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Gapful numbers

From Rosetta Code
Task
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.


Evenly divisible   means divisible with   no   remainder.


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 tasks


Also see



AWK[edit]

 
# syntax: GAWK -f GAPFUL_NUMBERS.AWK
# converted from C++
BEGIN {
show_gapful(100,30)
show_gapful(1000000,15)
show_gapful(1000000000,10)
show_gapful(7123,25)
exit(0)
}
function is_gapful(n, m) {
m = n
while (m >= 10) {
m = int(m / 10)
}
return(n % ((n % 10) + 10 * (m % 10)) == 0)
}
function show_gapful(n, count,i) {
printf("first %d gapful numbers >= %d:",count,n)
for (i=0; i<count; n++) {
if (is_gapful(n)) {
printf(" %d",n)
i++
}
}
printf("\n")
}
 
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

C[edit]

 
#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;
}
 

Output :

[email protected]:~/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

C++[edit]

#include <iostream>
 
bool gapful(int n) {
int m = n;
while (m >= 10)
m /= 10;
return n % ((n % 10) + 10 * (m % 10)) == 0;
}
 
void show_gapful_numbers(int n, int count) {
std::cout << "First " << count << " gapful numbers >= " << n << ":\n";
for (int i = 0; i < count; ++n) {
if (gapful(n)) {
if (i != 0)
std::cout << ", ";
std::cout << n;
++i;
}
}
std::cout << '\n';
}
 
int main() {
show_gapful_numbers(100, 30);
show_gapful_numbers(1000000, 15);
show_gapful_numbers(1000000000, 10);
return 0;
}
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

C#[edit]

 
using System;
 
namespace GapfulNumbers
{
class Program
{
static void Main(string[] args)
{
Console.WriteLine("The first 30 gapful numbers are: ");
/* Starting at 100, find 30 gapful numbers */
FindGap(100, 30);
 
Console.WriteLine("The first 15 gapful numbers > 1,000,000 are: ");
FindGap(1000000, 15);
 
Console.WriteLine("The first 10 gapful numbers > 1,000,000,000 are: ");
FindGap(1000000000, 10);
 
Console.Read();
}
 
public static int firstNum(int n)
{
/*Divide by ten until the leading digit remains.*/
while (n >= 10)
{
n /= 10;
}
return (n);
}
 
public static int lastNum(int n)
{
/*Modulo gives you the last digit. */
return (n % 10);
}
 
static void FindGap(int n, int gaps)
{
int count = 0;
while (count < gaps)
{
 
/* We have to convert our first and last digits to strings to concatenate.*/
string concat = firstNum(n).ToString() + lastNum(n).ToString();
/* And then convert our concatenated string back to an integer. */
int i = Convert.ToInt32(concat);
 
/* Modulo with our new integer and output the result. */
if (n % i == 0)
{
Console.Write(n + " ");
count++;
n++;
}
else
{
n++;
continue;
}
}
}
}
}
 
 
 
Output:
The first 30 gapful numbers are:
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 

The first 15 gapful numbers > 1,000,000 are:
1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 

The first 10 gapful numbers > 1,000,000,000 are:
1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032


Crystal[edit]

Translation of: Ruby

With lazy iterator

struct Int
def gapful?
a = self.to_s.chars.map(&.to_i)
self % (a.first*10 + a.last) == 0
end
end
 
specs = {100 => 30, 1_000_000 => 15, 1_000_000_000 => 10, 7123 => 25}
 
specs.each do |start, count|
puts "first #{count} gapful numbers >= #{start}:"
puts (start..).each.select(&.gapful?).first(count).to_a, "\n"
end

Alternative

struct Int
def gapful?
a = self.to_s.chars.map(&.to_i)
self % (a.first*10 + a.last) == 0
end
end
 
specs = {100 => 30, 1_000_000 => 15, 1_000_000_000 => 10, 7123 => 25}
 
specs.each do |start, count|
puts "first #{count} gapful numbers >= #{start}:"
i, gapful = 0, [] of Int32
(start..).each { |n| n.gapful? && (gapful << n; i += 1); break if i == count }
puts gapful, "\n"
end
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]

