# Eban numbers

Eban numbers
You are encouraged to solve this task according to the task description, using any language you may know.

Definition

An   eban   number is a number that has no letter   e   in it when the number is spelled in ENglish.

Or more literally,   spelled numbers that contain the letter   e   are banned.

The American version of spelling numbers will be used here   (as opposed to the British).

2,000,000,000   is two billion,   not   two milliard.

Only numbers less than   one sextillion   (1021)   will be considered in/for this task.

This will allow optimizations to be used.

Task
•   show all eban numbers   ≤   1,000   (in a horizontal format),   and a count
•   show all eban numbers between   1,000   and   4,000   (inclusive),   and a count
•   show a count of all eban numbers up and including           10,000
•   show a count of all eban numbers up and including         100,000
•   show a count of all eban numbers up and including      1,000,000
•   show a count of all eban numbers up and including    10,000,000
•   show all output here.

See also

## Go

`package main import "fmt" type Range struct {    start, end uint64    print      bool} func main() {    rgs := []Range{        {2, 1000, true},        {1000, 4000, true},        {2, 1e4, false},        {2, 1e5, false},        {2, 1e6, false},        {2, 1e7, false},        {2, 1e8, false},        {2, 1e9, false},    }    for _, rg := range rgs {        if rg.start == 2 {            fmt.Printf("eban numbers up to and including %d:\n", rg.end)        } else {            fmt.Printf("eban numbers between %d and %d (inclusive):\n", rg.start, rg.end)        }        count := 0        for i := rg.start; i <= rg.end; i += 2 {            b := i / 1000000000            r := i % 1000000000            m := r / 1000000            r = i % 1000000            t := r / 1000            r %= 1000            if m >= 30 && m <= 66 {                m %= 10            }            if t >= 30 && t <= 66 {                t %= 10            }            if r >= 30 && r <= 66 {                r %= 10            }            if b == 0 || b == 2 || b == 4 || b == 6 {                             if m == 0 || m == 2 || m == 4 || m == 6 {                    if t == 0 || t == 2 || t == 4 || t == 6 {                        if r == 0 || r == 2 || r == 4 || r == 6 {                            if rg.print {                                fmt.Printf("%d ", i)                            }                            count++                        }                    }                }            }        }        if rg.print {            fmt.Println()        }        fmt.Println("count =", count, "\n")    }}`
Output:
```eban numbers up to and including 1000:
2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66
count = 19

eban numbers between 1000 and 4000 (inclusive):
2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000
count = 21

eban numbers up to and including 10000:
count = 79

eban numbers up to and including 100000:
count = 399

eban numbers up to and including 1000000:
count = 399

eban numbers up to and including 10000000:
count = 1599

eban numbers up to and including 100000000:
count = 7999

eban numbers up to and including 1000000000:
count = 7999
```

## J

` Filter =: (#~`)(`:6) itemAmend =: (29&< *. <&67)`(,: 10&|)}iseban =: [: *./ 0 2 4 6 e.~ [: itemAmend [: |: (4#1000)&#:     (;~ #) iseban Filter >: i. 1000┌──┬─────────────────────────────────────────────────────┐│19│2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66│└──┴─────────────────────────────────────────────────────┘    NB. INPUT are the correct integers, head and tail shown   ({. , {:) INPUT =: 1000 + i. 30011000 4000    (;~ #)  iseban Filter INPUT┌──┬────────────────────────────────────────────────────────────────────────────────────────────────────────┐│21│2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000│└──┴────────────────────────────────────────────────────────────────────────────────────────────────────────┘   (, ([: +/ [: iseban [: >: i.))&> 10000 * 10 ^ i. +:2 10000   79100000  399   1e6  399   1e7 1599 `

## Julia

I changed your program but not the output. You had used the common cut-n-paste error repeating the 10000 println("eban numbers up to and including 10000:

` function iseban(n::Integer)    b, r = divrem(n, oftype(n, 10 ^ 9))    m, r = divrem(r, oftype(n, 10 ^ 6))    t, r = divrem(r, oftype(n, 10 ^ 3))    m, t, r = (30 <= x <= 66 ? x % 10 : x for x in (m, t, r))    return all(in((0, 2, 4, 6)), (b, m, t, r))end println("eban numbers up to and including 1000:")println(join(filter(iseban, 1:100), ", ")) println("eban numbers between 1000 and 4000 (inclusive):")println(join(filter(iseban, 1000:4000), ", ")) println("eban numbers up to and including 10000: ", count(iseban, 1:10000))println("eban numbers up to and including 100000: ", count(iseban, 1:100000))println("eban numbers up to and including 1000000: ", count(iseban, 1:1000000))println("eban numbers up to and including 10000000: ", count(iseban, 1:10000000))println("eban numbers up to and including 100000000: ", count(iseban, 1:100000000))println("eban numbers up to and including 1000000000: ", count(iseban, 1:1000000000)) `
Output:
```eban numbers up to and including 1000:
2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 66
eban numbers between 1000 and 4000 (inclusive):
2000, 2002, 2004, 2006, 2030, 2032, 2034, 2036, 2040, 2042, 2044, 2046, 2050, 2052, 2054, 2056, 2060, 2062, 2064, 2066, 4000

eban numbers up to and including 10000: 79
eban numbers up to and including 10000: 399
eban numbers up to and including 10000: 399
eban numbers up to and including 10000: 1599
eban numbers up to and including 10000: 7999
eban numbers up to and including 10000: 7999```

## Kotlin

Translation of: Go
`// Version 1.3.21 typealias Range = Triple<Int, Int, Boolean> fun main() {    val rgs = listOf<Range>(        Range(2, 1000, true),        Range(1000, 4000, true),        Range(2, 10_000, false),        Range(2, 100_000, false),        Range(2, 1_000_000, false),        Range(2, 10_000_000, false),        Range(2, 100_000_000, false),        Range(2, 1_000_000_000, false)    )    for (rg in rgs) {        val (start, end, prnt) = rg        if (start == 2) {            println("eban numbers up to and including \$end:")        } else {            println("eban numbers between \$start and \$end (inclusive):")        }        var count = 0        for (i in start..end step 2) {            val b = i / 1_000_000_000            var r = i % 1_000_000_000            var m = r / 1_000_000            r = i % 1_000_000            var t = r / 1_000            r %= 1_000            if (m >= 30 && m <= 66) m %= 10            if (t >= 30 && t <= 66) t %= 10            if (r >= 30 && r <= 66) r %= 10            if (b == 0 || b == 2 || b == 4 || b == 6) {                if (m == 0 || m == 2 || m == 4 || m == 6) {                    if (t == 0 || t == 2 || t == 4 || t == 6) {                        if (r == 0 || r == 2 || r == 4 || r == 6) {                            if (prnt) print("\$i ")                            count++                        }                    }                }            }        }        if (prnt) println()        println("count = \$count\n")    }}`
Output:
```Same as Go example.
```

## Perl

### Exhaustive search

A couple of 'e'-specific optimizations keep the running time reasonable.

`use strict;use warnings;use feature 'say';use Lingua::EN::Numbers qw(num2en); sub comma { reverse ((reverse shift) =~ s/(.{3})/\$1,/gr) =~ s/^,//r } sub e_ban {    my(\$power) = @_;    my @n;    for (1..10**\$power) {        next unless 0 == \$_%2;        next if \$_ =~ // or /.\$/ or /..\$/ or /...\$/ or /.....\$/;        push @n, \$_ unless num2en(\$_) =~ /e/;    }    @n;} my @OK = e_ban(my \$max = 7); my @a = grep { \$_ <= 1000 } @OK;say "Number of eban numbers up to and including 1000: @{[1+\$#a]}";say join(', ',@a);say ''; my @b = grep { \$_ >= 1000 && \$_ <= 4000 } @OK;say "Number of eban numbers between 1000 and 4000 (inclusive): @{[1+\$#b]}";say join(', ',@b);say ''; for my \$exp (4..\$max) {    my \$n = + grep { \$_ <= 10**\$exp } @OK;    printf "Number of eban numbers and %10s: %d\n", comma(10**\$exp), \$n;}`
Output:
```eban numbers up to and including 1000:
2, 4, 6, 30, 32, 34, 36, 40, 42, 44, 46, 50, 52, 54, 56, 60, 62, 64, 66

eban numbers between 1000 and 4000 (inclusive):
2000, 2002, 2004, 2006, 2030, 2032, 2034, 2036, 2040, 2042, 2044, 2046, 2050, 2052, 2054, 2056, 2060, 2062, 2064, 2066, 4000

Number of eban numbers up to     10,000: 79
Number of eban numbers up to    100,000: 399
Number of eban numbers up to  1,000,000: 399
Number of eban numbers up to 10,000,000: 1599```

### Algorithmically generate / count

Alternately, a partial translation of Perl 6. Does not need to actually generate the e-ban numbers to count them. Display counts up to 10**21.

