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



11l[edit]

Translation of: Python
L(start, n) [(100, 30), (1'000'000, 15), (1'000'000'000, 10)]
   print("\nFirst "n‘ gapful numbers from ’start)
   [Int] l
   L(x) start..
      I x % (Int(String(x)[0]) * 10 + (x % 10)) == 0
         l.append(x)
         I l.len == n
            L.break
   print(l)
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 1000000
[1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065]

First 10 gapful numbers from 1000000000
[1000000000, 1000000001, 1000000005, 1000000008, 1000000010, 1000000016, 1000000020, 1000000027, 1000000030, 1000000032]

Ada[edit]

with Ada.Text_IO;  use Ada.Text_IO;

procedure Gapful_Numbers is

   function Divisor (N : in Positive) return Positive is
      NN : Positive := N;
   begin
      while NN >= 10 loop
         NN := NN / 10;
      end loop;
      return 10 * NN + (N mod 10);
   end Divisor;

   function Is_Gapful (N : in Positive) return Boolean is
   begin
      return N mod Divisor (N) = 0;
   end Is_Gapful;

   procedure Find_Gapful (Count : in Positive; From : in Positive) is
      Found : Natural := 0;
   begin
      for Candidate in From .. Positive'Last loop
         if Is_Gapful (Candidate) then
            Put (Candidate'Image);
            Found := Found + 1;
            exit when Found = Count;
         end if;
      end loop;
      New_Line;
   end Find_Gapful;

begin
   Put_Line ("First 30 gapful numbers over 100:");
   Find_Gapful (From => 100, Count => 30);
   New_Line;

   Put_Line ("First 15 gapful numbers over 1_000_000:");
   Find_Gapful (From => 1_000_000, Count => 15);
   New_Line;

   Put_Line ("First 10 gapful numbers over 1_000_000_000:");
   Find_Gapful (From => 1_000_000_000, Count => 10);
   New_Line;
end Gapful_Numbers;
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 1_000_000:
 1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065

First 10 gapful numbers over 1_000_000_000:
 1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032

ALGOL 68[edit]

BEGIN # find some gapful numbers - numbers divisible by f*10 + b #
      # where f is the first digit and b is the final digit      #
    # unary GAPFUL operator - returns TRUE  if n is gapful       #
    #                                 FALSE otherwise            #
    OP   GAPFUL = ( INT n )BOOL:
         BEGIN
            INT abs n  = ABS n;
            INT back   = abs n MOD 10; # final digit             #
            INT front := abs n OVER 10;
            WHILE front > 9 DO front OVERAB 10 OD;
            abs n MOD ( ( front * 10 ) + back ) = 0
         END; # GAPFUL #
    # dyadic GAPFUL operator - returns an array of n gapful      #
    # numbers starting from first                                #
    PRIO GAPFUL = 9;
    OP   GAPFUL = ( INT n, INT first )[]INT:
         BEGIN
            [ 1 : n ]INT result;
            INT g pos := 0;
            FOR i FROM first WHILE g pos < n DO
                IF GAPFUL i THEN result[ g pos +:= 1 ] := i FI
            OD;
            result
         END; # GAPFUL #
    # prints a sequence of gapful numbers                        #
    PROC print gapful = ( []INT seq, INT start )VOID:
         BEGIN
            print( ( "First ", whole( ( UPB seq + 1 ) - LWB seq, 0 )
                   , " gapful numbers starting from ", whole( start, 0 )
                   , ":", newline
                   )
                 );
            FOR i FROM LWB seq TO UPB seq DO print( ( " ", whole( seq[ i ], 0 ) ) ) OD;
            print( ( newline ) )
         END; # print gapful #
    print gapful( 30 GAPFUL            100,           100 );
    print gapful( 15 GAPFUL      1 000 000,     1 000 000 );
    print gapful( 10 GAPFUL  1 000 000 000, 1 000 000 000 )
END
Output:
First 30 gapful numbers starting 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 starting from 1000000:
 1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065
First 10 gapful numbers starting from 1000000000:
 1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032

AppleScript[edit]

on isGapful(n)
    set units to n mod 10
    set temp to n div 10
    repeat until (temp < 10)
        set temp to temp div 10
    end repeat
    
    return (n mod (temp * 10 + units) = 0)
end isGapful

-- Task code:
on getGapfuls(n, q)
    set collector to {}
    repeat until ((count collector) = q)
        if (isGapful(n)) then set end of collector to n
        set n to n + 1
    end repeat
    return collector
end getGapfuls

local output, astid
set output to {}
set astid to AppleScript's text item delimiters
set AppleScript's text item delimiters to "  "
set end of output to "First 30 gapful numbers ≥ 100:" & linefeed & getGapfuls(100, 30)
set end of output to "First 15 gapful numbers ≥ 1,000,000:" & linefeed & getGapfuls(1000000, 15)
set end of output to "First 10 gapful numbers ≥ 100,000,000:" & linefeed & getGapfuls(1.0E+9, 10)
set AppleScript's text item delimiters to linefeed & linefeed
set output to output as text
set AppleScript's text item delimiters to astid
return output
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:
1.0E+9  1.000000001E+9  1.000000005E+9  1.000000008E+9  1.00000001E+9  1.000000016E+9  1.00000002E+9  1.000000027E+9  1.00000003E+9  1.000000032E+9"

Applesoft BASIC[edit]

Translation of: C

Numbers are printed digit by digit so that we can see what they actually are because large numbers would normally be printed in scientific notation.

