# Inconsummate numbers in base 10

Inconsummate numbers in base 10 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.

A consummate number is a non-negative integer that can be formed by some integer N divided by the the digital sum of N.

For instance

47 is a consummate number.

```   846 / (8 + 4 + 6) = 47
```

On the other hand, there are integers that can not be formed by a ratio of any integer over its digital sum. These numbers are known as inconsummate numbers.

62 is an inconsummate number. There is no integer ratio of an integer to its digital sum that will result in 62.

The base that a number is expressed in will affect whether it is inconsummate or not. This task will be restricted to base 10.

• Write a routine to find inconsummate numbers in base 10;
• Use that routine to find and display the first fifty inconsummate numbers.

Stretch
• Use that routine to find and display the one thousandth inconsummate number.

## 11l

Translation of: Python
```F digitalsum(num)
‘ Return sum of digits of a number in base 10 ’
R sum(String(num).map(d -> Int(d)))

F generate_inconsummate(max_wanted)
‘ generate the series of inconsummate numbers up to max_wanted ’
V minimum_digitsums = (1..14).map(i -> (Int64(10) ^ i, Int((Int64(10) ^ i - 1) / (9 * i))))
V limit = 20 * Int(min(minimum_digitsums.filter(p -> p[1] > @max_wanted).map(p -> p[0])))
V arr = [1] [+] [0] * (limit - 1)

L(dividend) 1 .< limit
V (quo, rem) = divmod(dividend, digitalsum(dividend))
I rem == 0 & quo < limit
arr[quo] = 1
[Int] r
L(flag) arr
I flag == 0
r.append(L.index)
R r

L(n) generate_inconsummate(100000)
V i = L.index
I i < 50
print(f:‘{n:6}’, end' I (i + 1) % 10 == 0 {"\n"} E ‘’)
E I i == 999
print("\nThousandth inconsummate number: "n)
E I i == 9999
print("\nTen-thousanth inconsummate number: "n)
E I i == 99999
print("\nHundred-thousanth inconsummate number: "n)
L.break```
Output:
```    62    63    65    75    84    95   161   173   195   216
261   266   272   276   326   371   372   377   381   383
386   387   395   411   416   422   426   431   432   438
441   443   461   466   471   476   482   483   486   488
491   492   493   494   497   498   516   521   522   527

Thousandth inconsummate number: 6996

Ten-thousanth inconsummate number: 59853

Hundred-thousanth inconsummate number: 536081
```

## Action!

Translation of: PL/M

and based on the Algol 68 sample. As with the PL/M sample, the limit of 16 bit arithmetic means only the basic task can be handled.

```;;; find some incomsummate numbers: integers that cannot be expressed as
;;;      an integer divided by the sum of its digits

PROC Main()

CARD i, tn, hn, th, tt, sumD, d, n, dRatio, count, maxSum, v, maxNumber

; table of numbers that can be formed by n / digit sum n
DEFINE MAX_C = "999"
BYTE ARRAY consummate(MAX_C+1)
FOR i = 0 TO MAX_C DO
consummate( i ) = 0
OD

; calculate the maximum number we must consider
v = MAX_C / 10;
maxSum = 9;
WHILE v > 0 DO
maxSum ==+ 9
v      ==/ 10
OD
maxNumber = maxSum * MAX_C

; construct the digit sums of the numbers up to maxNumber
; and find the consumate numbers, we start the loop from 10 to avoid
; having to deal with 0-9
consummate( 1 ) = 1
tn = 1 hn = 0 th = 0 tt = 0
FOR n = 10 TO maxNumber STEP 10 DO
sumD = tt + th + hn + tn
FOR d = n TO n + 9 DO
IF d MOD sumD = 0 THEN
; d is comsummate
dRatio = d / sumD
IF dRatio <= MAX_C THEN
consummate( dRatio ) = 1
FI
FI
sumD ==+ 1
OD
tn ==+ 1
IF tn > 9 THEN
tn = 0
hn ==+ 1
IF hn > 9 THEN
hn = 0
th ==+ 1
IF th > 9 THEN
th = 0
tt ==+ 1
FI
FI
FI
OD

count = 0
PrintE( "The first 50 inconsummate numbers:" )
i = 0
WHILE i < MAX_C AND count < 50 DO
i ==+ 1
IF consummate( i ) = 0 THEN
count ==+ 1
Put(' )
IF i <   10 THEN Put(' ) FI
IF i <  100 THEN Put(' ) FI
IF i < 1000 THEN Put(' ) FI
PrintC( i )
IF count MOD 10 = 0 THEN PutE() FI
FI
OD

