In the Anno video game series there are 6 games with a 7th one announced for early 2019. Their titles always feature a year in a specific pattern:

Anno 1602, Anno 1503, Anno 1701, Anno 1404, Anno 2070, Anno 2205, Anno 1800

  • The digital sum is always 9.
  • The years are four digits long.
  • They contain at least one zero.

Within these constrains there exist 109 possible titles:

[1008,1017,1026,1035,1044,1053,1062,1071,1080,1107,1170,1206,1260,1305,1350,1404,1440,1503,1530,1602,1620,1701,1710,1800,2007,2016,2025,2034,2043,2052,2061,2070,2106,2160,2205,2250,2304,2340,2403,2430,2502,2520,2601,2610,2700,3006,3015,3024,3033,3042,3051,3060,3105,3150,3204,3240,3303,3330,3402,3420,3501,3510,3600,4005,4014,4023,4032,4041,4050,4104,4140,4203,4230,4302,4320,4401,4410,4500,5004,5013,5022,5031,5040,5103,5130,5202,5220,5301,5310,5400,6003,6012,6021,6030,6102,6120,6201,6210,6300,7002,7011,7020,7101,7110,7200,8001,8010,8100,9000]

Your objective is to list them all in any reasonable form in the fewest number of bytes.

  • How flexible is the output format? Is this acceptable? – Luis Mendo Oct 15 at 14:50
  • 1
    @LuisMendo Yes, that's fine with me. – Laikoni Oct 15 at 14:54
  • Are lists of digits allowed? – Erik the Outgolfer Oct 15 at 18:18
  • 1
    @aslum I assume you mean a lot of spaces, not just one, right? Comment markdown doesn't allow for a good representation of that. And I would assume that's allowed, given that Luis's format above is allowed. ;-) – Erik the Outgolfer Oct 15 at 20:20
  • 1
    @EriktheOutgolfer I'd say no to lists of digits because they really do not look like years anymore. – Laikoni Oct 15 at 21:50

51 Answers 51

R, 59 51 bytes

Outputs the valid numbers as the names of a list of 201's. Why 201? Because ASCII 0 is 48, and 4*48+9 is... yeah. Saved 6 bytes by aliasing ^ to Map and another 2 by using 1:9e3 as range.

"^"=Map;x=sum^utf8ToInt^grep(0,1:9e3,,,T);x[x==201]

Try it online!

Explanation

# Create list of sums of ASCII char values of numbers,
# with the original numbers as the names of the list
x <- Map(sum,
  # Create a list from the strings where each element is the string split 
  # into ASCII char values
  Map(utf8ToInt,
      # Find all numbers between 1 and 9e3 that contain a zero
      # Return the matched values as a vector of strings (6th T arg)
      grep(pattern=0,x=1:9000,value=TRUE)
  )
)
# Pick out elements with value 201 (i.e. 4-digits that sum to 9)
# This implicitly only picks out elements with 4 digits, since 3-digit 
# sums to 9 won't have this ASCII sum, letting us use the 1:9e3 range
x[x==201] 
  • 3
    ah, grep, why do I never remember that it casts to character... – Giuseppe Oct 15 at 14:23

Perl 6, 35 33 bytes

-2 bytes thanks to Jo King

{grep {.ords.sum==201&&/0/},^1e4}

Try it online!

Python 2, 67 66 64 bytes

print[y for y in range(9001)if('0'in`y`)*sum(map(ord,`y`))==201]

Try it online!


Saved:

  • -1 byte, thanks to Luis felipe De jesus Munoz
  • -2 bytes, thanks to Kevin Cruijssen

Jelly, 11 bytes

9ȷṢ€æ.ẹ9ṫ19

Try it online!

How it works

9ȷṢ€æ.ẹ9ṫ19  Main link. No arguments.