 100 START = 100
 110 COUNT = 30
 120  GOSUB 230"GENERATE GAPS"
 130  PRINT 
 140 START = 1000000
 150 COUNT = 15
 160  GOSUB 230"GENERATE GAPS"
 170  PRINT 
 180 START = 1000000000
 190 COUNT = 15
 200  GOSUB 230"GENERATE GAPS"
 210  PRINT 
 220  END 
 230  GOSUB 340"PRINT START"
 240  PRINT " :"M$
 250  LET C = 0
 260  FOR START = START TO 1E38
 270      LET I$ =  STR$ (START)
 280      LET N =  VAL ( LEFT$ (I$,1) +  STR$ (START -  INT (START / B) * B))
 290      LET GAPFUL = START -  INT (START / N) * N = 0
 300      IF GAPFUL THEN  GOSUB 370"PRINT GAPFUL"
 310      IF C = COUNT THEN  RETURN
 320  NEXT START
 330  STOP
 340  LET M$ =  CHR$ (13)
 350  PRINT M$"FIRST "COUNT" ";
 360  PRINT "GAPFUL NUMBERS >=";
 370  LET C = C + 1
 380  LET B = 10
 390  LET P$ = ""
 400  FOR P = START TO 0 STEP 0
 410      LET Q =  INT (P / B)
 420      LET P$ =  STR$ (P - Q * B) + P$
 430      LET P = Q
 440  NEXT P
 450  PRINT MID$(M$ + " " + P$, 2 - (POS(0) + LEN(P$) + 1 > PEEK(33))) ;
 460  RETURN

Arturo[edit]

gapful?: function [n][
    s: to :string n
    divisor: to :integer (first s) ++ last s
    0 = n % divisor
]

specs: [100  30, 1000000  15, 1000000000 10, 7123 25]

loop specs [start,count][
    print "----------------------------------------------------------------"
    print ["first" count "gapful numbers starting from" start]
    print "----------------------------------------------------------------"
    i: start
    took: 0
    while [took < count][
        if gapful? i [
            prints i
            prints " "
            took: took + 1
        ]
        i: i + 1
    ]
    print "\n"
]
Output:
----------------------------------------------------------------
first 30 gapful numbers starting 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 starting from 1000000 
----------------------------------------------------------------
1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 

----------------------------------------------------------------
first 10 gapful numbers starting from 1000000000 
----------------------------------------------------------------
1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032 

----------------------------------------------------------------
first 25 gapful numbers starting from 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

AutoHotkey[edit]

Gapful_numbers(Min, Qty){
	counter:= 0, output := ""
	while (counter < Qty){
		n := A_Index+Min-1
		d := SubStr(n, 1, 1) * 10 + SubStr(n, 0)
		if (n/d = Floor(n/d))
			output .= ++counter ": " n "`t" n " / " d " = " Format("{:d}", n/d) "`n"
	}
	return output
}
Examples:
MsgBox, 262144, , % Gapful_numbers(100, 30)
MsgBox, 262144, , % Gapful_numbers(1000000, 15)
MsgBox, 262144, , % Gapful_numbers(1000000000, 10)
Output:
1: 100	100 / 10 = 10
2: 105	105 / 15 = 7
3: 108	108 / 18 = 6
4: 110	110 / 10 = 11
5: 120	120 / 10 = 12
6: 121	121 / 11 = 11
7: 130	130 / 10 = 13
8: 132	132 / 12 = 11
9: 135	135 / 15 = 9
10: 140	140 / 10 = 14
11: 143	143 / 13 = 11
12: 150	150 / 10 = 15
13: 154	154 / 14 = 11
14: 160	160 / 10 = 16
15: 165	165 / 15 = 11
16: 170	170 / 10 = 17
17: 176	176 / 16 = 11
18: 180	180 / 10 = 18
19: 187	187 / 17 = 11
20: 190	190 / 10 = 19
21: 192	192 / 12 = 16
22: 195	195 / 15 = 13
23: 198	198 / 18 = 11
24: 200	200 / 20 = 10
25: 220	220 / 20 = 11
26: 225	225 / 25 = 9
27: 231	231 / 21 = 11
28: 240	240 / 20 = 12
29: 242	242 / 22 = 11
30: 253	253 / 23 = 11
---------------------------
1: 1000000	1000000 / 10 = 100000
2: 1000005	1000005 / 15 = 66667
3: 1000008	1000008 / 18 = 55556
4: 1000010	1000010 / 10 = 100001
5: 1000016	1000016 / 16 = 62501
6: 1000020	1000020 / 10 = 100002
7: 1000021	1000021 / 11 = 90911
8: 1000030	1000030 / 10 = 100003
9: 1000032	1000032 / 12 = 83336
10: 1000034	1000034 / 14 = 71431
11: 1000035	1000035 / 15 = 66669
12: 1000040	1000040 / 10 = 100004
13: 1000050	1000050 / 10 = 100005
14: 1000060	1000060 / 10 = 100006
15: 1000065	1000065 / 15 = 66671
---------------------------
1: 1000000000	1000000000 / 10 = 100000000
2: 1000000001	1000000001 / 11 = 90909091
3: 1000000005	1000000005 / 15 = 66666667
4: 1000000008	1000000008 / 18 = 55555556
5: 1000000010	1000000010 / 10 = 100000001
6: 1000000016	1000000016 / 16 = 62500001
7: 1000000020	1000000020 / 10 = 100000002
8: 1000000027	1000000027 / 17 = 58823531
9: 1000000030	1000000030 / 10 = 100000003
10: 1000000032	1000000032 / 12 = 83333336

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


BASIC256[edit]

Translation of: Yabasic
function is_gapful(n)
	m = n
	l = n mod 10
	while (m >= 10)
		m = int(m / 10)
	end while
	return (m * 10) + l
end function

subroutine muestra_gapful(n, gaps)
	inc = 0
	print "Primeros "; gaps; " números gapful >= "; n
	while inc < gaps
		if n mod is_gapful(n) = 0 then
			print " " ; n ;
			inc = inc + 1
		end if
		n = n + 1
	end while
	print chr(10)
end subroutine

call muestra_gapful(100, 30)
call muestra_gapful(1000000, 15)
call muestra_gapful(1000000000, 10)
call muestra_gapful(7123,25)
end
Output:
Igual que la entrada de Yabasic.