RETURN```
Output:
```The first 50 inconsummate numbers:
62   63   65   75   84   95  161  173  195  216
261  266  272  276  326  371  372  377  381  383
386  387  395  411  416  422  426  431  432  438
441  443  461  466  471  476  482  483  486  488
491  492  493  494  497  498  516  521  522  527
```

## ALGOL 68

```BEGIN # find some incomsummate numbers: integers that cannot be expressed as #
#      an integer divided by the sum of its digits                     #
# table of numbers that can be formed by n / digit sum n                 #
[ 0 : 999 999  ]BOOL consummate;
FOR i FROM LWB consummate TO UPB consummate DO
consummate[ i ] := FALSE
OD;
# calculate the maximum number we must consider to find consummate       #
# numbers up to UPB consummate - which is 9 * the number of digits in    #
# UPB consummate                                                         #
INT max sum := 9;
INT v       := UPB consummate;
WHILE ( v OVERAB 10 ) > 0 DO max sum +:= 9 OD;
INT max number = UPB consummate * max sum;
# construct the digit sums of the numbers up to max number               #
# and find the consumate numbers, we start the loop from 10 to avoid     #
# having to deal with 0-9                                                #
consummate[ 1 ] := TRUE;
INT tn := 1, hn := 0, th := 0, tt := 0, ht := 0, mi := 0, tm := 0;
FOR n FROM 10 BY 10 TO max number DO
INT sumd := tm + mi + ht + tt + th + hn + tn;
FOR d FROM n TO n + 9 DO
IF d MOD sumd = 0 THEN
# d is comsummate                                            #
IF INT d ratio = d OVER sumd;
d ratio <= UPB consummate
THEN
consummate[ d ratio ] := TRUE
FI
FI;
sumd +:= 1
OD;
IF ( tn +:= 1 ) > 9 THEN
tn := 0;
IF ( hn +:= 1 ) > 9 THEN
hn  := 0;
IF ( th +:= 1 ) > 9 THEN
th  := 0;
IF ( tt +:= 1 ) > 9 THEN
tt  := 0;
IF ( ht +:= 1 ) > 9 THEN
ht  := 0;
IF ( mi +:= 1 ) > 9 THEN
mi  := 0;
tm +:= 1
FI
FI
FI
FI
FI
FI
OD;
INT count := 0;
print( ( "The first 50 inconsummate numbers:", newline ) );
FOR i TO UPB consummate WHILE count < 100 000 DO
IF NOT consummate[ i ] THEN
IF ( count +:= 1 ) < 51 THEN
print( ( whole( i, -6 ) ) );
IF count MOD 10 = 0 THEN print( ( newline ) ) FI
ELIF count = 1 000 OR count = 10 000 OR count = 100 000 THEN
print( ( "Inconsummate number ", whole( count, -6 )
, ": ", whole( i, -8 ), newline
)
)
FI
FI
OD
END```
Output:
```The first 50 inconsummate numbers:
62    63    65    75    84    95   161   173   195   216
261   266   272   276   326   371   372   377   381   383
386   387   395   411   416   422   426   431   432   438
441   443   461   466   471   476   482   483   486   488
491   492   493   494   497   498   516   521   522   527
Inconsummate number   1000:     6996
Inconsummate number  10000:    59853
Inconsummate number 100000:   536081
```

## APL

```task←{
gen ← ⍳~(⊢(/⍨)⌊=⊢)∘(⊢÷(+/⍎¨∘⍕)¨)∘⍳∘(⊢×9×≢∘⍕)
incons ← gen 9999
⎕←'The first 50 inconsummate numbers:'
⎕←5 10⍴50↑incons
⎕←'The 1000th inconsummate number:',incons[1000]
}
```
Output:
```The first 50 inconsummate numbers:
62  63  65  75  84  95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527
The 1000th inconsummate number: 6996```