9ȷ           Set the left argument and the return value to 9000.
  Ṣ€         Sort the digits of each integer in [1, ..., 9000].
    æ.       Perform the dot product of each digit list and the left argument,
             which gets promoted from 9000 to [9000].
             Overflowing digits get summed without multiplying, so we essentially
             map the digit list [a, b, c, d] to (9000a + b + c + d).
      ẹ9     Find all 1-based indices of 9.
             Note that 9000a + b + c + d == 9 iff a == 0 and b + c + d == 9.
        ṫ19  Tail 19; discard the first 18 indices.

PowerShell, 50 49 bytes

999..1e4-match0|?{([char[]]"$_"-join'+'|iex)-eq9}

Try it online!

Constructs a range from 999 to 10000, then uses inline -match as a filter to pull out those entries that regex match against 0. This leaves us with 1000, 1001, 1002, etc. We then pipe that into a Where-Object clause where we take the current number as a string "$_", cast it as a char-array, -join those characters together with + and Invoke-Expression (similar to eval) to come up with their digit sum. We check whether that is -equal to 9, and if so it's passed on the pipeline. At program completion, those numbers are picked up from the pipeline and implicitly output.

JavaScript (ES6), 78 73 bytes

Saved 2 bytes thanks to @KevinCruijssen

Returns a space-separated string.

f=(n=9e3)=>n>999?f(n-9)+(eval([...n+''].join`+`)&/0/.test(n)?n+' ':''):''

Try it online!

How?

We iterate over the range \$[1008..9000]\$ with an increment of \$9\$, ignoring numbers that don't have a \$0\$.

All these numbers are multiples of \$9\$, so the sum of their digits is guaranteed to be a multiple of \$9\$ as well.

Because valid numbers have at least one \$0\$, they have no more than two \$9\$'s, which means that the sum of the remaining digits is at most \$18\$. Therefore, it's enough to test if the sum of the digits is odd.

Hence the test:

(eval([...n + ''].join`+`) & /0/.test(n)
  • You can save a byte changing the 1008 to 999, since it doesn't contain a 0 anyway, and 999+9 = 1008. – Kevin Cruijssen Oct 15 at 15:13
  • Or even 2 bytes by changing it to f=(n=9e3)=>n<1e3?'':(eval([...n+''].join`+`)<10&/0/.test(n)?[n,,]:'')+f(n-9) (does contain a trailing comma though, so f=(n=9e3)=>n<1e3?'':(eval([...n+''].join`+`)<10&/0/.test(n)?n+' ':'')+f(n-9) with space delimiter including trailing space might look prettier) – Kevin Cruijssen Oct 15 at 15:17
  • @KevinCruijssen Thanks! I'm actually trying to update this for a while, but I have like 500B/s of Internet bandwidth where I am tonight. :/ – Arnauld Oct 15 at 16:42
  • I know the feeling.. Lately our internet at home is crap for some reason.. Can't download anything above 10 MB, and sometimes have to refresh videos or pages with 10+ images a few times before it completely loads.. Really annoying when I'm working from home on Mondays/Tuesdays.. >.> Tomorrow someone comes to fix it (and I'm not leaving him until it's fixed xD) – Kevin Cruijssen Oct 15 at 16:44

JavaScript (Node.js), 89 bytes

[...Array(9e3)].map(_=>i++,i=1e3).filter(a=>(s=[...a+""]).sort()[0]<1&eval(s.join`+`)==9)

Try it online!

  • -4 bytes thanks to @ETHproductions

JavaScript (Node.js), 129 127 126 124 115 114 111 110 105 97 93 92 90 bytes

[...Array(9e3)].map(f=(_,i)=>eval(s=[...(i+=1e3)+""].sort().join`+`)-9|s[0]?0:i).filter(f)

Try it online!

