Gapful numbers
You are encouraged to solve this task according to the task description, using any language you may know.
Numbers (positive integers expressed in base ten) that are (evenly) divisible by the number formed by the first and last digit are known as gapful numbers.
All one─ and two─digit numbers have this property and are trivially excluded. Only numbers ≥ 100 will be considered for this Rosetta Code task.
- Example
187 is a gapful number because it is evenly divisible by the number 17 which is formed by the first and last decimal digits of 187.
About 7.46% of positive integers are gapful.
- Task
-
- Generate and show all sets of numbers (below) on one line (horizontally) with a title, here on this page
- Show the first 30 gapful numbers
- Show the first 15 gapful numbers ≥ 1,000,000
- Show the first 10 gapful numbers ≥ 1,000,000,000
- Related task
- Also see
-
- The OEIS entry: A108343 gapful numbers.
- numbersaplenty gapful numbers
C
<lang C>
- include<stdio.h>
void generateGaps(unsigned long long int start,int count){
int counter = 0;
unsigned long long int i = start;
char str[100];
printf("\nFirst %d Gapful numbers >= %llu :\n",count,start);
while(counter<count){
sprintf(str,"%llu",i);
if((i%(10*(str[0]-'0') + i%10))==0L){
printf("\n%3d : %llu",counter+1,i);
counter++;
}
i++;
}
}
int main() {
unsigned long long int i = 100; int count = 0; char str[21];
generateGaps(100,30);
printf("\n");
generateGaps(1000000,15);
printf("\n");
generateGaps(1000000000,15);
printf("\n");
return 0;
} </lang> Output :
abhishek_ghosh@Azure:~/doodles$ ./a.out First 30 Gapful numbers >= 100 : 1 : 100 2 : 105 3 : 108 4 : 110 5 : 120 6 : 121 7 : 130 8 : 132 9 : 135 10 : 140 11 : 143 12 : 150 13 : 154 14 : 160 15 : 165 16 : 170 17 : 176 18 : 180 19 : 187 20 : 190 21 : 192 22 : 195 23 : 198 24 : 200 25 : 220 26 : 225 27 : 231 28 : 240 29 : 242 30 : 253 First 15 Gapful numbers >= 1000000 : 1 : 1000000 2 : 1000005 3 : 1000008 4 : 1000010 5 : 1000016 6 : 1000020 7 : 1000021 8 : 1000030 9 : 1000032 10 : 1000034 11 : 1000035 12 : 1000040 13 : 1000050 14 : 1000060 15 : 1000065 First 15 Gapful numbers >= 1000000000 : 1 : 1000000000 2 : 1000000001 3 : 1000000005 4 : 1000000008 5 : 1000000010 6 : 1000000016 7 : 1000000020 8 : 1000000027 9 : 1000000030 10 : 1000000032 11 : 1000000035 12 : 1000000039 13 : 1000000040 14 : 1000000050 15 : 1000000053
Factor
<lang factor>USING: formatting kernel lists lists.lazy math math.functions math.text.utils sequences ;
- gapful? ( n -- ? )
dup 1 digit-groups [ first ] [ last 10 * + ] bi divisor? ;
30 100 15 1,000,000 10 1,000,000,000 [
2dup lfrom [ gapful? ] lfilter ltake list>array "%d gapful numbers starting at %d:\n%[%d, %]\n\n" printf
] 2tri@</lang>
- Output:
30 gapful numbers starting at 100:
{ 100, 105, 108, 110, 120, 121, 130, 132, 135, 140, 143, 150, 154, 160, 165, 170, 176, 180, 187, 190, 192, 195, 198, 200, 220, 225, 231, 240, 242, 253 }
15 gapful numbers starting at 1000000:
{ 1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065 }
10 gapful numbers starting at 1000000000:
{ 1000000000, 1000000001, 1000000005, 1000000008, 1000000010, 1000000016, 1000000020, 1000000027, 1000000030, 1000000032 }
Fōrmulæ
In this page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Go
<lang go>package main
import "fmt"
func commatize(n uint64) string {
s := fmt.Sprintf("%d", n)
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
return s
}
func main() {
starts := []uint64{1e2, 1e6, 1e7, 1e9, 7123}
counts := []int{30, 15, 15, 10, 25}