`use strict;use warnings;use bigint;use feature 'say';use Lingua::EN::Nums2Words 'num2word';use List::AllUtils 'sum'; sub comma { reverse ((reverse shift) =~ s/(.{3})/\$1,/gr) =~ s/^,//r } sub nban {    my (\$n, @numbers) = @_;    grep { lc(num2word(\$_)) !~ /[\$n]/i } @numbers;} sub enumerate {    my (\$n, \$upto) = @_;    my @ban = nban(\$n, 1 .. 99);    my @orders;    for my \$o (2 .. \$upto) {        push @orders, [nban(\$n, map { \$_ * 10**\$o } 1 .. 9)];    }    for my \$oom (@orders) {        next unless +@\$oom;        my @these;        for my \$num (@\$oom) {            push @these, \$num, map { \$_ + \$num } @ban;        }       push @ban, @these;    }    unshift @ban, 0 if nban(\$n, 0);    @ban} sub count {    my (\$n, \$upto) = @_;    my @orders;    for my \$o (2 .. \$upto) {        push @orders, [nban(\$n, map { \$_ * 10**\$o } 1 .. 9)];    }    my @count = scalar nban(\$n, 1 .. 99);    for my \$o ( 0 .. \$#orders - 1 ) {        push @count, sum(@count) * (scalar @{\$orders[\$o]}) + (scalar @{\$orders[\$o]});    }    ++\$count if nban(\$n, 0);    for my \$m ( 0 .. \$#count - 1 ) {        next unless scalar \$orders[\$m];        if (nban(\$n, 10**(\$m+2))) { \$count[\$m]++; \$count[\$m + 1]-- }    }    map { sum( @count[0..\$_] ) } 0..\$#count;} for my \$t ('e') {    my @bans  = enumerate(\$t, 4);    my @count = count(\$t, my \$max = 21);     my @j = grep { \$_ <= 10 } @bans;    unshift @count, @{[1+\$#j]};     say "\n============= \$t-ban: =============";    my @a = grep { \$_ <= 1000 } @bans;    say "\$t-ban numbers up to 1000: @{[1+\$#a]}";    say '[', join(' ',@a), ']';    say '';     my @b = grep { \$_ >= 1000 && \$_ <= 4000 } @bans;    say "\$t-ban numbers between 1,000 & 4,000 (inclusive): @{[1+\$#b]}";    say '[', join(' ',@b), ']';    say '';     say "Counts of \$t-ban numbers up to ", lc(num2word(10**\$max));     for my \$exp (1..\$max) {        my \$nu = \$count[\$exp-1];        printf "Up to and including %23s: %s\n", lc(num2word(10**\$exp)), comma(\$nu);    }}`
```============= e-ban: =============
e-ban numbers up to 1000: 19
[2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66]

e-ban numbers between 1,000 & 4,000 (inclusive): 21
[2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000]

Counts of e-ban numbers up to one sextillion
Up to and including                     ten: 3
Up to and including             one hundred: 19
Up to and including            one thousand: 19
Up to and including            ten thousand: 79
Up to and including    one hundred thousand: 399
Up to and including             one million: 399
Up to and including             ten million: 1,599
Up to and including     one hundred million: 7,999
Up to and including             one billion: 7,999
Up to and including             ten billion: 31,999
Up to and including     one hundred billion: 159,999
Up to and including            one trillion: 159,999
Up to and including            ten trillion: 639,999
Up to and including    one hundred trillion: 3,199,999
Up to and including         one quadrillion: 3,199,999
Up to and including         ten quadrillion: 12,799,999
Up to and including one hundred quadrillion: 63,999,999
Up to and including         one quintillion: 63,999,999
Up to and including         ten quintillion: 255,999,999
Up to and including one hundred quintillion: 1,279,999,999
Up to and including          one sextillion: 1,279,999,999```

## Perl 6

Works with: Rakudo version 2018.12

Modular approach, very little is hard coded. Change the \$upto order-of-magnitude limit to adjust the search/display ranges. Change the letter(s) given to the enumerate / count subs to modify which letter(s) to disallow.

Will handle multi-character 'bans'. Demonstrate for e-ban, t-ban and subur-ban.

Directly find :

Considering numbers up to 1021, as the task directions suggest.