Explanation

[...Array(9e3)].map(f=(_,i)=>eval(s=[...(i+=1e3)+""].sort().join`+`)-9|s[0]?0:i).filter(f)
[...Array(9e3)].map(f=(_,i)=>                                                  )           // Create a 9000-length array and loop over it; store the loop body
                                    [...(i+=1e3)+""]                                       // Add 1000 to the index and split it into an array of characters (17 -> ["1", "0", "1", "7"])
                                                    .sort()                                // Sort the array of characters in ascending order by their code points ("0" will always be first) (["1", "0", "1", "7"] -> ["0", "1", "1", "7"])
                                  s=                       .join`+`                        // Join them together with "+" as the separator (["0", "1", "1", "7"] -> "0+0+2+9"); store the result
                             eval(                                 )-9                     // Evaluate and test if it's different than 9
                                                                       s[0]                // Take the first character of the string and implicitly test if it's different than "0"
                                                                      |    ?0              // If either of those tests succeeded, then the number doesn't meet challenge criteria - return a falsey value
                                                                             :i            // Otherwise, return the index
                                                                                .filter(f) // Filter out falsey values by reusing the loop body

First time doing code golf in JavaScript. I don't think I need to say it, but if I'm doing something wrong, please notify me in the comments below.

  • -3 bytes thanks to @Luis felipe De jesus Munoz

  • -6 bytes thanks to @Kevin Cruijssen

  • 1
    [...Array(9e3)] instead Array(9e3).fill() saves 2 bytes – Luis felipe De jesus Munoz Oct 15 at 15:03
  • 1
    .map(a=>+a) instead .map(Number) saves another byte – Luis felipe De jesus Munoz Oct 15 at 15:05
  • 1
    You can remove the space at (_, i) to save a byte, and s[0]+s[1]+s[2]+s[3] can be eval(s.join`+`) to save an additional 4 bytes. – Kevin Cruijssen Oct 15 at 15:29
  • 1
    Also, I'm pretty sure the || can be | in your answer. – Kevin Cruijssen Oct 15 at 15:41
  • 1
    If you use .map() only to generate the range, and keep the filtering separate, you can save 8 bytes: Try it online! – ETHproductions Oct 15 at 18:16

Python 2, 57 bytes

n=999
exec"n+=9\nif'0'in`n`>int(`n`,11)%10>8:print n\n"*n

Try it online!

2 bytes thanks to Dennis

Uses an exec loop to counts up n in steps of 9 as 1008, 1017, ..., 9981, 9990, printing those that meet the condition.

Only multiples of 9 can have digit sum 9, but multiples of 9 in this range can also have digits sum of 18 and 27. We rule these out with the condition int(`n`,11)%10>8. Interpreting n in base 11, its digit sum is equal to the number modulo 10, just like in base 10 a number equals its digit sum modulo 9. The digits sum of (9, 18, 27) correspond to (9, 8, 7) modulo 10, so taking those>8 works to filter out nines.

The number containing a zero is check with string membership. '0'in`n`. This condition is joined with the other one with a chained inequality, using that Python 2 treats strings as greater than numbers.

  • I like how heavily golfed Python seems to often have enormously long autogenerated executables... – J.Doe Oct 16 at 9:13

sed and grep (and seq), 72 64 63 bytes

seq 9e3|sed s/\\B/+/g|bc|grep -wn 9|sed s/:9//|grep 0|grep ....
  • Some of these aren't four digits long (but I'm not sure what the final grep is, so maybe I'm running it wrong?) – Sparhawk Oct 16 at 1:06
  • @Sparhawk: The last grep ensures that the number is 4 digits long – Thor Oct 16 at 15:58
  • @Thor Ah right. For some reason I parsed that as an ellipsis. – Sparhawk Oct 16 at 20:58

Haskell, 55 bytes

[i|i<-show<$>[1..5^6],201==sum(fromEnum<$>i),elem '0'i]

Thanks to @Laikoni, see the comments.