D[edit]

Translation of: Go
import std.conv;
import std.stdio;
 
string commatize(ulong n) {
auto s = n.to!string;
auto le = s.length;
for (int i = le - 3; i >= 1; i -= 3) {
s = s[0..i]
~ ","
~ s[i..$];
}
return s;
}
 
void main() {
ulong[] starts = [cast(ulong)1e2, cast(ulong)1e6, cast(ulong)1e7, cast(ulong)1e9, 7123];
int[] counts = [30, 15, 15, 10, 25];
for (int i = 0; i < starts.length; i++) {
int count = 0;
auto j = starts[i];
ulong pow = 100;
while (true) {
if (j < pow * 10) {
break;
}
pow *= 10;
}
writefln("First %d gapful numbers starting at %s:", counts[i], commatize(starts[i]));
while (count < counts[i]) {
auto fl = (j / pow) * 10 + (j % 10);
if (j % fl == 0) {
write(j, ' ');
count++;
}
j++;
if (j >= 10 * pow) {
pow *= 10;
}
}
writeln("\n");
}
}
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

Factor[edit]

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
] [email protected]
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 }

Frink[edit]

 
// Create function to calculate gapful number
gapful[num,totalCounter] :=
{
// Display a line explaining the current calculation.
println["First $totalCounter gapful numbers over $num:"]
// Start a counter to compare with the total count.
counter = 0
while counter < totalCounter
{
numStr = toString[num] // Convert the integer to a string
gapfulNumStr = left[numStr,1] + right[numStr,1] // Concatenate the first and last character of the number to form a two digit number
gapfulNumInt = parseInt[gapfulNumStr] // Turn the concatenated string back into an integer.
// If the concatenated two digit integer divides into the current num variable with no remainder, print it to the list and increase our counter
if num mod gapfulNumInt == 0
{
print[numStr + " "]
counter = counter + 1
}
// Increase the current number for the next cycle.
num = num + 1
}
println[] // Linkbreak
}
 
// Print the first 30 gapful numbers over 100, the top 15 over 1,000,000 and the first 10 over 1,000,000,000.
gapful[100,30]
gapful[1000000,15]
gapful[1000000000,10]
 
Output:
First 30 gapful numbers over 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 over 1000000:
1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 
First 10 gapful numbers over 1000000000:
1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032 

Fōrmulæ[edit]

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[edit]

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")
}
}
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 

Haskell[edit]

{-# LANGUAGE NumericUnderscores #-}
 
gapful :: Int -> Bool
gapful n = n `rem` firstLastDigit == 0
where
firstLastDigit = read [head asDigits, last asDigits]
asDigits = show n
 
main :: IO ()
main = do
putStrLn $ "\nFirst 30 Gapful numbers >= 100 :\n" ++ r 30 [100,101..]
putStrLn $ "\nFirst 15 Gapful numbers >= 1,000,000 :\n" ++ r 15 [1_000_000,1_000_001..]
putStrLn $ "\nFirst 10 Gapful numbers >= 1,000,000,000 :\n" ++ r 10 [1_000_000_000,1_000_000_001..]
where r n = show . take n . filter gapful
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 >= 1,000,000 :
[1000000,1000005,1000008,1000010,1000016,1000020,1000021,1000030,1000032,1000034,1000035,1000040,1000050,1000060,1000065]

First 10 Gapful numbers >= 1,000,000,000 :
[1000000000,1000000001,1000000005,1000000008,1000000010,1000000016,1000000020,1000000027,1000000030,1000000032]


Or, defining the predicate in applicative terms, and wrapping the output:

import Data.List.Split (chunksOf)
import Data.List (intercalate)
 
 
----------------------- GAPFUL NUMBERS ---------------------
 
isGapful :: Int -> Bool
isGapful = (0 ==) . (<*>) rem (read . (<*>) [head, last] . pure . show)
 