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 :

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


Clojure[edit]

(defn first-digit [n] (Integer/parseInt (subs (str n) 0 1)))

(defn last-digit [n] (mod n 10))

(defn bookend-number [n] (+ (* 10 (first-digit n)) (last-digit n)))

(defn is-gapful? [n] (and (>= n 100) (zero? (mod n (bookend-number n)))))

(defn gapful-from [n] (filter is-gapful? (iterate inc n)))

(defn gapful [] (gapful-from 1))

(defn gapfuls-in-range [start size] (take size (gapful-from start)))

(defn report-range [[start size]]
  (doall (map println
    [(format "First %d gapful numbers >= %d:" size start)
     (gapfuls-in-range start size)
     ""])))

(doall (map report-range [ [1 30] [1000000 15] [1000000000 10] ]))
Output:
First 30 gapful numbers >= 1:
(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)

COBOL[edit]

COBOL does not support suppressing the newline after each DISPLAY, so the numbers have to be on separate lines.

        IDENTIFICATION DIVISION.
        PROGRAM-ID. GAPFUL.
        
        DATA DIVISION.
        WORKING-STORAGE SECTION.
        01 COMPUTATION.
            02 N            PIC 9(10).
            02 N-DIGITS     REDEFINES N.
               03 ND        PIC 9 OCCURS 10 TIMES.
            02 DIV-CHECK    PIC 9(10)V9(2).
            02 DIV-PARTS    REDEFINES DIV-CHECK.
               03 DIV-INT   PIC 9(10).
               03 DIV-FRAC  PIC 9(2).
            02 GAP-AMOUNT   PIC 99.
            02 GAP-DSOR     PIC 99.
            02 FIRST-DIGIT  PIC 99.
        01 OUTPUT-FORMAT.
            02 N-OUT        PIC Z(10).
            
        PROCEDURE DIVISION.
        BEGIN.
            DISPLAY "First 30 gapful numbers >= 100:".
            MOVE 100 TO N. MOVE 30 TO GAP-AMOUNT.
            PERFORM CHECK-GAPFUL-NUMBER.
            
            DISPLAY " ".
            DISPLAY "First 15 gapful numbers >= 1000000:".
            MOVE 1000000 TO N. MOVE 15 TO GAP-AMOUNT.
            PERFORM CHECK-GAPFUL-NUMBER.
            
            DISPLAY " ".
            DISPLAY "First 10 gapful numbers >= 1000000000:".
            MOVE 1000000000 TO N. MOVE 10 TO GAP-AMOUNT.
            PERFORM CHECK-GAPFUL-NUMBER.
            STOP RUN.
        
        CHECK-GAPFUL-NUMBER.
            SET FIRST-DIGIT TO 1.
            INSPECT N TALLYING FIRST-DIGIT FOR LEADING '0'.
            COMPUTE GAP-DSOR = ND(FIRST-DIGIT) * 10 + ND(10).
            DIVIDE N BY GAP-DSOR GIVING DIV-CHECK.
            IF DIV-FRAC IS EQUAL TO 0
                MOVE N TO N-OUT
                DISPLAY N-OUT
                SUBTRACT 1 FROM GAP-AMOUNT.
            ADD 1 TO N.
            IF GAP-AMOUNT IS GREATER THAN 0 
                GO TO CHECK-GAPFUL-NUMBER.
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

Commodore BASIC[edit]

Translation of: Clojure

Numbers >= 1,000,000,000 are printed out in scientific notation and we can't see what they actually are, so I used 100,000,000 instead.

100 DEF FNFD(N) = VAL(MID$(STR$(N),2,1))
110 DEF FNLD(N) = N - 10 * INT(N/10)
120 DEF FNBE(N) = 10 * FNFD(N) + FNLD(N)
130 DEF FNGF(N) = (N >= 100) AND (N - FNBE(N)*INT(N/FNBE(N)) = 0)
140 READ S:IF S<0 THEN 260
150 READ C
160 PRINT"THE FIRST"C"GAPFUL NUMBERS >="S":"
170 I=S:F=0
180 IF NOT FNGF(I) THEN 220
190 PRINT I,
200 F=F+1
210 IF F>=C THEN 240
220 I=I+1
230 GOTO 180
240 PRINT:PRINT
250 GOTO 140
260 END
270 DATA 1,30, 1000000,15, 100000000,10
280 DATA -1
Output:
THE FIRST 30 GAPFUL NUMBERS >= 1 :
 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 >= 1000000 :
 1000000   1000005   1000008   1000010
 1000016   1000020   1000021   1000030
 1000032   1000034   1000035   1000040
 1000050   1000060   1000065

THE FIRST 10 GAPFUL NUMBERS >= 100000000 :
 100000000           100000005
 100000008           100000010
 100000016           100000020
 100000021           100000030
 100000032           100000035

Common Lisp[edit]

Translation of: Clojure
(defun first-digit (n) (parse-integer (string (char (write-to-string n) 0))))

(defun last-digit (n) (mod n 10))

(defun bookend-number (n) (+ (* 10 (first-digit n)) (last-digit n)))

(defun gapfulp (n) (and (>= n 100) (zerop (mod n (bookend-number n)))))

(defun gapfuls-in-range (start size)
  (loop for n from start
        with include = (gapfulp n)
        counting include into found
        if include collecting n
        until (= found size)))

(defun report-range (range)
  (destructuring-bind (start size) range
    (format t "The first ~a gapful numbers >= ~a:~% ~a~%~%" size start 
              (gapfuls-in-range start size))))

(mapcar #'report-range '((1 30) (1000000 15) (1000000000 10)))
Output:
The first 30 gapful numbers >= 1:
 
(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 >= 1000000:
 
(1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032
 1000034 1000035 1000040 1000050 1000060 1000065)
The first 10 gapful numbers >= 1000000000:
 
(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

Delphi[edit]

See Pascal.