`use Lingua::EN::Numbers::Cardinal; sub nban (\$seq, \$n = 'e') { (\$seq).map: { next if .&cardinal.contains(any(\$n.lc.comb)); \$_ } } sub enumerate (\$n, \$upto) {    my @ban = [nban(1 .. 99, \$n)],;    my @orders;    (2 .. \$upto).map: -> \$o {        given \$o % 3 { # Compensate for irregulars: 11 - 19            when 0  { @orders.push: [[flat (10**\$o X* 1..9).map: *.&nban(\$n)],                                     [flat (10**\$o X* 11..19).map: *.&nban(\$n)]]}            when 1  {  @orders.push: [flat (10**\$o X* 2..9).map: *.&nban(\$n)]   }            default {  @orders.push: [flat (10**\$o X* 1..9).map: *.&nban(\$n)]    }        }    }    my @those;    ^@orders .map: -> \$o {        @ban.push: [] and next unless +@orders[\$o];        my @these;        my @lower = |@orders[\$o];        my @upper = [];        if @orders[\$o].^name eq 'Array' {            @lower = |@orders[\$o];            @upper = |@orders[\$o];        }        @lower.map: -> \$m {            @these.push: \$m;            for ^@ban -> \$b {                next unless +@ban[\$b];                @these.push: \$_ for (flat @ban[\$b]) »+» \$m ;            }        }        @ban.push: @those.clone if +@those;        @those = ();        @upper.map: -> \$m {            @those.push: \$m;            for ^(@ban) -> \$b {                next unless +@ban[\$b];                @those.push: \$_ for (flat @ban[\$b]) »+» \$m ;            }        }        @ban.push: @these;    }    @ban.push: @those.clone if +@those;    @ban.unshift(0) if nban(0, \$n);    flat @ban.map: *.flat;} sub count (\$n, \$upto) {    my @orders;    (2 .. \$upto).map: -> \$o {        given \$o % 3 { # Compensate for irregulars: 11 - 19            when 0  { @orders.push: [[flat (10**\$o X* 1 .. 9).map: *.&nban(\$n)],                                     [flat (10**\$o X* 11..19).map: *.&nban(\$n)]]}            when 1  {  @orders.push: [flat (10**\$o X* 2 .. 9).map: *.&nban(\$n)] }            default {  @orders.push: [flat (10**\$o X* 1 .. 9).map: *.&nban(\$n)] }        }    }    my @count  = +nban(1 .. 99, \$n);    my \$those;    ^@orders .map: -> \$o {        @count.push: 0 and next unless +@orders[\$o];        my @lower = |@orders[\$o];        my @upper = [];        if @orders[\$o].^name eq 'Array' {            @lower = |@orders[\$o];            @upper = |@orders[\$o];        }        my \$prev  = so (@lower.first( { \$_ ~~ /^ '1' '0'+ \$/ } ) // 0 );        my \$sum = @count.sum;        my \$these = +@lower * \$sum + @lower;        @count[1 + \$o] += \$those if \$those;        \$those = +@upper * \$sum + @upper;        \$these-- if \$prev;        @count[1 + \$o] += \$these;        ++@count[\$o]  if \$prev;    }    @count[*-1] += \$those if \$those and \$upto % 3 == 1;    ++@count if nban(0, \$n);    [\+] @count;} sub comma { \$^i.flip.comb(3).join(',').flip } #for < e o t tali subur tur ur cali i u > -> \$n { # All of themfor < e t subur > -> \$n { # An assortment for demonstration    my \$upto   = 21; # 1e21    my @bans   = enumerate(\$n, 4);    my @counts = count(\$n, \$upto);     # DISPLAY    my @k = @bans.grep: * < 1000;    my @j = @bans.grep: 1000 <= * <= 4000;    put "\n============= {\$n}-ban: =============\n" ~        "{\$n}-ban numbers up to 1000: {[email protected]}\n[{@k».&comma}]\n\n" ~        "{\$n}-ban numbers between 1,000 & 4,000: {[email protected]}\n[{@j».&comma}]\n" ~        "\nCounts of {\$n}-ban numbers up to {cardinal 10**\$upto}"        ;     my \$s = max (1..\$upto).map: { (10**\$_).&cardinal.chars };    @counts.unshift: @bans.first: * > 10, :k;    for ^\$upto -> \$c {        printf "Up to and including %{\$s}s: %s\n", cardinal(10**(\$c+1)), comma(@counts[\$c]);    }}`
Output:
```============= e-ban: =============
e-ban numbers up to 1000: 19
[2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66]