## AppleScript

```on inconsummateNumbers(numberRequired)
set limit to numberRequired * 1000 -- Seems to be enough.
script o
property lst : makeList(limit, true)
end script

repeat with n from 11 to limit -- 1 to 10 obviously can't be inconsummate.
set digitSum to n mod 10
set temp to n div 10
repeat while (temp > 9)
set digitSum to digitSum + temp mod 10
set temp to temp div 10
end repeat
set digitSum to digitSum + temp

tell (n div digitSum) to if (it = n / digitSum) then set o's lst's item it to false
end repeat

set i to 0
repeat with n from 11 to limit
if (o's lst's item n) then
set i to i + 1
set o's lst's item i to n
if (i = numberRequired) then return o's lst's items 1 thru i
end if
end repeat

return {}
end inconsummateNumbers

on makeList(limit, filler)
if (limit < 1) then return {}
script o
property lst : {filler}
end script

set counter to 1
repeat until (counter + counter > limit)
set o's lst to o's lst & o's lst
set counter to counter + counter
end repeat
if (counter < limit) then set o's lst to o's lst & o's lst's items 1 thru (limit - counter)
return o's lst
end makeList

on join(lst, delim)
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to delim
set txt to lst as text
set AppleScript's text item delimiters to astid
return txt
end join

script o
property inconsummates : inconsummateNumbers(100000)
end script
set output to {"First 50 inconsummate numbers:"}
set row to {}
repeat with i from 1 to 50
set row's end to text -5 thru -1 of ("   " & o's inconsummates's item i)
if (i mod 10 = 0) then
set output's end to join(row, "")
set row to {}
end if
end repeat
set output's end to linefeed & "1,000th inconsummate number: " & o's inconsummates's item 1000
set output's end to "10,000th inconsummate number: " & o's inconsummates's item 10000
set output's end to "100,000th inconsummate number: " & o's inconsummates's end
return join(output, linefeed)

```
Output:
```"First 50 inconsummate numbers:
62   63   65   75   84   95  161  173  195  216
261  266  272  276  326  371  372  377  381  383
386  387  395  411  416  422  426  431  432  438
441  443  461  466  471  476  482  483  486  488
491  492  493  494  497  498  516  521  522  527

1,000th inconsummate number: 6996
10,000th inconsummate number: 59853
100,000th inconsummate number: 536081"
```

## BASIC

```10 DEFINT A-Z
20 M=999
30 DIM C(M)
40 Z=M*9*(LEN(STR\$(M))-1)
50 FOR I=10 TO Z
60 J=I:S=0
70 S=S+J MOD 10:J=J\10:IF J GOTO 70
80 IF I MOD S=0 THEN J=I\S:IF J<=M THEN C(J)=-1
90 NEXT
100 J=0
110 FOR I=10 TO M
120 IF J=50 THEN END
130 IF NOT C(I) THEN J=J+1:PRINT I,
140 NEXT
```
Output:
``` 62            63            65            75            84
95            161           173           195           216
261           266           272           276           326
371           372           377           381           383
386           387           395           411           416
422           426           431           432           438
441           443           461           466           471
476           482           483           486           488
491           492           493           494           497
498           516           521           522           527```

## Delphi

Works with: Delphi version 6.0

```function IsCosumate(N: integer): boolean;
{Test is number is consumate = V / SumDigits(V) = N}
var I, V: integer;
begin
Result:=True;
for I:=1 to High(integer) do
begin
{Test values for V that are multiples of N}
V:= I * N;
{Satisfy - V / SumDigits(V) = N? then exit}
if V = (N * SumDigits(V)) then exit;
{This value derrive by trial and error}
if V>=29000000 then break;
end;
Result:=False;
end;

procedure InconsummateNumbers(Memo: TMemo);
var S: string;
var N,I: integer;
begin
{Display first 50 inconsummate numbers}
N:=1; S:='';
for I:=1 to 50 do
begin
while IsCosumate(N) do Inc(N);
S:=S+Format('%5D',[N]);
if (I mod 10)=0 then S:=S+CRLF_Char;
Inc(N);
end;
{Then display 1,000th, 10,000th and 100,000th}
N:=1;
for I:=1 to 100000 do
begin
while IsCosumate(N) do Inc(N);
if I=1000 then Memo.Lines.Add(  Format('  1,000th = %7.0n',[N+0.0]));
if I=10000 then Memo.Lines.Add( Format(' 10,000th = %7.0n',[N+0.0]));
if I=100000 then Memo.Lines.Add(Format('100,000th = %7.0n',[N+0.0]));
Inc(N);
end;
end;
```
Output:
```   62   63   65   75   84   95  161  173  195  216
261  266  272  276  326  371  372  377  381  383
386  387  395  411  416  422  426  431  432  438
441  443  461  466  471  476  482  483  486  488
491  492  493  494  497  498  516  521  522  527