 
---------------------------- TEST --------------------------
main :: IO ()
main =
mapM_
(putStrLn . showSample)
[ "First 30 gapful numbers >= 100"
, "First 15 Gapful numbers >= 1000000"
, "First 10 Gapful numbers >= 1000000000"
]
 
showSample :: String -> String
showSample k =
let ws = words k
in k <> ":\n\n" <>
unlines
(fmap (intercalate ", " . fmap show) $
chunksOf 5 $ take (read (ws !! 1)) [read (ws !! 5) :: Int ..])
Output:
First 30 gapful numbers >= 100:

100, 101, 102, 103, 104
105, 106, 107, 108, 109
110, 111, 112, 113, 114
115, 116, 117, 118, 119
120, 121, 122, 123, 124
125, 126, 127, 128, 129

First 15 Gapful numbers >= 1000000:

1000000, 1000001, 1000002, 1000003, 1000004
1000005, 1000006, 1000007, 1000008, 1000009
1000010, 1000011, 1000012, 1000013, 1000014

First 10 Gapful numbers >= 1000000000:

1000000000, 1000000001, 1000000002, 1000000003, 1000000004
1000000005, 1000000006, 1000000007, 1000000008, 1000000009

J[edit]

 
gapful =: 0 = (|~ ({.,{:)&.(10&#.inv))
 
task =: 100&$: :(dyad define) NB. MINIMUM task TALLY
gn =. y {. (#~ gapful&>) x + i. y * 25
assert 0 ~: {: gn
'The first ' , (": y) , ' gapful numbers exceeding ' , (":<:x) , ' are ' , (":gn)
)
 
   task 30
The first 30 gapful numbers exceeding 99 are 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
   1e6 task 15
The first 15 gapful numbers exceeding 999999 are 1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065
   1e9 task 10
The first 10 gapful numbers exceeding 999999999 are 1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032

Java[edit]

Translation of: D
import java.util.List;
 
public class GapfulNumbers {
private static String commatize(long n) {
StringBuilder sb = new StringBuilder(Long.toString(n));
int le = sb.length();
for (int i = le - 3; i >= 1; i -= 3) {
sb.insert(i, ',');
}
return sb.toString();
}
 
public static void main(String[] args) {
List<Long> starts = List.of((long) 1e2, (long) 1e6, (long) 1e7, (long) 1e9, (long) 7123);
List<Integer> counts = List.of(30, 15, 15, 10, 25);
for (int i = 0; i < starts.size(); ++i) {
int count = 0;
Long j = starts.get(i);
long pow = 100;
while (j >= pow * 10) {
pow *= 10;
}
System.out.printf("First %d gapful numbers starting at %s:\n", counts.get(i), commatize(starts.get(i)));
while (count < counts.get(i)) {
long fl = (j / pow) * 10 + (j % 10);
if (j % fl == 0) {
System.out.printf("%d ", j);
count++;
}
j++;
if (j >= 10 * pow) {
pow *= 10;
}
}
System.out.println('\n');
}
}
}
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 

JavaScript[edit]

Windows command line version[edit]

// Function to construct a new integer from the first and last digits of another
function gapfulness_divisor (number) {
var digit_string = number.toString(10)
var digit_count = digit_string.length
var first_digit = digit_string.substring(0, 1)
var last_digit = digit_string.substring(digit_count - 1)
return parseInt(first_digit.concat(last_digit), 10)
}
 
// Divisibility test to determine gapfulness
function is_gapful (number) {
return number % gapfulness_divisor(number) == 0
}
 
// Function to search for the least gapful number greater than a given integer
function next_gapful (number) {
do {
++number
} while (!is_gapful(number))
return number
}
 
// Constructor for a list of gapful numbers starting from given lower bound
function gapful_numbers (start, amount) {
var list = [], count = 0, number = start
if (amount > 0 && is_gapful(start)) {
list.push(start)
}
while (list.length < amount) {
number = next_gapful(number)
list.push(number)
}
return list
}
 