Erlang[edit]

Translation of: Clojure

The implementation of the gapfulness check is straightforward, but this solution also uses an implementation of lazy streams to simplify producing the final output lists.

The gapfulness implementation:

-module(gapful).
-export([first_digit/1, last_digit/1, bookend_number/1, is_gapful/1]).

first_digit(N) ->
  list_to_integer(string:slice(integer_to_list(N),0,1)).

last_digit(N) -> N rem 10.

bookend_number(N) -> 10 * first_digit(N) + last_digit(N).

is_gapful(N) -> (N >= 100) and (0 == N rem bookend_number(N)).

The streams implementation:

-module(stream).
-export([yield/1, naturals/0, naturals/1, filter/2, take/2, to_list/1]).

yield(F) when is_function(F) -> F().

naturals() -> naturals(1).
naturals(N) -> fun() -> {N, naturals(N+1)} end.

filter(Pred, Stream) ->
  fun() -> do_filter(Pred, Stream) end.

do_filter(Pred, Stream) ->
  case yield(Stream) of
    {X, Xs} ->
      case Pred(X) of
        true -> {X, filter(Pred, Xs)};
        false -> do_filter(Pred, Xs)
      end;
    halt -> halt
  end.

take(N, Stream) when N >= 0 ->
  fun() ->
    case yield(Stream) of
      {X, Xs} ->
        case N of
          0 -> halt;
          _ -> {X, take(N - 1, Xs)}
        end;
      halt -> halt
    end
  end.

to_list(Stream) -> to_list(Stream, []).
to_list(Stream, Acc) ->
  case yield(Stream) of
    {X, Xs} -> to_list(Xs, [X|Acc]);
    halt -> lists:reverse(Acc)
  end.

The main program that puts them together:

-module(gapful_demo).
-mode(compile).

report_range([Start, Size]) ->
 io:fwrite("The first ~w gapful numbers >= ~w:~n~w~n~n", [Size, Start,
  stream:to_list(stream:take(Size, stream:filter(fun gapful:is_gapful/1,
  stream:naturals(Start))))]).

main(_) -> lists:map(fun report_range/1, [[1,30],[1000000,15],[1000000000,10]]).
Output:
The first 30 gapful numbers >= 1:
[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 >= 1000000:
[1000000,1000005,1000008,1000010,1000016,1000020,1000021,1000030,1000032,1000034,1000035,1000040,1000050,1000060,1000065]

The first 10 gapful numbers >= 1000000000:
[1000000000,1000000001,1000000005,1000000008,1000000010,1000000016,1000000020,1000000027,1000000030,1000000032]

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
] 2tri@
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 }

Forth[edit]

Developed with Gforth 0.7.9

variable                  cnt
: Int>Str                 s>d <# #s #>  ;
: firstDigit              C@  [char] 0 -   ;
: lastDigit               + 1- c@ [char] 0 - ;
: cnt++                   cnt dup  @ 1+ dup rot ! ;
: GapfulNumber?           dup dup Int>Str
                          2dup drop firstDigit 10 * 
                          -rot lastDigit +
                          /mod drop  0= ;
: main                    0 cnt ! 2dup  
                          cr ." First " . ." gapful numbers  >= " .
                          begin  dup cnt @ -
                          while swap GapfulNumber? 
                                if dup cr cnt++  . ." : " . then
                                1+ swap
                          repeat 2drop ;

100        30 main cr
1000000    15 main cr
1000000000 10 main cr
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 10 gapful numbers  >= 1000000000 
1 : 1000000000 
2 : 1000000001 
3 : 1000000005 
4 : 1000000008 
5 : 1000000010 
6 : 1000000016 
7 : 1000000020 
8 : 1000000027 
9 : 1000000030 
10 : 1000000032 

FreeBASIC[edit]

function is_gapful( n as uinteger ) as boolean
    if n<100 then return false
    dim as string ns = str(n)
    dim as uinteger gap = 10*val(mid(ns,1,1)) + val(mid(ns,len(ns),1))
    if n mod gap = 0 then return true else return false
end function

dim as ulongint i = 100
dim as ushort c
print "The first thirty gapful numbers:"
while c<30
    if is_gapful(i) then
        c += 1
        print i;" ";
    end if
    i += 1
wend
print : print
i = 1000000 : c = 0
print "The first fifteen gapful numbers above 999,999:"
while c<15
    if is_gapful(i) then
        c += 1
        print i;" ";
    end if
    i += 1
wend
print : print
i = 1000000000 : c = 0
print "The first ten gapful numbers above 999,999,999:"
while c<10
    if is_gapful(i) then
        c += 1
        print i;" ";
    end if
    i += 1
wend
print
Output:
The first thirty 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 

The first fifteen gapful numbers above 999,999
1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 

The first ten gapful numbers above 999,999,999
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]

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. 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 storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.