Readable:

import Data.Char (digitToInt)

[i | i <- show <$> [1000..9999]
   , sum (digitToInt <$> i) == 9
   , '0' `elem` i
   ]
  • 2
    Welcome to PPCG and Haskell golfing in particular! You can save a few bytes by dropping (-48+) and comparing the sum against 201 instead of 9. Incidentally this also allows you to use 1 instead of 1000 for the range. – Laikoni Oct 16 at 12:10
  • Also your previous version without main=print was fine as per this consensus on Meta. – Laikoni Oct 16 at 12:13
  • 9999 can be 5^6 instead. – Laikoni Oct 16 at 13:38
  • 1
    Ha, there's always another byte to shave! Thanks :-) – mb21 Oct 16 at 13:56

R, 82 bytes

write((x=t(expand.grid(1:9,0:9,0:9,0:9)))[,colSums(x)==9&!apply(x,2,all)],1,4,,"")

Try it online!

Generates a matrix x of all possible 4-digit numbers, excluding leading zeros, going down columns. Then filters for column (digital) sums of 9 and containing zero, i.e., not all are nonzero. write prints down the columns, so we write to stdout with a width of 4 and a separator of "".

Outgolfed by J.Doe

  • Nice answer! I came up with a different route... – J.Doe Oct 15 at 14:17

Japt, 20 18 bytes.

-2 bytes thanks to @Shaggy and @ETHproductions

A³òL² f_=ì)x ¥9«Z×

A³òL² f_=ì)x ¥9«Z×  Full program
A³òL²               Range [1000, 10000]
      f_            Filter by : 
        =ì)         Convert to array 
           x ¥9     Sum equal to 9?
               «    And 
                Z×  Product not 0

Try it online!

  • This is actually 28 bytes. Using a literal integer instead is 22 bytes but A³ò9000 f_ìx ¥9©ZsøT gets you back down to 20. – Shaggy Oct 15 at 17:28
  • 1
    You can save 1 byte by using ì instead of s and ¬, which has to be done in the filter: f_=ì)x ¥9.... Then you can save another by checking if the product of Z is zero with «Z×: Try it online! – ETHproductions Oct 15 at 17:59

Java 8, 128 117 115 bytes

v->{int i=109,r[]=new int[i],n=i;for(;i>0;n++)if((n+"").chars().sum()==201&(n+"").contains("0"))r[--i]=n;return r;}

-11 bytes thanks to @nwellnhof.

Try it online.

Explanation:

v->{                              // Method with empty unused parameter & int-array return
  int i=109,                      //  Index-integer, starting at 109
      r[]=new int[i],             //  Result-array of size 109
      n=i;                        //  Number integer, starting at 109
   for(;i>0;                      //  Loop as long as `i` is not 0 yet:
       n++)                       //    After every iteration, increase `n` by 1
     if((n+"").chars().sum()==201 //   If the sum of the unicode values of `n` is 201,
                                  //   this means there are four digits, with digit-sum = 9
        &(n+"").contains("0"))    //   and `n` contains a 0:
       r[--i                      //    Decrease `i` by 1 first
            ]=n;                  //    And put `n` in the array at index `i`
  return r;}                      //  Return the array as result
  • 1
    What about chars().sum()==201? – nwellnhof Oct 15 at 15:33
  • @nwellnhof Ah, of course. Thanks! – Kevin Cruijssen Oct 15 at 15:37

R, 85 bytes

(just competing for the best abuse of R square brackets ... :P )

`[`=`for`;i[a<-0:9,j[a,k[a,w[a,if(sum(s<-c(i,j,k,w))==9&any(!s)&i)write(s,1,s='')]]]]

Try it online!

  • 1
    Holy for loops, Batman! – BLT Oct 18 at 17:43

Pip, 18 bytes

{0Na&$+a=9}FIm,t*m

Use an ouput-format flag such as -p to get readable output. Try it online!

{0Na&$+a=9}FIm,t*m
             m,t*m  Range from 1000 to 10*1000
{         }FI       Filter on this function:
 0Na                 There is at least one 0 in the argument
    &                and
     $+a             The sum of the argument
        =9           equals 9

Wolfram Language (Mathematica), 56 55 bytes

Select[9!!~Range~9999,Tr@#==Times@@#+9&@*IntegerDigits]

Try it online!