1,000th =   6,996
10,000th =  59,853
100,000th = 536,081

Elapsed Time: 3.564 Sec.

```

## FreeBASIC

Translation of: Action!
```Dim As Boolean consummate(0 To 999999)
For j As Integer = Lbound(consummate) To Ubound(consummate)
consummate(j) = False
Next j
consummate(1) = True

Dim As Integer maxSum = 9
Dim As Integer ub = Ubound(consummate)
While ub > 10
maxSum += 9
ub /= 10
Wend

Dim As Integer maxNumber = Ubound(consummate) * maxSum

Dim As Integer tn = 1, hn = 0, th = 0, tt = 0, ht = 0, mi = 0, tm = 0
Dim As Integer sumD, n, d, dRatio
For n = 10 To maxNumber Step 10
sumD = tm + mi + ht + tt + th + hn + tn
For d = n To n + 9
If d Mod sumD = 0 Then    ' d is comsummate
dRatio = d / sumD
If dRatio <= Ubound(consummate) Then consummate(dRatio) = True
End If
sumD += 1
Next d
tn += 1
If tn > 9 Then
tn = 0
hn += 1
If hn > 9 Then
hn  = 0
th += 1
If th > 9 Then
th  = 0
tt += 1
If tt > 9 Then
tt  = 0
ht += 1
If ht > 9 Then
ht  = 0
mi += 1
If mi > 9 Then
mi  = 0
tm += 1
End If
End If
End If
End If
End If
End If
Next n

Dim As Integer i = 0, count = 0
Print "The first 50 inconsummate numbers:"
While i < Ubound(consummate) And count < 100000
i += 1
If Not consummate(i) Then
count += 1
If count < 51 Then
Print Using "######"; i;
If count Mod 10 = 0 Then Print
Elseif count = 1000 Or count = 10000 Or count = 100000 Then
Print Using !"\nInconsummate number ######: ######"; count; i;
End If
End If
Wend
Sleep
```
Output:
```The first 50 inconsummate numbers:
62    63    65    75    84    95   161   173   195   216
261   266   272   276   326   371   372   377   381   383
386   387   395   411   416   422   426   431   432   438
441   443   461   466   471   476   482   483   486   488
491   492   493   494   497   498   516   521   522   527

Inconsummate number   1000:   6996
Inconsummate number  10000:  59853
Inconsummate number 100000: 536081```

## J

With:

```dsum=:  [: +/"(1) 10&#.inv NB. digital sum
incons=: (-. (% dsum))&(1+i.) (90* 10 >.&.^. ]) NB. exclude many digital sum ratios
```

We get:

```   5 10\$incons 1000
62  63  65  75  84  95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527
999 { incons 10000
6996
9999 { incons 100000
59853
99999 { incons 1000000
536081
```

## jq

Works with jq, the C implementation of jq

```def digitSum:
def add(s): reduce s as \$_ (0; .+\$_);
add(tostring | explode[] | . - 48);

# Maximum ratio for 6 digit numbers is 100,000
def cons:
reduce range(1; 1000000) as \$i ([];
(\$i | digitSum) as \$ds
| (\$i/\$ds) as \$ids
| if \$ids|floor == \$ids then .[\$ids] = true else . end);

def lpad(\$len): tostring | (\$len - length) as \$l | (" " * \$l)[:\$l] + .;

cons as \$cons
| reduce range(1; \$cons|length) as \$i ([];
if (\$cons[\$i]|not) then . + [\$i] else . end)
| "First \(\$n) inconsummate numbers in base 10:",
(.[0:50] | _nwise(10) | map(lpad(3)) | join(" ")),
"\nThe \(\$nth)th:",
.[\$nth - 1];

Output:
```First 50 inconsummate numbers in base 10:
62  63  65  75  84  95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527

The 1000th:
6996
```

## Nim

```import std/[math, strformat, sugar]

const
D = 4                 # Sufficient to find the one thousandth first, as required.
N = 10^D              # Size of sieve.
M = 9 * (D + 1) * N   # Maximum value for the number to divide.

func digitalSum(n: Natural): int =
## Return the digital sum of "n".
var n = n
while n != 0:
inc result, n mod 10
n = n div 10