// Formatter for a comma-separated list of gapful numbers
function single_line_gapfuls (start, amount) {
var list = gapful_numbers(start, amount)
return list.join(", ")
}
 
// Windows console output wrapper
function print(message) {
WScript.StdOut.WriteLine(message)
}
 
// Main algorithm
 
function print_gapfuls_with_header(start, amount) {
print("First " + start + " gapful numbers starting at " + amount)
print(single_line_gapfuls(start, amount))
}
 
print_gapfuls_with_header(100, 30)
print_gapfuls_with_header(1000000, 15)
print_gapfuls_with_header(1000000000, 10)
Output:
First 100 gapful numbers starting at 30
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 1000000 gapful numbers starting at 15
1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065
First 1000000000 gapful numbers starting at 10
1000000000, 1000000001, 1000000005, 1000000008, 1000000010, 1000000016, 1000000020, 1000000027, 1000000030, 1000000032

ES6[edit]

(() => {
'use strict';
 
// ------------------ GAPFUL NUMBERS -------------------
 
// isGapful :: Int -> Bool
const isGapful = n =>
compose(
x => 0 === n % x,
JSON.parse,
concat,
ap([head, last]),
x => [x],
JSON.stringify
)(n);
 
// ----------------------- TEST ------------------------
const main = () =>
unlines([
'First 30 gapful numbers >= 100',
'First 15 Gapful numbers >= 1E6',
'First 10 Gapful numbers >= 1E9'
].map(k => {
const
ws = words(k),
mn = [1, 5].map(
i => JSON.parse(ws[i])
);
return k + ':\n\n' + showList(
take(mn[0])(
filterGen(isGapful)(
enumFrom(mn[1])
)
)
);
}));
 
 
// ----------------- GENERIC FUNCTIONS -----------------
 
// ap (<*>) :: [(a -> b)] -> [a] -> [b]
const ap = fs =>
// The sequential application of each of a list
// of functions to each of a list of values.
xs => fs.flatMap(f => xs.map(x => f(x)));
 
 
// chunksOf :: Int -> [a] -> [[a]]
const chunksOf = n =>
xs => enumFromThenTo(0)(n)(
xs.length - 1
).reduce(
(a, i) => a.concat([xs.slice(i, (n + i))]),
[]
);
 
 
// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
const compose = (...fs) =>
// A function defined by the right-to-left
// composition of all the functions in fs.
fs.reduce(
(f, g) => x => f(g(x)),
x => x
);
 
 
// concat :: [[a]] -> [a]
// concat :: [String] -> String
const concat = xs => (
ys => 0 < ys.length ? (
ys.every(Array.isArray) ? (
[]
) : ''
).concat(...ys) : ys
)(list(xs));
 
 
// enumFrom :: Enum a => a -> [a]
function* enumFrom(x) {
// A non-finite succession of enumerable
// values, starting with the value x.
let v = x;
while (true) {
yield v;
v = succ(v);
}
}
 
 
// enumFromThenTo :: Int -> Int -> Int -> [Int]
const enumFromThenTo = x1 =>
x2 => y => {
const d = x2 - x1;
return Array.from({
length: Math.floor(y - x2) / d + 2
}, (_, i) => x1 + (d * i));
};
 
 
// filterGen :: (a -> Bool) -> Gen [a] -> [a]
const filterGen = p => xs => {
function* go() {
let x = xs.next();
while (!x.done) {
let v = x.value;
if (p(v)) {
yield v;
}
x = xs.next();
}
}
return go(xs);
};
 
 
// head :: [a] -> a
const head = xs => (
ys => ys.length ? (
ys[0]
) : undefined
)(list(xs));
 
 
// last :: [a] -> a
const last = xs => (
// The last item of a list.
ys => 0 < ys.length ? (
ys.slice(-1)[0]
) : undefined
)(list(xs));
 
 
// list :: StringOrArrayLike b => b -> [a]
const list = xs =>
// xs itself, if it is an Array,
// or an Array derived from xs.
Array.isArray(xs) ? (
xs
) : Array.from(xs || []);
 