We test the range from 9!! = 945 to 9999, since there are no results between 945 and 999. Maybe there's a shorter way to write a number between 9000 and 10007, as well.

Tr@#==Times@@#+9& applied to {a,b,c,d} tests if a+b+c+d == a*b*c*d+9, which ends up being equivalent to The Anno Condition.

  • In retrospect, 9!! isn't any shorter than 999 or something, but it beats 1000. – Misha Lavrov Oct 16 at 0:37
  • What is 9!! ? In guessing it isnt related to factorials. – Robert Fraser Oct 16 at 6:49
  • @RobertFraser Double factorial: 9*7*5*3*1. – Misha Lavrov Oct 16 at 14:16

Ruby, 46 42 41 bytes

?9.upto(?9*4){|x|x.sum==201&&x[?0]&&p(x)}

Try it online!

How it works:

  • Iterate on strings ranging from '9' to '9999'
  • Check that sum of ASCII values is 201
  • Check if string contains a zero (without regex, a regex would be 1 byte longer)

(Thanks Laikoni for -2 bytes)

  • 1
    9*3 can be just 9, because checking against 201 already requires 4 digit numbers. – Laikoni Oct 16 at 0:01

05AB1E, 15 13 bytes

₄4°Ÿʒ0å}ʒSO9Q

-2 bytes thanks to @Emigna.

Try it online.

Explanation:

₄4°Ÿ             # List in the range [1000,10000]
    ʒ  }         # Filter this list by:
     0å          #  The number contains a 0
        ʒ        # Filter the filtered list further by:
         SO9Q    #  The sum of its digits is exactly 9

13-bytes alternative:

₄4°ŸʒW_sSO9Q*

Try it online.

Explanation:

₄4°Ÿ             # List in the range [1000,10000]
    ʒ            # Filter this list by:
     W_          #  Take the minimum digit (without popping), and check if it's a 0
            *    #  And
       sSO9Q     #  Where the sum of its digits is exactly 9

And yet another 13-bytes alternative:

4°Lʒ0å}ʒÇOт·-

Credit for this alternative goes to @Mr.Xcoder.

Try it online.

Explanation:

4°L              # List in the range [1,10000]
   ʒ  }          # Filter this list by:
    0å           #  The number contains a 0
       ʒ         # Filter the filtered list further by:
        Ç        #  Convert the digits to unicode values
         O       #  Take the sum
          т·-    #  Subtract 200 (Only 1 is truthy in 05AB1E, and the sum of digits
                 #                as unicode values equaling 9 is 201)
  • 1
    ₄4°Ÿʒ0å}ʒSO9Q. Splitting filters are usually shorter – Emigna Oct 15 at 15:58
  • @Emigna Ah, I was looking for a shorter way for the range, but completely forgot about . Thanks. And you're indeed right that multiple loose filters (at the end) are shorter. Will also add it to one of my tip answers. Thanks for both bytes! – Kevin Cruijssen Oct 15 at 16:12
  • A strictly hypothetical solution is ∞4úʒ0å}ʒSO9Q (12 bytes), but I don't think it should count as valid so I'll leave it here as-is. – Mr. Xcoder Oct 16 at 4:55
  • 1
    And my other 13-byter (inspired by the ord sum == 201 trick) is 4°Lʒ0å}ʒÇOт·-. Leaving this here, maybe someone can golf it further – Mr. Xcoder Oct 16 at 5:06
  • @Mr.Xcoder Added it to the answer. Found another 13-bytes alternative as well: ₄4°ŸʒW_sSO9Q*. – Kevin Cruijssen Oct 16 at 7:23

Octave, 49 bytes

6 bytes saved using a more convenient output format as suggested by J.Doe.

Thanks to @Laikoni for a correction.

y=dec2base(x=1e3:9999,10)'-48;x(sum(y)==9>all(y))

Try it online!