for i := 0; i < len(starts); i++ {
count := 0
j := starts[i]
pow := uint64(100)
for {
if j < pow*10 {
break
}
pow *= 10
}
fmt.Printf("First %d gapful numbers starting at %s:\n", counts[i], commatize(starts[i]))
for count < counts[i] {
fl := (j/pow)*10 + (j % 10)
if j%fl == 0 {
fmt.Printf("%d ", j)
count++
}
j++
if j >= 10*pow {
pow *= 10
}
}
fmt.Println("\n")
}
}</lang>
- Output:
First 30 gapful numbers starting at 100: 100 105 108 110 120 121 130 132 135 140 143 150 154 160 165 170 176 180 187 190 192 195 198 200 220 225 231 240 242 253 First 15 gapful numbers starting at 1,000,000: 1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 First 15 gapful numbers starting at 10,000,000: 10000000 10000001 10000003 10000004 10000005 10000008 10000010 10000016 10000020 10000030 10000032 10000035 10000040 10000050 10000060 First 10 gapful numbers starting at 1,000,000,000: 1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032 First 25 gapful numbers starting at 7,123: 7125 7140 7171 7189 7210 7272 7275 7280 7296 7350 7373 7420 7425 7474 7488 7490 7560 7575 7630 7632 7676 7700 7725 7770 7777
Julia
<lang julia>using Lazy, Formatting
firstlast(a) = 10 * a[end] + a[1] isgapful(n) = (d = digits(n); length(d) < 3 || (m = firstlast(d)) != 0 && mod(n, m) == 0) gapfuls(start) = filter(isgapful, Lazy.range(start))
for (x, n) in [(100, 30), (1_000_000, 15), (1_000_000_000, 10)]
println("First $n gapful numbers starting at ", format(x, commas=true), ":\n",
take(n, gapfuls(x)))
end
</lang>
- Output:
First 30 gapful numbers starting at 100: (100 105 108 110 120 121 130 132 135 140 143 150 154 160 165 170 176 180 187 190 192 195 198 200 220 225 231 240 242 253) First 15 gapful numbers starting at 1,000,000: (1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065) First 10 gapful numbers starting at 1,000,000,000: (1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032)
Pascal
Now using using en passant updated MOD-values. Only recognizable for huge amounts of tests 100|74623687 ( up to 1 billion )-> takes 1.845s instead of 11.25s <lang pascal>program gapful; {$IFDEF FPC}
{$MODE DELPHI}{$OPTIMIZATION ON,ALL}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
uses
sysutils,// IntToStr strUtils;// Numb2USA aka commatize
const
cIdx = 5;
starts : array [0..cIdx-1] of Uint64
= (100,1000*1000, 10*1000*1000,1000*1000*1000, 7123);
counts : array [0..cIdx-1] of Uint64 = (30, 15,15, 10, 25);
//100| 74623687 => 1000*1000*1000
//100| 746236131 => 10*1000*1000*1000
//100|7462360431 =>100*1000*1000*1000
Base = 10;
var
ModsHL : array[0..99] of NativeUint; Pow10 : Uint64; //global, seldom used countLmt : NativeUint;//Uint64; only for extreme counting
procedure OutHeader(i: NativeInt); Begin
writeln('First ',counts[i],', gapful numbers starting at ',
Numb2USA(IntToStr(starts[i])));
end;
procedure OutNum(n:Uint64); Begin
write(' ',n);
end;
procedure InitMods(n:Uint64;H_dgt:NativeUint); //calculate first mod of n, when it reaches n var
i,j : NativeInt;
Begin
j := H_dgt; //= H_dgt+i For i := 0 to Base-1 do Begin ModsHL[j] := n MOD j; inc(n); inc(j); end;
end;
procedure InitMods2(n:Uint64;H_dgt,L_Dgt:NativeUint); //calculate first mod of n, when it reaches n //beware, that the lower n are reached in the next base round var
i,j : NativeInt;
Begin
j := H_dgt; n := n-L_Dgt; For i := 0 to L_Dgt-1 do Begin ModsHL[j] := (n+base) MOD j; inc(n); inc(j); end; For i := L_Dgt to Base-1 do Begin ModsHL[j] := n MOD j; inc(n); inc(j); end;
end;