 
// showList :: [a] -> String
const showList = xs =>
unlines(
chunksOf(5)(xs).map(
ys => '\t' + ys.join(',')
)
) + '\n';
 
 
// succ :: Enum a => a -> a
const succ = x => {
const t = typeof x;
return 'number' !== t ? (() => {
const [i, mx] = [x, maxBound(x)].map(fromEnum);
return i < mx ? (
toEnum(x)(1 + i)
) : Error('succ :: enum out of range.')
})() : x < Number.MAX_SAFE_INTEGER ? (
1 + x
) : Error('succ :: Num out of range.')
};
 
 
// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = n =>
// The first n elements of a list,
// string of characters, or stream.
xs => 'GeneratorFunction' !== xs
.constructor.constructor.name ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));
 
 
// toEnum :: a -> Int -> a
const toEnum = e =>
// The first argument is a sample of the type
// allowing the function to make the right mapping
x => ({
'number': Number,
'string': String.fromCodePoint,
'boolean': Boolean,
'object': v => e.min + v
} [typeof e])(x);
 
 
// unlines :: [String] -> String
const unlines = xs =>
// A single string formed by the intercalation
// of a list of strings with the newline character.
xs.join('\n');
 
 
// words :: String -> [String]
const words = s =>
// List of space-delimited sub-strings.
s.split(/\s+/);
 
// MAIN ---
return main();
})();
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 >= 1E6:

    1000000,1000005,1000008,1000010,1000016
    1000020,1000021,1000030,1000032,1000034
    1000035,1000040,1000050,1000060,1000065

First 10 Gapful numbers >= 1E9:

    1000000000,1000000001,1000000005,1000000008,1000000010
    1000000016,1000000020,1000000027,1000000030,1000000032

Julia[edit]

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
 
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)

Kotlin[edit]

Translation of: Java
private fun commatize(n: Long): String {
val sb = StringBuilder(n.toString())
val le = sb.length
var i = le - 3
while (i >= 1) {
sb.insert(i, ',')
i -= 3
}
return sb.toString()
}
 
fun main() {
val starts = listOf(1e2.toLong(), 1e6.toLong(), 1e7.toLong(), 1e9.toLong(), 7123.toLong())
val counts = listOf(30, 15, 15, 10, 25)
for (i in starts.indices) {
var count = 0
var j = starts[i]
var pow: Long = 100
while (j >= pow * 10) {
pow *= 10
}
System.out.printf(
"First %d gapful numbers starting at %s:\n",
counts[i],
commatize(starts[i])
)
while (count < counts[i]) {
val fl = j / pow * 10 + j % 10
if (j % fl == 0L) {
System.out.printf("%d ", j)
count++
}
j++
if (j >= 10 * pow) {
pow *= 10
}
}
println('\n')
}
}
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 

Lambdatalk[edit]

 
{def gapfuls
{lambda {:n :i :N}
{if {>= :i :N}
then
else {if {= {% :n {W.first :n}{W.last :n}} 0}
then :n {gapfuls {+ :n 1} {+ :i 1} :N}
else {gapfuls {+ :n 1}  :i  :N}}}}}
-> gapfuls
 
{gapfuls 100 0 30}
-> 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
 
{gapfuls 1000000 0 15}
-> 1000000 1000005 1000008 1000010 1000016 1000020 1000021
1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065
 
{gapfuls 1000000000 0 10}
-> 1000000000 1000000001 1000000005 1000000008 1000000010
1000000016 1000000020 1000000027 1000000030 1000000032
 

Lua[edit]

Translation of: C
function generateGaps(start, count)
local counter = 0
local i = start
 
print(string.format("First %d Gapful numbers >= %d :", count, start))
 
while counter < count do
local str = tostring(i)
local denom = 10 * tonumber(str:sub(1, 1)) + (i % 10)
if i % denom == 0 then
print(string.format("%3d : %d", counter + 1, i))
counter = counter + 1
end
i = i + 1
end
end
 
generateGaps(100, 30)
print()
 
generateGaps(1000000, 15)
print()
 
generateGaps(1000000000, 15)
print()
Output:
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

min[edit]

Works with: min version 0.19.6
(() 0 shorten) :new
(((10 mod) (10 div)) cleave) :moddiv
((dup 0 ==) (pop new) 'moddiv 'cons linrec) :digits
(digits ((last 10 *) (first +)) cleave) :flnum
(mod 0 ==) :divisor?
(dup flnum divisor?) :gapful?
 
