# 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.

## 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
```

## Pascal

### Free Pascal

Inconsummate numbers are not a divisor of a niven number.
Therefore I tried a solution niven number.
There is only a small increase in the needed factor in count of Inconsummate numbers

```program Inconsummate;

{\$IFDEF FPC}
{\$MODE DELPHI}{\$OPTIMIZATION ON,ALL}{\$CODEALIGN proc=8,loop=1}
{\$ENDIF}

uses
SysUtils;

const
base = 10;

type
//  tNum = 0..250 * 1000;//    6996
//  tNum = 0..260 * 10000;//  59837
//  tNum = 0..290 * 100000;//536081
tNum = 0..319 * 1000000;//5073249

const
cntbasedigits = 16;//trunc(ln(High(tNum)) / ln(base)) + 1;

type
tSumDigit = record
sdDigits: array[0..cntbasedigits - 1] of byte;
sdSumDig: uint32;
sdNumber: tNum;
sdDiv: tNum;
sdIsNiven: boolean;
end;
var
isN: array[0..High(tNUm) div 1 + 1] of boolean;

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
pD: pbyte;
i, d, s: uint32;
begin
i := 0;
pD := @sd.sdDigits[0];
with sd do
begin
s := sdSumDig;
Inc(sdNumber);
repeat
d := pD[i];
Inc(d);
Inc(s);
//base-1 times the repeat is left here
if d < base then
begin
pD[i] := d;
BREAK;
end
else
begin
pD[i] := 0;
Dec(s, base);
Inc(i);
end;
until i > high(sdDigits);
sdSumDig := s;
i := sdNumber div s;
sdDiv := i;
sdIsNiven := (sdNUmber - i * s) = 0;
end;
end;

var
MySumDig: tSumDigit;
lnn: tNum;
Limit, cnt: integer;

begin
{\$IFNDEF FPC}
cntbasedigits := trunc(ln(High(tNum)) / ln(base)) + 1;
{\$ENDIF}
MySumDig := InitSumDigit(0);
cnt := 0;
with MySumDig do
repeat
IncSumDigit(MySumDig);
if sdIsNiven then
isN[sdDiv] := True;
until sdnumber > High(tNum) - 1;

limit := 10;
for lnn := 1 to High(isN) - 1 do
if not (isN[lnn]) then
begin
Inc(cnt);
Write(lnn: 5);
if (cnt = limit) then
begin
writeln;
Inc(limit, 10);
end;
if cnt >= 50 then
BREAK;
end;
writeln;

limit := 100;
for lnn := lnn + 1 to High(isN) - 1 do
if not (isN[lnn]) then
begin
Inc(cnt);
if cnt = limit then
begin
Writeln(limit: 10, lnn: 10);
limit *= 10;
if limit > 1000 * 1000 then
EXIT;
end;
end;
writeln;
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

100       936
1000      6996
10000     59853
100000    536081
1000000   5073249

Real time: 3.342 s CPU share: 99.16 %
```

## 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 = 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: 375410
```

## 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.

```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 I think the Python version is in fact wrong for the 100,000th number since if you enumerate up to 10,000 you get the 10,000th inconsummate number to be 42,171 rather than 59,853.

The problem seems to be that the minimum divisor for (say) 6 digit numbers is not 999999/54 = 18518 but 109999/37 = 2972. I've corrected for that in the following translation.

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

var generateInconsummate = Fn.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 gi = Fiber.new(generateInconsummate)
var incons = List.filled(50, 0)
var incons1k
var incons10k
var incons100k
System.print("First 50 inconsummate numbers in base 10:")

for (i in 1..100000) {
var j = gi.call(100000)
if (i <= 50) {
incons[i-1] = j
} else if (i == 1000) {
incons1k = j
} else if (i == 10000) {
incons10k = j
} else if (i == 100000) {
incons100k = j
}
}
Fmt.tprint("\$3d", incons, 10)
Fmt.print("\nOne thousandth \$,d", incons1k)
Fmt.print("Ten thousandth \$,d", incons10k)
Fmt.print("100 thousandth \$,d", incons100k)```
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
```