procedure Main(TestNum:Uint64;Cnt:NativeUint); var
LmtNextNewHiDgt: Uint64; tmp,LowDgt,GapNum : NativeUint;
Begin
countLmt := cnt; Pow10 := Base*Base; LmtNextNewHiDgt := Base*Pow10; while LmtNextNewHiDgt <= TestNum do Begin Pow10 := LmtNextNewHiDgt; LmtNextNewHiDgt *= Base; end; LowDgt := TestNum MOD Base; GapNum := TestNum DIV Pow10; LmtNextNewHiDgt := (GapNum+1)*Pow10; GapNum := Base*GapNum; IF LowDgt <> 0 then InitMods2(TestNum,GapNum,LowDgt) else InitMODS(TestNum,GapNum);
GapNum += LowDgt; repeat
// if TestNum MOD (GapNum) = 0 then
if ModsHL[GapNum]=0 then
Begin
tmp := countLmt-1;
IF tmp < 32 then
OutNum(TestNum);
countLmt := tmp;
// Test and BREAK only if something has changed
IF tmp = 0 then
BREAK;
end;
tmp := Base + ModsHL[GapNum];
//translate into "if-less" version 3.35s -> 1.85s
//bad branch prediction :-(
//if tmp >= GapNum then tmp -= GapNum;
tmp -= (-ORD(tmp >=GapNum) AND GapNum);
ModsHL[GapNum]:= tmp;
TestNum += 1;
tmp := LowDgt+1;
GapNum+=1;
IF tmp >= Base then
Begin
tmp := 0;
GapNum -= Base;
end;
LowDgt := tmp;
//next Hi Digit
if TestNum >= LmtNextNewHiDgt then
Begin
LowDgt := 0;
GapNum +=Base;
LmtNextNewHiDgt += Pow10;
//next power of 10
if GapNum >= Base*Base then
Begin
Pow10 *= Base;
LmtNextNewHiDgt := 2*Pow10;
GapNum := Base;
end;
initMods(TestNum,GapNum);
end;
until false;
end;
var
i : integer;
Begin
for i := 0 to High(starts) do Begin OutHeader(i); Main(starts[i],counts[i]); writeln(#13#10); end;
end.</lang>
- Output:
First 30, gapful numbers starting at 100 100 105 108 110 120 121 130 132 135 140 143 150 154 160 165 170 176 180 187 190 192 195 198 200 220 225 231 240 242 253 First 15, gapful numbers starting at 1,000,000 1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 First 15, gapful numbers starting at 10,000,000 10000000 10000001 10000003 10000004 10000005 10000008 10000010 10000016 10000020 10000030 10000032 10000035 10000040 10000050 10000060 First 10, gapful numbers starting at 1,000,000,000 1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032 First 25, gapful numbers starting at 7,123 7125 7140 7171 7189 7210 7272 7275 7280 7296 7350 7373 7420 7425 7474 7488 7490 7560 7575 7630 7632 7676 7700 7725 7770 7777 _____ First 74623687, gapful numbers starting at 100 999998976 999999000 999999090 999999091 999999099 999999152 999999165 999999180 999999270 999999287 999999333 999999354 999999355 999999360 999999448 999999450 999999456 999999540 999999545 999999612 999999630 999999720 999999735 999999810 999999824 999999900 999999925 999999936 999999938 999999990 1000000000 real 0m1,845s start | count //100| 74623687 => 1000*1000*1000 //100| 746236131 => 10*1000*1000*1000 //100|7462360431 =>100*1000*1000*1000
only counting
<lang pascal>program gapful; {$IFDEF FPC}
{$MODE DELPHI}{$OPTIMIZATION ON,ALL}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF} uses
sysutils,// IntToStr strUtils;// Numb2USA aka commatize
var
LCMsHL : array of NativeInt;
function GCD(a, b: Int64): Int64; var
temp: Int64;
begin
while b <> 0 do begin temp := b; b := a mod b; a := temp end; result := a
end;
function LCM(a, b: Int64): Int64; begin
LCM := (a DIV GCD(a,b)) * b;
end;
procedure InitLCM(Base:NativeInt); var
i : integer;
Begin
For i := Base to (Base*Base-1) do LCMsHL[i] := LCM(i,Base);
end;
function CountGapFul(H_Digit,Base:NativeInt;PotBase:Uint64):Uint64; //Counts gapfulnumbers [n*PotBase..(n+1)*PotBase -1] ala [100..199] var
EndDgt,Dgt : NativeInt; P,k,lmt,sum,dSum: UInt64;
begin
P := PotBase*H_Digit;
lmt := P+PotBase-1;
Dgt := H_Digit*Base;