e-ban numbers between 1,000 & 4,000: 21
[2,000 2,002 2,004 2,006 2,030 2,032 2,034 2,036 2,040 2,042 2,044 2,046 2,050 2,052 2,054 2,056 2,060 2,062 2,064 2,066 4,000]

Counts of e-ban numbers up to one sextillion
Up to and including                     ten: 3
Up to and including             one hundred: 19
Up to and including            one thousand: 19
Up to and including            ten thousand: 79
Up to and including    one hundred thousand: 399
Up to and including             one million: 399
Up to and including             ten million: 1,599
Up to and including     one hundred million: 7,999
Up to and including             one billion: 7,999
Up to and including             ten billion: 31,999
Up to and including     one hundred billion: 159,999
Up to and including            one trillion: 159,999
Up to and including            ten trillion: 639,999
Up to and including    one hundred trillion: 3,199,999
Up to and including         one quadrillion: 3,199,999
Up to and including         ten quadrillion: 12,799,999
Up to and including one hundred quadrillion: 63,999,999
Up to and including         one quintillion: 63,999,999
Up to and including         ten quintillion: 255,999,999
Up to and including one hundred quintillion: 1,279,999,999
Up to and including          one sextillion: 1,279,999,999

============= t-ban: =============
t-ban numbers up to 1000: 56
[0 1 4 5 6 7 9 11 100 101 104 105 106 107 109 111 400 401 404 405 406 407 409 411 500 501 504 505 506 507 509 511 600 601 604 605 606 607 609 611 700 701 704 705 706 707 709 711 900 901 904 905 906 907 909 911]

t-ban numbers between 1,000 & 4,000: 0
[]

Counts of t-ban numbers up to one sextillion
Up to and including                     ten: 7
Up to and including             one hundred: 9
Up to and including            one thousand: 56
Up to and including            ten thousand: 56
Up to and including    one hundred thousand: 56
Up to and including             one million: 57
Up to and including             ten million: 392
Up to and including     one hundred million: 393
Up to and including             one billion: 2,801
Up to and including             ten billion: 19,600
Up to and including     one hundred billion: 19,601
Up to and including            one trillion: 140,000
Up to and including            ten trillion: 140,000
Up to and including    one hundred trillion: 140,000
Up to and including         one quadrillion: 140,001
Up to and including         ten quadrillion: 980,000
Up to and including one hundred quadrillion: 980,001
Up to and including         one quintillion: 7,000,000
Up to and including         ten quintillion: 7,000,000
Up to and including one hundred quintillion: 7,000,000
Up to and including          one sextillion: 7,000,000

============= subur-ban: =============
subur-ban numbers up to 1000: 35
[1 2 5 8 9 10 11 12 15 18 19 20 21 22 25 28 29 50 51 52 55 58 59 80 81 82 85 88 89 90 91 92 95 98 99]

subur-ban numbers between 1,000 & 4,000: 0
[]

Counts of subur-ban numbers up to one sextillion
Up to and including                     ten: 6
Up to and including             one hundred: 35
Up to and including            one thousand: 35
Up to and including            ten thousand: 35
Up to and including    one hundred thousand: 35
Up to and including             one million: 36
Up to and including             ten million: 215
Up to and including     one hundred million: 1,259
Up to and including             one billion: 1,259
Up to and including             ten billion: 1,259
Up to and including     one hundred billion: 1,259
Up to and including            one trillion: 1,259
Up to and including            ten trillion: 1,259
Up to and including    one hundred trillion: 1,259
Up to and including         one quadrillion: 1,259
Up to and including         ten quadrillion: 1,259
Up to and including one hundred quadrillion: 1,259
Up to and including         one quintillion: 1,259
Up to and including         ten quintillion: 1,259
Up to and including one hundred quintillion: 1,259
Up to and including          one sextillion: 1,259```

Note that the limit to one sextillion is somewhat arbitrary and is just to match the task parameters.