# Fill a sieve to find consummate numbers.
var isConsummate: array[1..N, bool]
for n in 1..M:
let ds = n.digitalSum
if n mod ds == 0:
let q = n div ds
if q <= N:
isConsummate[q] = true

# Build list of inconsummate numbers.
let inconsummateNums = collect:
for n, c in isConsummate:
if not c: n

echo "First 50 inconsummate numbers in base 10:"
for i in 0..49:
stdout.write &"{inconsummateNums[i]:3}"
stdout.write if i mod 10 == 9: '\n' else: ' '

echo &"\nOne thousandth is {inconsummateNums[999]}."
```
Output:
```First 50 inconsummate numbers in base 10:
62  63  65  75  84  95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527

One thousandth is 6996.
```

## Pascal

### Free Pascal

```program Inconsummate;
{\$IFDEF FPC}
{\$MODE DELPHI}{\$OPTIMIZATION ON,ALL}{\$CODEALIGN proc=8,loop=1}
{\$ENDIF}
uses
sysutils;
const
base = 10;
DgtSumLmt = base*base*base*base*base;
type
tDgtSum = array[0..DgtSumLmt-1] of byte;
var
DgtSUm : tDgtSum;
max: Uint64;
procedure Init(var ds:tDgtSum);
var
i,l,k0,k1: NativeUint;
Begin
For i := 0 to base-1 do
ds[i] := i;
k0 := base;
repeat
k1 := k0-1;
For i := 1 to base-1 do
For l := 0 to k1 do
begin
ds[k0] := ds[l]+i;
inc(k0);
end;
until k0 >= High(ds);
end;

function GetSumOfDecDigits(n:Uint64):NativeUint;
var
r,d: NativeUint;
begin
result := 0;
repeat
r := n DIv DgtSumLmt;
d := n-r* DgtSumLmt;
result +=DgtSUm[d];
n := r;
until r = 0;
end;

function OneTest(n:Nativeint):Boolean;
var
i,d : NativeInt;
begin
result := true;
d := n;
For i := 1 TO 121 DO
begin
IF GetSumOfDecDigits(n)= i then
Begin
if i > max then
max := i;
Exit(false);
end;
n +=d;
end;
end;

var
d,
cnt,lmt: Uint64;
begin
Init(DgtSUm);
cnt := 0;
For d := 1 to 527 do//5375540350 do
begin
if OneTest(d) then
begin
inc(cnt);
write(d:5);
if cnt mod 10 = 0 then writeln;
end;
end;
writeln;
writeln('Count      Number(count) Maxfactor needed');
cnt := 0;
max := 0;
lmt := 10;
For d := 1 to 50332353 do // 5260629551 do
begin
if OneTest(d) then
begin
inc(cnt);
if cnt = lmt then
begin
writeln(cnt:10,d:12,max:5);
lmt *=10;
end;
end;
end;
writeln(cnt);
end.
```
@TIO.RUN:
```   62   63   65   75   84   95  161  173  195  216
261  266  272  276  326  371  372  377  381  383
386  387  395  411  416  422  426  431  432  438
441  443  461  466  471  476  482  483  486  488
491  492  493  494  497  498  516  521  522  527

Count      Number(count) Maxfactor needed
10         216   27
100         936   36
1000        6996   36
10000       59853   54
100000      536081   63
1000000     5073249   69
10000000    50332353   81
10000000
Real time: 23.898 s

@Home AMD 5600G 4.4 Ghz:
Count      Number(count) Maxfactor needed
10         216   27
100         936   36
1000        6996   36
10000       59853   54
100000      536081   63
1000000     5073249   69
10000000    50332353   81 //real	0m6,395s
100000000   517554035   87
1000000000  5260629551   96
1000000000

real  15m54,915s
```

## Perl

```use v5.36;
use List::Util <sum max>;

sub table (\$c, @V) { my \$t = \$c * (my \$w = 2 + length max @V); ( sprintf( ('%'.\$w.'d')x@V, @V) ) =~ s/.{1,\$t}\K/\n/gr }

my \$upto = 1e6;
my @cons = (0) x \$upto;
for my \$i (1..\$upto) {
my \$r = \$i / sum split '', \$i;
\$cons[\$r] = 1 if \$r eq int \$r;
}
my @incons = grep { \$cons[\$_]==0 } 1..\$#cons;

say 'First fifty inconsummate numbers (in base 10):';
say table 10, @incons[0..49];
say "One thousandth: " . \$incons[999];
```
Output:
```First fifty inconsummate numbers (in base 10):
62   63   65   75   84   95  161  173  195  216
261  266  272  276  326  371  372  377  381  383
386  387  395  411  416  422  426  431  432  438
441  443  461  466  471  476  482  483  486  488
491  492  493  494  497  498  516  521  522  527