In this page you can see the program(s) related to this task and their results.

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 

Groovy[edit]

Translation of: Java
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()
    }

    static void main(String[] args) {
        List<Long> starts = [(long) 1e2, (long) 1e6, (long) 1e7, (long) 1e9, (long) 7123]
        List<Integer> counts = [30, 15, 15, 10, 25]
        for (int i = 0; i < starts.size(); ++i) {
            println("First ${counts.get(i)} gapful numbers starting at ${commatize(starts.get(i))}")

            long j = starts.get(i)
            long pow = 100
            while (j >= pow * 10) {
                pow *= 10
            }

            int count = 0
            while (count < counts.get(i)) {
                long fl = ((long) (j / pow)) * 10 + (j % 10)
                if (j % fl == 0) {
                    print("$j ")
                    count++
                }
                if (++j >= 10 * pow) {
                    pow *= 10
                }
            }

            println()
            println()
        }
    }
}
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 (intercalate)
import Data.List.Split (chunksOf)

---------------------- 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

jq[edit]

The following program works with both jq and gojq; for very large integers, the latter should be used.

# emit a stream of gapful numbers greater than or equal to $start, 
# which is assumed to be an integer
def gapful($start):
    range($start; infinite)
    | . as $i
    | tostring as $s
    | (($s[:1] + $s[-1:]) | tonumber) as $x
    | select($i % $x == 0);

"First 30 gapful numbersstarting from 100:",
([limit(30;gapful(100))] | join(" ")),
"First 15 gapful numbers starting from 1,000,000:",
([limit(15;gapful(1000000))] | join(" ")),
"First 10 gapful numbers starting from 10^9:",
([limit(10;gapful(pow(10;9)))] | join(" "))
Output:
First 30 gapful numbers starting 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 starting from 1,000,000:
1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065
First 10 gapful numbers starting from 10^9:
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

[edit]

Translation of: Clojure
to bookend_number :n
  output sum product 10 first :n last :n
end

to gapful? :n
  output and greaterequal? :n 100 equal? 0 modulo :n bookend_number :n
end

to gapfuls_in_range :start :size
  localmake "gapfuls []
  do.while [
    if (gapful? :start) [
      make "gapfuls (lput :start gapfuls)
    ]
    make "start sum :start 1
  ] [less? (count :gapfuls) :size]
  output :gapfuls
end

to report_range :start :size
  print (word "|The first | :size "| gapful numbers >= | :start "|:|)
  print gapfuls_in_range :start :size
  (print)
end

foreach [ [1 30] [1000000 15] [1000000000 10] ] [
  apply "report_range ?
]
Output:
The first 30 gapful numbers >= 1:
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 >= 1000000:
1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065

The first 10 gapful numbers >= 1000000000:
1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032

LOLCODE[edit]

Translation of: Clojure
HAI 1.2

HOW IZ I FurstDigit YR Numbr
  I HAS A Digit
  IM IN YR Loop NERFIN YR Dummy WILE DIFFRINT Numbr AN 0
    Digit R MOD OF Numbr AN 10
    Numbr R QUOSHUNT OF Numbr AN 10
  IM OUTTA YR Loop
  FOUND YR Digit
IF U SAY SO

HOW IZ I LastDigit YR Numbr
  FOUND YR MOD OF Numbr AN 10
IF U SAY SO

HOW IZ I Bookend YR Numbr
  FOUND YR SUM OF PRODUKT OF I IZ FurstDigit YR Numbr MKAY AN 10 AN I IZ LastDigit YR Numbr MKAY
IF U SAY SO

HOW IZ I CheckGapful YR Numbr
  I HAS A Bookend ITZ I IZ Bookend YR Numbr MKAY
  I HAS A BigEnuff ITZ BOTH SAEM Numbr AN BIGGR OF Numbr AN 100
  FOUND YR BOTH OF BigEnuff AN BOTH SAEM 0 AN MOD OF Numbr AN Bookend
IF U SAY SO

HOW IZ I FindGapfuls YR Start AN YR HowMany
  I HAS A Numbr ITZ Start
  I HAS A Anser ITZ A BUKKIT
  I HAS A Found ITZ 0
  IM IN YR Loop UPPIN YR Dummy WILE DIFFRINT Found AN HowMany
    I IZ CheckGapful YR Numbr MKAY
    O RLY?
      YA RLY
        Anser HAS A SRS Found ITZ Numbr
        Found R SUM OF Found AN 1
    OIC
    Numbr R SUM OF Numbr AN 1
  IM OUTTA YR Loop
  FOUND YR Anser
IF U SAY SO

HOW IZ I Report YR Start AN YR HowMany
  VISIBLE "The furst " !
  VISIBLE HowMany !
  VISIBLE " Gapful numbrs starting with " !
  VISIBLE Start !
  VISIBLE "::"
  I HAS A Anser ITZ I IZ FindGapfuls YR Start AN YR HowMany MKAY
  IM IN YR Loop UPPIN YR Index TIL BOTH SAEM Index AN HowMany
    DIFFRINT Index AN 0
    O RLY?
      YA RLY
        VISIBLE ", " !
    OIC
    VISIBLE Anser'Z SRS Index !
  IM OUTTA YR Loop
  VISIBLE ""
  VISIBLE ""
IF U SAY SO

I IZ Report YR 1 AN YR 30 MKAY
I IZ Report YR 1000000 AN YR 15 MKAY
I IZ Report YR 1000000000 AN YR 10 MKAY
KTHXBYE
Output:
The furst 30 Gapful numbrs starting with 1:
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 furst 15 Gapful numbrs starting with 1000000:
1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065