sum := (PotBase-1) DIV dgt +1;
For EndDgt := 1 to Base-1 do
Begin
inc(Dgt);
//search start
//first value divisible by dgt
k := p-(p MOD dgt)+ dgt;
//value divisible by dgt ending in the right digit
while (k mod Base) <> EndDgt do
inc(k,dgt);
IF k> lmt then
continue;
//one found +1
//count the occurences in (lmt-k)
dSum := (lmt-k) DIV LCMsHL[dgt] +1;
inc(sum,dSum);
//writeln(dgt:5,k:21,dSum:21,Sum:21);
end;
//writeln(p:21,Sum:21);
CountGapFul := sum;
end;
procedure Main(Base:NativeUInt); var
i : NativeUInt; pot,total,lmt: Uint64;//High(Uint64) = 2^64-1
Begin
lmt := High(pot) DIV Base;
pot := sqr(Base);//"100" in Base
setlength(LCMsHL,pot);
InitLCM(Base);
total := 0;
repeat
IF pot > lmt then
break;
For i := 1 to Base-1 do //ala 100..199 ,200..299,300..399,..,900..999
inc(total,CountGapFul(i,base,pot));
pot *= Base;
writeln('Total [',sqr(Base),'..',Numb2USA(IntToStr(pot)),'] : ',Numb2USA(IntToStr(total+1)));
until false;
setlength(LCMsHL,0);
end;
BEGIN
Main(10); Main(100);
END.</lang>
- Output:
Base :10 Total [100..1,000] : 77 Total [100..10,000] : 765 Total [100..100,000] : 7,491 Total [100..1,000,000] : 74,665 Total [100..10,000,000] : 746,286 Total [100..100,000,000] : 7,462,438 Total [100..1,000,000,000] : 74,623,687 Total [100..10,000,000,000] : 746,236,131 Total [100..100,000,000,000] : 7,462,360,431 Total [100..1,000,000,000,000] : 74,623,603,381 Total [100..10,000,000,000,000] : 746,236,032,734 Total [100..100,000,000,000,000] : 7,462,360,326,234 Total [100..1,000,000,000,000,000] : 74,623,603,260,964 Total [100..10,000,000,000,000,000] : 746,236,032,608,141 Total [100..100,000,000,000,000,000] : 7,462,360,326,079,810 Total [100..1,000,000,000,000,000,000] : 74,623,603,260,796,424 Total [100..10,000,000,000,000,000,000] : 746,236,032,607,962,357 Base :100 Total [10000..1,000,000] : 6,039 Total [10000..100,000,000] : 251,482 Total [10000..10,000,000,000] : 24,738,934 Total [10000..1,000,000,000,000] : 2,473,436,586 Total [10000..100,000,000,000,000] : 247,343,160,115 Total [10000..10,000,000,000,000,000] : 24,734,315,489,649 Total [10000..1,000,000,000,000,000,000] : 2,473,431,548,401,507
Perl
<lang perl>use strict; use warnings; use feature 'say';
sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
sub is_gapful { my $n = shift; 0 == $n % join(, (split //, $n)[0,-1]) }
use constant Inf => 1e10; for ([1e2, 30], [1e6, 15], [1e9, 10], [7123, 25]) {
my($start, $count) = @$_;
printf "\nFirst $count gapful numbers starting at %s:\n", comma $start;
my $n = 0; my $g = ;
$g .= do { $n < $count ? (is_gapful($_) and ++$n and "$_ ") : last } for $start .. Inf;
say $g;
}</lang>
- Output:
First 30 gapful numbers starting at 100: 100 105 108 110 120 121 130 132 135 140 143 150 154 160 165 170 176 180 187 190 192 195 198 200 220 225 231 240 242 253 First 15 gapful numbers starting at 1,000,000: 1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 First 10 gapful numbers starting at 1,000,000,000: 1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032 First 25 gapful numbers starting at 7,123: 7125 7140 7171 7189 7210 7272 7275 7280 7296 7350 7373 7420 7425 7474 7488 7490 7560 7575 7630 7632 7676 7700 7725 7770 7777
Perl 6
Also test starting on a number that doesn't start with 1. Required to have titles, may as well make 'em noble. :-)
<lang perl6>use Lingua::EN::Numbers;
for (1e2, 30, 1e6, 15, 1e9, 10, 7123, 25)».Int -> $start, $count {
put "\nFirst $count gapful numbers starting at {comma $start}:\n" ~
<Sir Lord Duke King>.pick ~ ": ", ~