One thousandth: 6996
```

## Phix

```with javascript_semantics
--constant limit = 1_000*250    -- NB: 250 magic'd out of thin air
--constant limit = 10_000*269   -- NB: 269 magic'd out of thin air
constant limit = 100_000*290    -- NB: 290 magic'd out of thin air
sequence consummate = repeat(false,limit),
inconsummate = {}

function digital_sum(integer i)
integer res = 0
while i do
res += remainder(i,10)
i = floor(i/10)
end while
return res
end function

for d=1 to limit do
integer ds = digital_sum(d)
if rmdr(d,ds)=0 then
integer q = floor(d/ds)
if q<=limit then consummate[q] = true end if
end if
end for
for d=1 to limit do
if not consummate[d] then inconsummate &= d end if
end for

printf(1,"First 50 inconsummate numbers in base 10:\n%s\n",
join_by(inconsummate[1..50],1,10," ",fmt:="%3d"))
printf(1,"One thousandth %,d\n", inconsummate[1_000])
printf(1,"Ten thousandth %,d\n", inconsummate[10_000])
printf(1,"100 thousandth %,d\n", inconsummate[100_000])
```
Output:
```First 50 inconsummate numbers in base 10:
62  63  65  75  84  95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527

One thousandth 6,996
Ten thousandth 59,853
100 thousandth 536,081
```

## PL/M

Based on the Algol 68 sample but only finds the first 50 Inconsummate numbers as it would require more than 16 bit arithmetic t0 go much further.
Calculating the digit sums as here appears to be faster than using MOD and division.

Works with: 8080 PL/M Compiler
... under CP/M (or an emulator)
```100H: /* FIND SOME INCOMSUMMATE NUMBERS: INTEGERS THAT CANNOT BE EXPRESSED   */
/*      AS AN INTEGER DIVIDED BY THE SUM OF ITS DIGITS                 */

DECLARE FALSE LITERALLY '0', TRUE LITERALLY '0FFH';

/* CP/M SYSTEM CALL AND I/O ROUTINES                                      */
BDOS:      PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
PR\$CHAR:   PROCEDURE( C ); DECLARE C BYTE;    CALL BDOS( 2, C );  END;
PR\$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S );  END;
PR\$NL:     PROCEDURE;   CALL PR\$CHAR( 0DH ); CALL PR\$CHAR( 0AH ); END;
PR\$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH  */
DECLARE V ADDRESS, N\$STR ( 6 )BYTE, W BYTE;
V = N;
W = LAST( N\$STR );
N\$STR( W ) = '\$';
N\$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N\$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR\$STRING( .N\$STR( W ) );
END PR\$NUMBER;
/* END SYSTEM CALL AND I/O ROUTINES                                       */

DECLARE MAX\$C        LITERALLY '999', /* MAXIMUM NUMBER WE WILL TEST      */
MAX\$C\$PLUS\$1 LITERALLY '1000';  /* MAX\$C + 1 FOR ARRAY BOUNDS     */

DECLARE ( I, J, MAX\$SUM, V, MAX\$NUMBER, COUNT ) ADDRESS;

/* TABLE OF NUMBERS THAT CAN BE FORMED BY N / DIGIT SUM N                 */
DECLARE CONSUMMATE ( MAX\$C\$PLUS\$1 ) ADDRESS;
DO I = 0 TO LAST( CONSUMMATE ); CONSUMMATE( I ) = FALSE; END;

/* CALCULATE THE MAXIMUM NUMBER WE MUST CONSIDER TO FIND CONSUMMATE       */
/* NUMBERS UP TO LAST(CONSUMMATE)- WHICH IS 9 * THE NUMBER OF DIGITS IN   */
/* LAST(CONSUMMATE)                                                       */
MAX\$SUM = 9;
V       = LAST( CONSUMMATE );
DO WHILE ( V := V / 10 ) > 0; MAX\$SUM = MAX\$SUM + 9; END;
MAX\$NUMBER = LAST( CONSUMMATE ) * MAX\$SUM;