(
 :target :n 0 :count
"$1 gapful numbers starting at $2:" (target n) => % puts!
(count target <) (
(n gapful?) (
count succ @count
n print! " " print!
) when
n succ @n
) while
newline
) :show-gapfuls
 
100 30 show-gapfuls newline
1000000 15 show-gapfuls newline
1000000000 10 show-gapfuls
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

newLISP[edit]

; Part 1: Useful functions
 
;; Create an integer out of the first and last digits of a given integer
(define (first-and-last-digits number)
(local (digits first-digit last-digit)
(set 'digits (format "%d" number))
(set 'first-digit (first digits))
(set 'last-digit (last digits))
(int (append first-digit last-digit))))
 
;; Divisbility test
(define (divisible-by? num1 num2)
(zero? (% num1 num2)))
 
;; Gapfulness test
(define (gapful? number)
(divisible-by? number (first-and-last-digits number)))
 
;; Increment until a gapful number is found
(define (next-gapful-after number)
(do-until (gapful? number)
(++ number)))
 
;; Return a list of gapful numbers beyond some (excluded) lower limit.
(define (gapful-numbers quantity lower-limit)
(let ((gapfuls '()) (number lower-limit))
(dotimes (counter quantity)
(set 'number (next-gapful-after number))
(push number gapfuls))
(reverse gapfuls)))
 
;; Format a list of numbers together into decimal notation.
(define (format-many numbers)
(map (curry format "%d") numbers))
 
;; Format a list of integers on one line with commas
(define (format-one-line numbers)
(join (format-many numbers) ", "))
 
;; Display a quantity of gapful numbers beyond some (excluded) lower limit.
(define (show-gapfuls quantity lower-limit)
(println "The first " quantity " gapful numbers beyond " lower-limit " are:")
(println (format-one-line (gapful-numbers quantity lower-limit))))
 
; Part 2: Complete the task
(show-gapfuls 30 99)
(show-gapfuls 15 999999)
(show-gapfuls 10 999999999)
(exit)
Output:
The first 30 gapful numbers beyond 99 are:
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
The first 15 gapful numbers beyond 999999 are:
1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065
The first 10 gapful numbers beyond 999999999 are:
1000000000, 1000000001, 1000000005, 1000000008, 1000000010, 1000000016, 1000000020, 1000000027, 1000000030, 1000000032

Pascal[edit]

Translation of: Go
Works with: Free Pascal
Works with: Delphi

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

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.
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[edit]

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.
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[edit]

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;
}
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

Phix[edit]

Translation of: Go
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
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

Plain English[edit]

To run:
Start up.
Show the gapful numbers at various spots.
Wait for the escape key.
Shut down.
 
A digit is a number.
 
To get a digit of a number (last):
Privatize the number.
Divide the number by 10 giving a quotient and a remainder.
Put the remainder into the digit.
 
To get a digit of a number (first):
Privatize the number.
Loop.
Divide the number by 10 giving a quotient and a remainder.
Put the quotient into the number.
If the number is 0, put the remainder into the digit; exit.
Repeat.
 
To make a number from the first and last digits of another number:
Get a digit of the other number (first).
Get another digit of the other number (last).
Put the digit times 10 plus the other digit into the number.
 
To decide if a number is gapful:
Make another number from the first and last digits of the number.
If the number is evenly divisible by the other number, say yes.
Say no.
 
To show a number of the gapful numbers starting from another number:
Privatize the other number.
Put 0 into a gapful counter.
Loop.
If the other number is gapful, write "" then the other number then " " on the console without advancing; bump the gapful counter.
If the gapful counter is the number, break.
Bump the other number.
Repeat.
Write "" then the return byte on the console.
 