  • I don't know anything about Octave, but it looks like you can leave the disp off... – J.Doe Oct 15 at 14:48
  • @J.Doe OP has confirmed that that output format is acceptable. Thanks for the suggestion! – Luis Mendo Oct 15 at 14:57

Dart,  103 100  96 bytes

f()=>List.generate(9001,(i)=>'$i').where((i)=>i.contains('0')&&i.runes.fold(0,(p,e)=>p+e)==201);

  • -3 bytes by setting the value in the array to string, making the conversion once and not twice
  • -4 bytes by using runes instead of codeUnits
  • Pretty self-explanatory. generates a list of 9001 (0-9000) cells with the cell's index as value, filters the ones containing a 0 then the one having an ASCII sum of 201 (The result if all the ASCII characters sum to 9). These conditions implictly include that the year is 4 digits long because using 2 ASCII numbers (and the 0), you cannot reach 201.

    Try it on Dartpad!

    • Welcome to PPCG. :) – Laikoni Oct 16 at 13:36
    • 1
      Thanks ! Been lurking for a while, can finally participate – Elcan Oct 16 at 13:43

    Bash (with seq, grep), 39 bytes

    seq 0 9 1e4|egrep '([0-4].*){3}'|grep 0
    

    Try it online!

    • @xobzoo suggested seq 0 9 1e4|awk '/([0-4].*){3}/&&/0/' to save two bytes. – Timtech Oct 21 at 3:03

    K (ngn/k), 22 bytes

    55_&(|/~a)&9=+/a:!4#10
    

    Try it online!

    • Nice! 55_&9=+/y*|/'~y:!4#10 for 21? – streetster Oct 17 at 23:21
    • 1
      @streetster thanks :) the ' in |/' looks wrong. the result includes 1116, 1125, 1134, etc which are not supposed to be there – ngn Oct 18 at 6:43

    APL (Dyalog Unicode), 23 bytes

    55↓⍸(×⌿<9=+⌿)10⊥⍣¯1⍳9e3
    

    Try it online!

    • that use of < is clever. you can make it even shorter with ⎕io←0 and ( )10⊥⍣¯1⍳9e3 -> ( )¨,⍳4⍴10 – ngn Nov 1 at 16:14

    PHP, 69, 87 bytes 74 bytes

    for($i=999;$i<9001;$i++){echo((array_sum(str_split($i))==9&strpos($i,"0")!=0)?$i:" ");} for($i=999;$i++<1e4;)echo!strpos($i,48)|array_sum(str_split($i))-9?" ":$i;

    Note this puts a space for every "failed" number, leading to some kind of funky spacing. This can be changed to comma separation, but will add another 4 characters: ?$i.",":""

    Got bigger because I wasn't checking for 0. Derp. Shortened by 13 by Titus!

    • 2
      I don't really know PHP, but does this code ensure that each year contains a zero? – Laikoni Oct 16 at 0:13
    • This code does not check for zero in the number. – krzysiej Oct 16 at 11:29
    • 1
      13 bytes shorter: for($i=999;$i++<1e4;)echo!strpos($i,48)|array_sum(str_split($i))-9?" ":$i; – Titus Oct 18 at 15:29
    • Here´s another byte: ?"$i,":"" er ... now the other way round: ?"":"$i," – Titus Oct 18 at 21:06
    • Actually @Titus that adds a couple bytes. We don't need quotes around $i unless we're including a string w/ it. – aslum Oct 22 at 12:40

    APL(Dyalog), 33 29 bytes

    1e3+⍸(0∘∊∧9=+/)¨⍎¨∘⍕¨1e3+⍳9e3
    

    -4 bytes thanks to @Adam

    Try it online!