The furst 10 Gapful numbrs starting with 1000000000:
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

Mathematica / Wolfram Language[edit]

ClearAll[GapFulQ]
GapFulQ[n_Integer] := Divisible[n, FromDigits[IntegerDigits[n][[{1, -1}]]]]
i = 100;
res = {};
While[Length[res] < 30,
 If[GapFulQ[i], AppendTo[res, i]];
 i++
 ]
res
i = 10^6;
res = {};
While[Length[res] < 15,
 If[GapFulQ[i], AppendTo[res, i]];
 i++
 ]
res
i = 10^9;
res = {};
While[Length[res] < 10,
 If[GapFulQ[i], AppendTo[res, i]];
 i++
 ]
res
Output:
{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}
{1000000,1000005,1000008,1000010,1000016,1000020,1000021,1000030,1000032,1000034,1000035,1000040,1000050,1000060,1000065}
{1000000000,1000000001,1000000005,1000000008,1000000010,1000000016,1000000020,1000000027,1000000030,1000000032}

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

Nim[edit]

import strutils


func gapfulDivisor(n: Positive): Positive =
  ## Return the gapful divisor of "n".
  
  let last = n mod 10
  var first = n div 10
  while first > 9:
    first = first div 10
  result = 10 * first + last


iterator gapful(start: Positive): Positive =
  ## Yield the gapful numbers starting from "start".
  
  var n = start
  while true:
    let d = n.gapfulDivisor()
    if n mod d == 0: yield n
    inc n


proc displayGapfulNumbers(start, num: Positive) =
  ## Display the first "num" gapful numbers greater or equal to "start".
  
  echo "\nFirst $1 gapful numbers ⩾ $2:".format(num, start)
  var count = 0
  var line: string
  for n in gapful(start):
    line.addSep(" ")
    line.add($n)
    inc count
    if count == num: break
  echo line


when isMainModule:
  displayGapfulNumbers(100, 30)
  displayGapfulNumbers(1_000_000, 15)
  displayGapfulNumbers(1_000_000_000, 10)
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

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
{$IFDEF FPC}
  ,strUtils // Numb2USA aka commatize
{$ENDIF};

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

{$IFNDEF FPC}

function Numb2USA(const S: string): string;
var
  i, NA: Integer;
begin
  i := Length(S);
  Result := S;
  NA := 0;
  while (i > 0) do
  begin
    if ((Length(Result) - i + 1 - NA) mod 3 = 0) and (i <> 1) then
    begin
      insert(',', Result, i);
      inc(NA);
    end;
    Dec(i);
  end;
end;
{$ENDIF}

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 := 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 := 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 := tmp - (-ORD(tmp >= GapNum) and GapNum);
    ModsHL[GapNum] := tmp;

    TestNum := TestNum + 1;
    tmp := LowDgt + 1;

    inc(GapNum);
    if tmp >= Base then
    begin
      tmp := 0;
      GapNum := GapNum - Base;
    end;
    LowDgt := tmp;
    //next Hi Digit
    if TestNum >= LmtNextNewHiDgt then
    begin
      LowDgt := 0;
      GapNum := GapNum + Base;
      LmtNextNewHiDgt := LmtNextNewHiDgt + Pow10;
      //next power of 10
      if GapNum >= Base * Base then
      begin
        Pow10 := 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;
  {$IFNDEF LINUX}  readln; {$ENDIF}
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

PL/M[edit]

This can be compiled with the original 8080 PL/M compiler and run under CP/M or a clone or emulator. Just shows the first 30 Gapful numbers >= 100, as the original 8080 PL/M only supports at most (unsigned) 16-bit numbers.

100H: /* FIND SOME GAPFUL NUMBERS: NUMBERS DIVISIBLE BY 10F + L WHERE F IS */
      /*                           THE FIRST DIGIT AND L IS THE LAST DIGIT */
   BDOS: PROCEDURE( FN, ARG ); /* CP/M BDOS SYSTEM CALL */
      DECLARE FN BYTE, ARG ADDRESS;
      GOTO 5;
   END BDOS;
   PR$CHAR:   PROCEDURE( C ); DECLARE C BYTE;    CALL BDOS( 2, C ); END;
   PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
   PR$NL:     PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH );  END;
   PR$NUMBER: PROCEDURE( N );
      DECLARE N ADDRESS;
      DECLARE V ADDRESS, N$STR( 6 ) BYTE, W BYTE;
      V = N;
      W = LAST( N$STR );
      N$STR( W ) = '$';
      N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      DO WHILE( ( V := V / 10 ) > 0 );
         N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
      END;
      CALL PR$STRING( .N$STR( W ) );
   END PR$NUMBER;
   /* RETURNS TRUE IF N IS GAPFUL, FALSE OTHERWISE                         */
   IS$GAPFUL: PROCEDURE( N )BYTE;
      DECLARE N ADDRESS;
      DECLARE F ADDRESS;
      F = N / 10;
      DO WHILE ( F > 9 );
         F = F / 10;
      END;
      RETURN N MOD ( ( F * 10 ) + ( N MOD 10 ) ) = 0;
   END IS$GAPFUL;
   /* FIND THE FIRST 30 GAPFUL NUMBERS >= 100                              */
   CALL PR$STRING( .'FIRST 30 GAPFUL NUMBERS STARTING FROM 100:$' );
   CALL PR$NL; 
   DECLARE N ADDRESS, G$COUNT BYTE;
   G$COUNT = 0;
   N = 100;
   DO WHILE ( G$COUNT < 30 );
      IF IS$GAPFUL( N ) THEN DO;
         /* HAVE A GAPFUL NUMBER                                           */
         G$COUNT = G$COUNT + 1;
         CALL PR$CHAR( ' ' );
         CALL PR$NUMBER( N );
      END;
      N = N + 1;
   END;
   CALL PR$NL;