To show the gapful numbers at various spots:
Write "30 gapful numbers starting at 100:" on the console.
Show 30 of the gapful numbers starting from 100.
Write "15 gapful numbers starting at 1000000:" on the console.
Show 15 of the gapful numbers starting from 1000000.
Write "10 gapful numbers starting at 1000000000:" on the console.
Show 10 of the gapful numbers starting from 1000000000.
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

Python[edit]

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)))
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]

Raku[edit]

(formerly Perl 6)

Works with: Rakudo version 2019.07.1

Also test starting on a number that doesn't start with 1. Required to have titles, may as well make 'em noble. :-)

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];
}
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

REXX[edit]

/*REXX program computes and displays a series of gapful numbers starting at some number.*/
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
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

Ring[edit]

 
nr = 0
gapful1 = 99
gapful2 = 999999
gapful3 = 999999999
limit1 = 30
limit2 = 15
limit3 = 10
 
see "First 30 gapful numbers >= 100:" + nl
while nr < limit1
gapful1 = gapful1 + 1
gap1 = left((string(gapful1)),1)
gap2 = right((string(gapful1)),1)
gap = number(gap1 +gap2)
if gapful1 % gap = 0
nr = nr + 1
see "" + nr + ". " + gapful1 + nl
ok
end
see nl
 
see "First 15 gapful numbers >= 1000000:" + nl
nr = 0
while nr < limit2
gapful2 = gapful2 + 1
gap1 = left((string(gapful2)),1)
gap2 = right((string(gapful2)),1)
gap = number(gap1 +gap2)
if (nr < limit2) and gapful2 % gap = 0
nr = nr + 1
see "" + nr + ". " + gapful2 + nl
ok
end
see nl
 
see "First 10 gapful numbers >= 1000000000:" + nl
nr = 0
while nr < limit3
gapful3 = gapful3 + 1
gap1 = left((string(gapful3)),1)
gap2 = right((string(gapful3)),1)
gap = number(gap1 +gap2)
if (nr < limit2) and gapful3 % gap = 0
nr = nr + 1
see "" + nr + ". " + gapful3 + nl
ok
end
 
 
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

Ruby[edit]

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
 
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]

Sidef[edit]

Concept extended to other bases:

func is_gapful(n, base=10) {
n.is_div(base*floor(n / base**n.ilog(base)) + n%base)
}
 
var task = [
"(Required) The first %s gapful numbers (>= %s)", 30, 1e2, 10,
"(Required) The first %s gapful numbers (>= %s)", 15, 1e6, 10,
"(Required) The first %s gapful numbers (>= %s)", 10, 1e9, 10,
"(Extra) The first %s gapful numbers (>= %s)", 10, 987654321, 10,
"(Extra) The first %s gapful numbers (>= %s)", 10, 987654321, 12,
]
 
task.each_slice(4, {|title, n, from, b|
say sprintf("\n#{title} for base #{b}:", n, from.commify)
say (from..Inf -> lazy.grep{ is_gapful(_,b) }.first(n).join(' '))
})
Output:
(Required) The first 30 gapful numbers (>= 100) for base 10:
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

(Required) The first 15 gapful numbers (>= 1,000,000) for base 10:
1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065

(Required) The first 10 gapful numbers (>= 1,000,000,000) for base 10:
1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032

(Extra) The first 10 gapful numbers (>= 987,654,321) for base 10:
987654330 987654334 987654336 987654388 987654420 987654485 987654510 987654513 987654592 987654600

(Extra) The first 10 gapful numbers (>= 987,654,321) for base 12:
987654325 987654330 987654336 987654360 987654368 987654384 987654388 987654390 987654393 987654395

Swift[edit]

func isGapful(n: Int) -> Bool {
guard n > 100 else {
return true
}
 
let asString = String(n)
let div = Int("\(asString.first!)\(asString.last!)")!
 