    Scala (76 63 61 56 bytes)

    for(n<-0 to 9000;t=n+""if t.sum==201&t.min<49)println(t)
    

    Try it online

    • Thanks to Laikoni for the suggestions
    • Two more bytes shed after applying Jo King's comment
    • 1
      Welcome to PPCG! Do you have an idea what needs to be added to the header or footer section to get this code to run on TIO? Try it online! – Laikoni Oct 18 at 21:18
    • @Laikoni, didn't know I could run Scala in TIO. Fixed it. Thanks for the comment. – jrook Oct 18 at 22:40
    • 1
      It looks like t.sum==201 works instead of t.map(_.asDigit).sum==9. – Laikoni Oct 18 at 23:00
    • You may find our tips for golfing in Scala interesting. E.g. it looks like s"$n" can be n+"" and s"$t " can be t+" ". – Laikoni Oct 20 at 10:33
    • 1
      Since you're using the sum is 201 trick, the range doesn't need to start at 999 – Jo King Oct 24 at 11:16

    Tcl, 77 bytes

    time {if [incr i]>1e3&[regexp 0 $i]&9==[join [split $i ""] +] {puts $i}} 9999
    

    Try it online!

    Japt, 16 bytes

    Returns an array of digit arrays.

    L²õì l4 k_ת9aZx
    

    Test it


    Explanation

    L                    :100
     ²                   :Squared
      õ                  :Range [1,L²]
       ì                 :Convert each to a digit array
         l4              :Filter elements of length 4
            k_           :Remove each Z that returns truthy (not 0)
              ×          :  When reduced by multiplication
               ª         :  OR
                  Zx     :  When reduced by addition
                9a       :   And subtracted from 9
    
    • 1
      OP has ruled that digit arrays aren't valid output unfortunately :o( – Sok Oct 16 at 8:12

    APL(NARS), 45 chars, 90 bytes

    f←{⍵×⍳(0∊x)∧9=+/x←⍎¨⍕⍵}⋄f¨1e3..5e3⋄f¨5e3..9e3
    

    test afther some formatting:

    1008  1017  1026  1035  1044  1053  1062  1071  1080  1107  1170  1206  1260  
      1305  1350  1404  1440  1503  1530  1602  1620  1701  1710  1800  2007  2016  
      2025  2034  2043  2052  2061  2070  2106  2160  2205  2250  2304  2340  
      2403  2430  2502  2520  2601  2610  2700  3006  3015  3024  3033  3042  3051  
      3060  3105  3150  3204  3240  3303  3330  3402  3420  3501  3510  3600  
      4005  4014  4023  4032  4041  4050  4104  4140  4203  4230  4302  4320  4401  
      4410  4500 
    5004  5013  5022  5031  5040  5103  5130  5202  5220  5301  5310  5400  6003  
      6012  6021  6030  6102  6120  6201  6210  6300  7002  7011  7020  7101  7110  
      7200  8001  8010  8100  9000 
    

    possible alternative

    r←f;i;x
       r←⍬⋄i←1e3⋄→B
    A: r←r,i
    B: i+←1⋄→A×⍳(0∊x)∧9=+/x←⍎¨⍕i⋄→B×⍳i≤9e3
    

    Jelly, 13 bytes

    ȷ4µṢÄm3Ḍ)ẹ9ṫ4
    

    Try it online!

    How?

    ȷ4µṢÄm3Ḍ)ẹ9ṫ4 - Link: no arguments
    ȷ4            - literal 10^4 = 10000
      µ     )     - for each in range (1,2,3,...,10000): e.g. 3042       or  5211
       Ṣ          -   sort (given an integer makes digits)    [0,2,3,4]      [1,1,2,5]
        Ä         -   cumulative addition                     [0,2,5,9]      [1,2,4,9]
         m3       -   modulo 3 slice (1st,4th,7th...)         [0,9]          [1,9]
           Ḍ      -   convert from decimal digits             9              19
             ẹ9   - 1-based indices equal to nine             [9,99,999,1008,1017,...,8100,9000]
               ṫ4 - tail from the 4th index                   [1008,1017,...,8100,9000]
    

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