($start..*).grep( { $_ %% .comb[0, *-1].join } )[^$count];
}</lang>
- Output:
First 30 gapful numbers starting at 100: Sir: 100 105 108 110 120 121 130 132 135 140 143 150 154 160 165 170 176 180 187 190 192 195 198 200 220 225 231 240 242 253 First 15 gapful numbers starting at 1,000,000: Duke: 1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 First 10 gapful numbers starting at 1,000,000,000: King: 1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032 First 25 gapful numbers starting at 7,123: King: 7125 7140 7171 7189 7210 7272 7275 7280 7296 7350 7373 7420 7425 7474 7488 7490 7560 7575 7630 7632 7676 7700 7725 7770 7777
Phix
<lang Phix>constant starts = {1e2, 1e6, 1e7, 1e9, 7123},
counts = {30, 15, 15, 10, 25}
for i=1 to length(starts) do
integer count = counts[i],
j = starts[i],
pow = 100
while j>=pow*10 do pow *= 10 end while
printf(1,"First %d gapful numbers starting at %,d: ", {count, j})
while count do
integer fl = floor(j/pow)*10 + remainder(j,10)
if remainder(j,fl)==0 then
printf(1,"%d ", j)
count -= 1
end if
j += 1
if j>=10*pow then
pow *= 10
end if
end while
printf(1,"\n")
end for</lang>
- Output:
First 30 gapful numbers starting at 100: 100 105 108 110 120 121 130 132 135 140 143 150 154 160 165 170 176 180 187 190 192 195 198 200 220 225 231 240 242 253 First 15 gapful numbers starting at 1,000,000: 1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 First 15 gapful numbers starting at 10,000,000: 10000000 10000001 10000003 10000004 10000005 10000008 10000010 10000016 10000020 10000030 10000032 10000035 10000040 10000050 10000060 First 10 gapful numbers starting at 1,000,000,000: 1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032 First 25 gapful numbers starting at 7,123: 7125 7140 7171 7189 7210 7272 7275 7280 7296 7350 7373 7420 7425 7474 7488 7490 7560 7575 7630 7632 7676 7700 7725 7770 7777
Python
<lang python>from itertools import islice, count for start, n in [(100, 30), (1_000_000, 15), (1_000_000_000, 10)]:
print(f"\nFirst {n} gapful numbers from {start:_}")
print(list(islice(( x for x in count(start)
if (x % (int(str(x)[0]) * 10 + (x % 10)) == 0) )
, n)))</lang>
- Output:
First 30 gapful numbers from 100 [100, 105, 108, 110, 120, 121, 130, 132, 135, 140, 143, 150, 154, 160, 165, 170, 176, 180, 187, 190, 192, 195, 198, 200, 220, 225, 231, 240, 242, 253] First 15 gapful numbers from 1_000_000 [1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065] First 10 gapful numbers from 1_000_000_000 [1000000000, 1000000001, 1000000005, 1000000008, 1000000010, 1000000016, 1000000020, 1000000027, 1000000030, 1000000032]
REXX
<lang rexx>/*REXX program computes and displays gapful numbers and also palindromic gapful numbers.*/ numeric digits 20 /*ensure enough decimal digits gapfuls.*/ parse arg gapfuls /*obtain optional arguments from the CL*/ if gapfuls= then gapfuls= 30 25@7123 15@1000000 10@1000000000 /*assume defaults. */
do until gapfuls=; parse var gapfuls stuff gapfuls; call gapful stuff
end /*until*/
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ gapful: procedure; parse arg n '@' sp; #= 0; if sp== then sp= 100
say center(' 'n " gapful numbers starting at: " sp' ', 125, "═")
$= /*initialize $ list.*/
do j=sp until #==n /*SP: start point. */
parse var j a 2 -1 b /*get 1st & last dig*/
if j // (a||b) \== 0 then iterate /*perform ÷ into J.*/
#= # + 1; $= $ j /*bump #; append──►$*/
end /*j*/
say strip($); say; return</lang>
- output when using the default inputs:
(Shown at 5/6 size.)