This will quite happily count *-bans up to one hundred centillion. (10305) It takes longer, but still on the order of seconds, not minutes.

```Counts of e-ban numbers up to one hundred centillion
...
Up to and including one hundred centillion: 35,184,372,088,831,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999```

## REXX

Programming note:   REXX has no shortcuts for   if   statements, so the multiple   if   statements weren't combined into one.

`/*REXX program to display eban numbers (those that don't have an "e" their English name)*/numeric digits 20                                /*support some gihugic numbers for pgm.*/parse arg \$                                      /*obtain optional arguments from the cL*/if \$=''  then \$= '1 1000   1000 4000   1 -10000   1 -100000   1 -1000000   1 -10000000'       do k=1  by 2  to words(\$)                  /*step through the list of numbers.    */      call banE  word(\$, k),  word(\$, k+1)       /*process the numbers, from low──►high.*/      end   /*k*/exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/banE: procedure; parse arg x,y,_;  z= reverse(x) /*obtain the number to be examined.    */      tell= y>=0                                 /*Is HI non-negative?  Display eban #s.*/      #= 0                                       /*the count of  eban  numbers (so far).*/           do j=x  to abs(y)                     /*probably process a range of numbers. */           if hasE(j)  then iterate              /*determine if the number has an  "e". */           #= # + 1                              /*bump the counter of  eban  numbers.  */           if tell  then _= _  j                 /*maybe add to a list of eban numbers. */           end   /*j*/      if _\==''  then say strip(_)               /*display the list  (if there is one). */      say;     say #   ' eban numbers found for: '   x   " "   y;     say copies('═', 105)      return/*──────────────────────────────────────────────────────────────────────────────────────*/hasE: procedure; parse arg x;  z= reverse(x)     /*obtain the number to be examined.    */        do k=1  by 3                             /*while there're dec. digit to examine.*/        @= reverse( substr(z, k, 3) )            /*obtain 3 dec. digs (a period) from Z.*/        if @=='   '           then return 0      /*we have reached the "end" of the num.*/        uni= right(@, 1)                         /*get units dec. digit of this period. */        if uni//2==1          then return 1      /*if an odd digit, then not an eban #. */        if uni==8             then return 1      /*if an  eight,      "   "   "   "  "  */        tens=substr(@, 2, 1)                     /*get tens  dec. digit of this period. */        if tens==1            then return 1      /*if teens,        then not an eban #. */        if tens==2            then return 1      /*if twenties,       "   "   "   "  "  */        if tens>6             then return 1      /*if 70s, 80s, 90s,  "   "   "   "  "  */        hun= left(@, 1)                          /*get hundreds dec. dig of this period.*/        if hun==0             then iterate       /*if zero, then there is more of number*/        if hun\==' '          then return 1      /*any hundrEd (not zero) has an  "e".  */        end   /*k*/                              /*A "period" is a group of 3 dec. digs */     return 0                                    /*in the number, grouped from the right*/`
output   when using the default inputs:
```2 4 6 30 32 34 36 40 42 44 46 50 52 54 56 60 62 64 66

19  eban numbers found for:  1   1000
═════════════════════════════════════════════════════════════════════════════════════════════════════════
2000 2002 2004 2006 2030 2032 2034 2036 2040 2042 2044 2046 2050 2052 2054 2056 2060 2062 2064 2066 4000

21  eban numbers found for:  1000   4000
═════════════════════════════════════════════════════════════════════════════════════════════════════════

79  eban numbers found for:  1   -10000
═════════════════════════════════════════════════════════════════════════════════════════════════════════

399  eban numbers found for:  1   -100000
═════════════════════════════════════════════════════════════════════════════════════════════════════════

399  eban numbers found for:  1   -1000000
═════════════════════════════════════════════════════════════════════════════════════════════════════════

1599  eban numbers found for:  1   -10000000
═════════════════════════════════════════════════════════════════════════════════════════════════════════
```

## Yabasic

Translation of: Go
`data 2, 100, truedata 1000, 4000, truedata 2, 1e4, falsedata 2, 1e5, falsedata 2, 1e6, falsedata 2, 1e7, falsedata 2, 1e8, falseREM data 2, 1e9, false  // it takes a lot of timedata 0, 0, false do    read start, ended, printable    if not start break      if start = 2 then        Print "eban numbers up to and including ", ended    else        Print "eban numbers between ", start, " and ", ended, " (inclusive):"    endif    count = 0    for i = start to ended step 2        b = int(i / 1000000000)        r = mod(i, 1000000000)        m = int(r / 1000000)        r = mod(i, 1000000)        t = int(r / 1000)        r = mod(r, 1000)        if m >= 30 and m <= 66 m = mod(m, 10)        if t >= 30 and t <= 66 t = mod(t, 10)        if r >= 30 and r <= 66 r = mod(r, 10)        if b = 0 or b = 2 or b = 4 or b = 6 then                         if m = 0 or m = 2 or m = 4 or m = 6 then                if t = 0 or t = 2 or t = 4 or t = 6 then                    if r = 0 or r = 2 or r = 4 or r = 6 then                        if printable Print i;                        count = count + 1                    endif                endif            endif        endif    next    if printable Print    Print "count = ", count, "\n"loop`