EOF
Output:
FIRST 30 GAPFUL NUMBERS STARTING 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

PowerShell[edit]

Translation of: Clojure
function Get-FirstDigit {
  param ( [int] $Number )
  [int]$Number.ToString().Substring(0,1)
}

function Get-LastDigit {
  param ( [int] $Number )
  $Number % 10
}

function Get-BookendNumber {
  param ( [Int] $Number )
  10 * (Get-FirstDigit $Number) + (Get-LastDigit $Number)
}

function Test-Gapful {
  param ( [Int] $Number )
  100 -lt $Number -and 0 -eq $Number % (Get-BookendNumber $Number)
}

function Find-Gapfuls {
  param ( [Int] $Start, [Int] $Count )
  $result = @()

  While ($result.Count -lt $Count)  {
    If (Test-Gapful $Start) {
      $result += @($Start)
    }
    $Start += 1
  }
  return $result
}

function Search-Range {
  param ( [Int] $Start, [Int] $Count )
  Write-Output "The first $Count gapful numbers >= $($Start):"
  Write-Output( (Find-Gapfuls $Start $Count) -join ",")
  Write-Output ""
}

Search-Range 1 30
Search-Range 1000000 15
Search-Range 1000000000 10
Output:
The first 30 gapful numbers >= 1:
105,108,110,120,121,130,132,135,140,143,150,154,160,165,170,176,180,187,190,192,195,198,200,220,225,231,240,242,253,260

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

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

PureBasic[edit]

Procedure.b isGapNum(n.i)
  n1.i=n%10
  n2.i=Val(Left(Str(n),1))  
  If n%(n2*10+n1)=0
    ProcedureReturn #True
  Else
    ProcedureReturn #False
  EndIf
EndProcedure

Procedure PutGapNum(start.i,rep.i,lfi.i=10)
  n.i=start
  While rep
    If isGapNum(n)
      Print(Str(n)+" ")
      rep-1
      If rep%lfi=0 : PrintN("") : EndIf
    EndIf
    n+1 
  Wend
EndProcedure

OpenConsole()
PrintN("First 30 gapful numbers ≥ 100:")
PutGapNum(100,30)
PrintN(~"\nFirst 15 gapful numbers ≥ 1,000,000:")
PutGapNum(1000000,15,5)
PrintN(~"\nFirst 10 gapful numbers ≥ 1,000,000,000:")
PutGapNum(1000000000,10,5)
Input()
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

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 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
gapful: procedure; parse arg n "@" sp;  if sp==''  then sp= 100 /*get args; use default.*/
        say center(' 'n      " gapful numbers starting at: "     sp' ', 125, "═")
        $=;                   #= 0                              /*initialize the $ list.*/
               do j=sp  until #==n                              /*SP:  starting point.  */
               parse var   j   a  2  ''  -1  b                  /*get 1st and last digit*/
               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]

Scheme[edit]

Translation of: Clojure
(define (first-digit n) (string->number (string (string-ref (number->string n) 0))))
(define (last-digit n) (modulo n 10))
(define (bookend-number n) (+ (* 10 (first-digit n)) (last-digit n)))
(define (gapful? n) (and (>= n 100) (zero? (modulo n (bookend-number n)))))

(define (gapfuls-in-range start size)
  (let ((found 0) (result '()))
    (do ((n start (+ n 1))) ((>= found size) (reverse result))
      (if (gapful? n) 
        (begin (set! result (cons n result)) (set! found (+ found 1)))))))

(define (report-range range)
  (apply (lambda (start size) 
    (newline)
    (display "The first ")(display size)(display " gapful numbers >= ")
    (display start)(display ":")(newline)
    (display (gapfuls-in-range start size))(newline)) range))

(map report-range '((100 30) (1000000 15) (1000000000 10)))
Output:
The 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)

The first 15 gapful numbers >= 1000000:
(1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065)

The first 10 gapful numbers >= 1000000000:
(1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032)

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

SQL[edit]

Works with: ORACLE 19c

This is not a particularly efficient solution, but it gets the job done.

/*
This code is an implementation of gapful numbers in SQL ORACLE 19c 
p_start -- start point
p_count -- total number to be found
*/
with
  function gapful_numbers(p_start in integer, p_count in integer) return varchar2 is
    v_start   integer := p_start;
    v_count   integer := 0;
    v_res     varchar2(32767); 
  begin
    v_res := 'First '||p_count||' gapful numbers starting from '||p_start||': ';
    -- main cycle
    while true loop
      if mod(v_start,to_number(substr(v_start,1,1)||substr(v_start,-1))) = 0 then 
         v_res := v_res || v_start; 
         v_count := v_count + 1;
         exit when v_count = p_count;
         v_res := v_res || ', ';  
      end if;
      v_start := v_start + 1; 
    end loop;
    --
    return v_res;
    --
  end;

--Test
select gapful_numbers(100,30) as res from dual
union all 
select gapful_numbers(1000000,15) as res from dual
union all 
select gapful_numbers(1000000000,10) as res from dual;
/
Output:
First 30 gapful numbers starting 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 starting from 1000000: 1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065
First 10 gapful numbers starting from 1000000000: 1000000000, 1000000001, 1000000005, 1000000008, 1000000010, 1000000016, 1000000020, 1000000027, 1000000030, 1000000032

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

UNIX Shell[edit]