═══════════════════════════════════════════ 30 gapful numbers starting at: 100 ════════════════════════════════════════════ 100 105 108 110 120 121 130 132 135 140 143 150 154 160 165 170 176 180 187 190 192 195 198 200 220 225 231 240 242 253 ═══════════════════════════════════════════ 25 gapful numbers starting at: 7123 ═══════════════════════════════════════════ 7125 7140 7171 7189 7210 7272 7275 7280 7296 7350 7373 7420 7425 7474 7488 7490 7560 7575 7630 7632 7676 7700 7725 7770 7777 ═════════════════════════════════════════ 15 gapful numbers starting at: 1000000 ══════════════════════════════════════════ 1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 ════════════════════════════════════════ 10 gapful numbers starting at: 1000000000 ════════════════════════════════════════ 1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032
Ruby
<lang ruby>class Integer
def gapful? a = digits self % (a.last*10 + a.first) == 0 end
end
specs = {100 => 30, 1_000_000 => 15, 1_000_000_000 => 10, 7123 => 25}
specs.each do |start, num|
puts "first #{num} gapful numbers >= #{start}:"
p (start..).lazy.select(&:gapful?).take(num).to_a
end </lang>
- Output:
first 30 gapful numbers >= 100: [100, 105, 108, 110, 120, 121, 130, 132, 135, 140, 143, 150, 154, 160, 165, 170, 176, 180, 187, 190, 192, 195, 198, 200, 220, 225, 231, 240, 242, 253] first 15 gapful numbers >= 1000000: [1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065] first 10 gapful numbers >= 1000000000: [1000000000, 1000000001, 1000000005, 1000000008, 1000000010, 1000000016, 1000000020, 1000000027, 1000000030, 1000000032] first 25 gapful numbers >= 7123: [7125, 7140, 7171, 7189, 7210, 7272, 7275, 7280, 7296, 7350, 7373, 7420, 7425, 7474, 7488, 7490, 7560, 7575, 7630, 7632, 7676, 7700, 7725, 7770, 7777]
zkl
<lang zkl>fcn gapfulW(start){ //--> iterator
[start..].tweak(
fcn(n){ if(n % (10*n.toString()[0] + n%10)) Void.Skip else n })
}</lang> <lang zkl>foreach n,z in
( T( T(100, 30), T(1_000_000, 15), T(1_000_000_000, 10), T(7_123,25) )){
println("First %d gapful numbers starting at %,d:".fmt(z,n));
gapfulW(n).walk(z).concat(", ").println("\n");
}</lang>
- Output:
First 30 gapful numbers starting at 100: 100, 105, 108, 110, 120, 121, 130, 132, 135, 140, 143, 150, 154, 160, 165, 170, 176, 180, 187, 190, 192, 195, 198, 200, 220, 225, 231, 240, 242, 253 First 15 gapful numbers starting at 1,000,000: 1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065 First 10 gapful numbers starting at 1,000,000,000: 1000000000, 1000000001, 1000000005, 1000000008, 1000000010, 1000000016, 1000000020, 1000000027, 1000000030, 1000000032 First 25 gapful numbers starting at 7,123: 7125, 7140, 7171, 7189, 7210, 7272, 7275, 7280, 7296, 7350, 7373, 7420, 7425, 7474, 7488, 7490, 7560, 7575, 7630, 7632, 7676, 7700, 7725, 7770, 7777