(Void.Skip); n })
.walk(20).println();
[1..].filter(20,harshad).println(); [1001..].filter1(harshad).println();</lang> Walkers are zkl iterators. [a..b] is a Walker from a to b. Walkers can be tweaked to transform the sequence they are walking. In this case, ignore non Harshad numbers. Then tell the walker to get 20 items from that [modified] sequence.
In this case, filters are the better solution.
- Output:
L(1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,36,40,42) L(1,2,3,4,5,6,7,8,9,10,12,18,20,21,24,27,30,36,40,42) L(1002)
ZX Spectrum Basic
<lang zxbasic>10 LET k=0: LET n=0 20 IF k=20 THEN GO TO 60 30 LET n=n+1: GO SUB 1000 40 IF isHarshad THEN PRINT n;" ";: LET k=k+1 50 GO TO 20 60 LET n=1001 70 GO SUB 1000: IF NOT isHarshad THEN LET n=n+1: GO TO 70 80 PRINT '"First Harshad number larger than 1000 is ";n 90 STOP 1000 REM is Harshad? 1010 LET s=0: LET n$=STR$ n 1020 FOR i=1 TO LEN n$ 1030 LET s=s+VAL n$(i) 1040 NEXT i 1050 LET isHarshad=NOT FN m(n,s) 1060 RETURN 1100 DEF FN m(a,b)=a-INT (a/b)*b</lang>
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