Works with: Bourne Again Shell
Translation of: Clojure
first-digit() {
  printf '%s\n' "${1:0:1}"
}

last-digit() {
  printf '%s\n' $(( $1 % 10 ))
}

bookend-number() {
  printf '%s%s\n' "$(first-digit "$@")" "$(last-digit "$@")"
}

is-gapful() {
  (( $1 >= 100 && $1 % $(bookend-number "$1") == 0 ))
}

gapfuls-in-range() {
  local gapfuls=()
  local -i i found
  for (( i=$1, found=0; found < $2; ++i )); do
    if is-gapful "$i"; then
      if (( found )); then
        printf ' ';
      fi
      printf '%s' "$i"
      (( found++ ))
    fi
  done
  printf '\n'
}

report-ranges() {
  local range
  local -i start size
  for range; do
    IFS=, read start size <<<"$range"
    printf 'The first %d gapful numbers >= %d:\n' "$size" "$start"
    gapfuls-in-range "$start" "$size"
    printf '\n'
  done
}

report-ranges 1,30 1000000,15 1000000000,10
Output:
The first 30 gapful numbers >= 1:
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 >= 1000000:
1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065

The first 10 gapful numbers >= 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

Vlang[edit]

Translation of: go
fn commatize(n u64) string {
    mut s := n.str()
    le := s.len
    for i := le - 3; i >= 1; i -= 3 {
        s = '${s[0..i]},$s[i..]'
    }
    return s
}
 
fn main() {
    starts := [u64(1e2), u64(1e6), u64(1e7), u64(1e9), u64(7123)]
    counts := [30, 15, 15, 10, 25]
    for i in 0..starts.len {
        mut count := 0
        mut j := starts[i]
        mut pow := u64(100)
        for {
            if j < pow*10 {
                break
            }
            pow *= 10
        }
        println("First ${counts[i]} gapful numbers starting at ${commatize(starts[i])}:")
        for count < counts[i] {
            fl := (j/pow)*10 + (j % 10)
            if j%fl == 0 {
                print("$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 

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 

XBS[edit]

func isgapful(n:number):boolean{
	set s:string = tostring(n);
	set d = toint(`{s::at(0)}{s::at(?s-1)}`);
	send (n%d)==0
}

func findGapfulNumbers(start,amount){
	set gapful=[];
	set ind = start;
	while(true){
		if(isgapful(ind)){
			gapful::insert(ind);
		}
		ind++;
		if((?gapful)>=amount){
			stop;
		}
	}
	log(`First {amount} gapful ints at {start}: {gapful::join(", ")}`);
}

findGapfulNumbers(100,30);
findGapfulNumbers(1000000,15);
findGapfulNumbers(1000000000,15);
Output:
First 30 gapful ints 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 ints at 1000000: 1000000, 1000005, 1000008, 1000010, 1000016, 1000020, 1000021, 1000030, 1000032, 1000034, 1000035, 1000040, 1000050, 1000060, 1000065
First 15 gapful ints at 1000000000: 1000000000, 1000000001, 1000000005, 1000000008, 1000000010, 1000000016, 1000000020, 1000000027, 1000000030, 1000000032, 1000000035, 1000000039, 1000000040, 1000000050, 1000000053

XPL0[edit]

func Gapful(N0);        \Return 'true' if gapful number
int  N0, N, First, Last;
[N:= N0;
N:= N/10;
Last:= rem(0);
repeat  N:= N/10;
        First:= rem(0);
until   N = 0;
N:= First*10 + Last;
return rem(N0/N) = 0;
];

proc ShowGap(Start, Limit); \Display gapful numbers
int  Start, Limit, Count, N;
[Text(0, "First "); IntOut(0, Limit); Text(0, " gapful numbers starting from ");
IntOut(0, Start);  Text(0, ":^m^j");
Count:= 0;  N:= Start;
loop    [if Gapful(N) then
            [IntOut(0, N);  ChOut(0, ^ );
            Count:= Count+1;
            if Count >= Limit then quit;
            ];
        N:= N+1;
        ];
CrLf(0);
];

[ShowGap(100, 30);
ShowGap(1_000_000, 15);
ShowGap(1_000_000_000, 10);
]
Output:
First 30 gapful numbers starting 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 starting from 1000000:
1000000 1000005 1000008 1000010 1000016 1000020 1000021 1000030 1000032 1000034 1000035 1000040 1000050 1000060 1000065 
First 10 gapful numbers starting from 1000000000:
1000000000 1000000001 1000000005 1000000008 1000000010 1000000016 1000000020 1000000027 1000000030 1000000032 

Yabasic[edit]

sub is_gapful(n)
    m = n
	l = mod(n, 10)
    while (m >= 10) 
        m = int(m / 10)
    wend
    return (m * 10) + l
end sub

sub muestra_gapful(n, gaps)
    inc = 0
    print "Primeros ", gaps, " numeros gapful >= ", n
    while inc < gaps       
        if mod(n, is_gapful(n)) = 0 then
            print " " , n , 
            inc = inc + 1
        end if
        n = n + 1
    wend
    print chr$(10)
end sub

muestra_gapful(100, 30)
muestra_gapful(1000000, 15)
muestra_gapful(1000000000, 10)
muestra_gapful(7123,25)
end
Output:
Primeros 30 numeros 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

Primeros 15 numeros gapful >= 1000000
 1000000         1000005         1000008         1000010         1000016         1000020         1000021         1000030         1000032         1000034         1000035         1000040         1000050         1000060         1000065

Primeros 10 numeros gapful >= 1000000000
 1000000000      1000000001      1000000005      1000000008      1000000010      1000000016      1000000020      1000000027      1000000030      1000000032

Primeros 25 numeros gapful >= 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

---Program done, press RETURN---


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