/* CONSTRUCT THE DIGIT SUMS OF THE NUMBERS UP TO MAX\$NUMBER               */
/* AND FIND THE CONSUMATE NUMBERS, WE START THE LOOP FROM 10 TO AVOID     */
/* HAVING TO DEAL WITH 0-9                                                */
DO;
DECLARE ( D, N, TN, HN, TH, TT, SUMD ) ADDRESS;
CONSUMMATE( 1 ) = TRUE;
TT, TH, HN = 0; TN = 1;
DO N = 10 TO MAX\$NUMBER BY 10;
SUMD = TT + TH + HN + TN;
DO D = N TO N + 9;
IF D MOD SUMD = 0 THEN DO;
/* D IS COMSUMMATE                                            */
IF ( D\$RATIO := D / SUMD ) <= LAST( CONSUMMATE ) THEN DO;
CONSUMMATE( D\$RATIO ) = TRUE;
END;
END;
SUMD = SUMD + 1;
END;
IF ( TN := TN + 1 ) > 9 THEN DO;
TN = 0;
IF ( HN := HN + 1 ) > 9 THEN DO;
HN = 0;
IF ( TH := TH + 1 ) > 9 THEN DO;
TH = 0;
TT = TT + 1;
END;
END;
END;
END;
END;

COUNT = 0;
CALL PR\$STRING( .'THE FIRST 50 INCONSUMMATE NUMBERS:\$' );CALL PR\$NL;
I = 0;
DO WHILE ( I := I + 1 ) <= LAST( CONSUMMATE ) AND COUNT < 50;
IF NOT CONSUMMATE( I ) THEN DO;
IF I <   10 THEN CALL PR\$CHAR( ' ' );
IF I <  100 THEN CALL PR\$CHAR( ' ' );
IF I < 1000 THEN CALL PR\$CHAR( ' ' );
CALL PR\$CHAR( ' ' );
CALL PR\$NUMBER( I );
IF ( COUNT := COUNT + 1 ) MOD 10 = 0 THEN CALL PR\$NL;
END;
END;

EOF```
Output:
```THE FIRST 50 INCONSUMMATE NUMBERS:
62   63   65   75   84   95  161  173  195  216
261  266  272  276  326  371  372  377  381  383
386  387  395  411  416  422  426  431  432  438
441  443  461  466  471  476  482  483  486  488
491  492  493  494  497  498  516  521  522  527
```

## Python

```''' Rosetta code rosettacode.org/wiki/Inconsummate_numbers_in_base_10 '''

def digitalsum(num):
''' Return sum of digits of a number in base 10 '''
return sum(int(d) for d in str(num))

def generate_inconsummate(max_wanted):
''' generate the series of inconsummate numbers up to max_wanted '''
minimum_digitsums = [(10**i, int((10**i - 1) / (9 * i)))
for i in range(1, 15)]
limit = 20 * min(p[0] for p in minimum_digitsums if p[1] > max_wanted)
arr = [1] + [0] * (limit - 1)

for dividend in range(1, limit):
quo, rem = divmod(dividend, digitalsum(dividend))
if rem == 0 and quo < limit:
arr[quo] = 1
for j, flag in enumerate(arr):
if flag == 0:
yield j

for i, n in enumerate(generate_inconsummate(100000)):
if i < 50:
print(f'{n:6}', end='\n' if (i + 1) % 10 == 0 else '')
elif i == 999:
print('\nThousandth inconsummate number:', n)
elif i == 9999:
print('\nTen-thousanth inconsummate number:', n)
elif i == 99999:
print('\nHundred-thousanth inconsummate number:', n)
break
```
Output:
```    62    63    65    75    84    95   161   173   195   216
261   266   272   276   326   371   372   377   381   383
386   387   395   411   416   422   426   431   432   438
441   443   461   466   471   476   482   483   486   488
491   492   493   494   497   498   516   521   522   527

Thousandth inconsummate number: 6996

Ten-thousanth inconsummate number: 59853

Hundred-thousanth inconsummate number:  536081
```