return n % div == 0
}
 
let first30 = (100...).lazy.filter(isGapful).prefix(30)
let mil = (1_000_000...).lazy.filter(isGapful).prefix(15)
let bil = (1_000_000_000...).lazy.filter(isGapful).prefix(15)
 
print("First 30 gapful numbers: \(Array(first30))")
print("First 15 >= 1,000,000: \(Array(mil))")
print("First 15 >= 1,000,000,000: \(Array(bil))")
Output:
First 30 gapful numbers: [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 >= 1,000,000: [1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065]
First 15 >= 1,000,000,000: [1000000000, 1000000001, 1000000005, 1000000008, 1000000010, 1000000016, 1000000020, 1000000027, 1000000030, 1000000032, 1000000035, 1000000039, 1000000040, 1000000050, 1000000053]

Tcl[edit]

proc ungap n {
if {[string length $n] < 3} {
return $n
}
return [string index $n 0][string index $n end]
}
 
proc gapful n {
return [expr {0 == ($n % [ungap $n])}]
}
 
## --> list of gapful numbers >= n
proc GFlist {count n} {
set r {}
while {[llength $r] < $count} {
if {[gapful $n]} {
lappend r $n
}
incr n
}
return $r
}
 
proc show {count n} {
puts "The first $count gapful >= $n: [GFlist $count $n]"
}
 
show 30 100
show 15 1000000
show 10 1000000000
 
Output:
The first 30 gapful >= 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
The first 15 gapful >= 1000000: 1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065
The first 10 gapful >= 1000000000: 1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032

Visual Basic .NET[edit]

Translation of: C#
Module Module1
 
Function FirstNum(n As Integer) As Integer
REM Divide by ten until the leading digit remains.
While n >= 10
n /= 10
End While
Return n
End Function
 
Function LastNum(n As Integer) As Integer
REM Modulo gives you the last digit.
Return n Mod 10
End Function
 
Sub FindGap(n As Integer, gaps As Integer)
Dim count = 0
While count < gaps
Dim i = FirstNum(n) * 10 + LastNum(n)
 
REM Modulo with our new integer and output the result.
If n Mod i = 0 Then
Console.Write("{0} ", n)
count += 1
End If
 
n += 1
End While
Console.WriteLine()
Console.WriteLine()
End Sub
 
Sub Main()
Console.WriteLine("The first 30 gapful numbers are: ")
FindGap(100, 30)
 
Console.WriteLine("The first 15 gapful numbers > 1,000,000 are: ")
FindGap(1000000, 15)
 
Console.WriteLine("The first 10 gapful numbers > 1,000,000,000 are: ")
FindGap(1000000000, 10)
End Sub
 
End Module
Output:
The first 30 gapful numbers are:
100 105 108 110 120 121 130 132 135 140 143 156 160 168 175 180 200 220 225 231 240 242 253 270 300 315 330 341 360 400

The first 15 gapful numbers > 1,000,000 are:
1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065

The first 10 gapful numbers > 1,000,000,000 are:
1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032

Wren[edit]

Translation of: Go
Library: Wren-fmt
import "/fmt" for Fmt
 
var starts = [1e2, 1e6, 1e7, 1e9, 7123]
var counts = [30, 15, 15, 10, 25]
for (i in 0...starts.count) {
var count = 0
var j = starts[i]
var pow = 100
while (true) {
if (j < pow * 10) break
pow = pow * 10
}
System.print("First %(counts[i]) gapful numbers starting at %(Fmt.dc(0, starts[i]))")
while (count < counts[i]) {
var fl = (j/pow).floor*10 + (j % 10)
if (j%fl == 0) {
System.write("%(j) ")
count = count + 1
}
j = j + 1
if (j >= 10*pow) pow = pow * 10
}
System.print("\n")
}
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 

zkl[edit]

fcn gapfulW(start){	//--> iterator
[start..].tweak(
fcn(n){ if(n % (10*n.toString()[0] + n%10)) Void.Skip else n })
}
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");
}
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