## Raku

Not really pleased with this entry. It works, but seems inelegant.

```my \$upto = 1000;

my @ratios = unique (^∞).race.map({(\$_ / .comb.sum).narrow})[^(\$upto²)].grep: Int;
my @incons = (sort keys (1..\$upto * 10) (-) @ratios)[^\$upto];

put "First fifty inconsummate numbers (in base 10):\n" ~ @incons[^50]».fmt("%3d").batch(10).join: "\n";
put "\nOne thousandth: " ~ @incons[999]
```
Output:
```First fifty inconsummate numbers (in base 10):
62  63  65  75  84  95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527

One thousandth: 6996```

## Wren

Library: Wren-math
Library: Wren-fmt

It appears to be more than enough to calculate ratios for all numbers up to 999,999 (which only takes about 0.4 seconds on my machine) to be sure of finding the 1,000th inconsummate number.

In fact it still finds the 1,000th number if you limit the search to the first 249,999 (but not 248,999) numbers which may be where the first magic number of '250' comes from in the Pascal/Phix entries.

```import "./math" for Int
import "./fmt" for Fmt

// Maximum ratio for 6 digit numbers is 100,000
var cons = List.filled(100001, false)
for (i in 1..999999) {
var ds = Int.digitSum(i)
var ids = i/ds
if (ids.isInteger) cons[ids] = true
}
var incons = []
for (i in 1...cons.count) {
}
System.print("First 50 inconsummate numbers in base 10:")
Fmt.tprint("\$3d", incons[0..49], 10)
Fmt.print("\nOne thousandth: \$,d", incons[999])
```
Output:
```First 50 inconsummate numbers in base 10:
62  63  65  75  84  95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527

One thousandth: 6,996
```

Alternatively and more generally:

Translation of: Python

...though much quicker as I'm using lower figures for the second component of the minDigitSums tuples.

```import "./math" for Int, Nums
import "./fmt" for Fmt

var generateInconsummate = Fiber.new { |maxWanted|
var minDigitSums = (2..14).map { |i| [10.pow(i), ((10.pow(i-2) * 11 - 1) / (9 * i - 17)).floor] }
var limit = Nums.min(minDigitSums.where { |p| p[1] > maxWanted }.map { |p| p[0] })
var arr = List.filled(limit, 0)
arr[0] = 1
for (dividend in 1...limit) {
var ds = Int.digitSum(dividend)
var quo = (dividend/ds).floor
var rem = dividend % ds
if (rem == 0 && quo < limit) arr[quo] = 1
}
for (j in 0...arr.count) {
if (arr[j] == 0) Fiber.yield(j)
}
}

var incons = List.filled(50, 0)
System.print("First 50 inconsummate numbers in base 10:")
for (i in 1..100000) {
var j = generateInconsummate.call(100000)
if (i <= 50) {
incons[i-1] = j
if (i == 50) Fmt.tprint("\$3d", incons, 10)
} else if (i == 1000) {
Fmt.print("\nOne thousandth \$,d", j)
} else if (i == 10000) {
Fmt.print("Ten thousandth \$,d", j)
} else if (i == 100000) {
Fmt.print("100 thousandth \$,d", j)
}
}
```
Output:
```First 50 inconsummate numbers in base 10:
62  63  65  75  84  95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527

One thousandth 6,996
Ten thousandth 59,853
100 thousandth 536,081
```

## XPL0

```func DigSum(Num);       \Return sum of digits of a number in base 10
int  Num, Sum;
[Sum:= 0;
repeat  Num:= Num/10;
Sum:= Sum + rem(0);
until   Num = 0;
return Sum;
];

func Consum(N);         \Return 'true' if N is a consummate number
int  N, M, P;
[M:= 1;
repeat  P:= M*N;
if P = N*DigSum(P) then return true;
M:= M+1;
until   P >= 1000000;   \8-)
return false;
];

int  C, N;
[C:= 0;  N:= 1;
Format(4, 0);
repeat  if not Consum(N) then
[C:= C+1;
if C <= 50 then
[RlOut(0, float(N));
if rem(C/10) = 0 then CrLf(0);
];
if C = 1000 then
[Text(0, "One thousandth inconsummate number: ");
IntOut(0, N);
CrLf(0);
];
];
N:= N+1;
until   C >= 1000;
]```
Output:
```  62  63  65  75  84  95 161 173 195 216
261 266 272 276 326 371 372 377 381 383
386 387 395 411 416 422 426 431 432 438
441 443 461 466 471 476 482 483 486 488
491 492 493 494 497 498 516 521 522 527
One thousandth inconsummate number: 6996
```