(Redirected from Largest product)
Largest five adjacent number is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Generate random 1000-digit number.
Find the five adjacent digits in the 1000-digit number that form the largest 5-digit number.

Extra credit

Find the five adjacent digits in the 1000-digit number that form the smallest 5-digit number.

## 11l

Translation of: Python
```V largeNum = [random:(1..9)] [+] (0.<999).map(i -> random:(0..9))

V (maxNum, minNum) = (0, 99999)

L(i) 996
V num = Int(largeNum[i.+5].join(‘’))
I num > maxNum
maxNum = num
E I num < minNum
minNum = num

print(‘Largest 5-digit number extracted from random 1000-digit number: ’maxNum)
print(‘Smallest 5-digit number extracted from random 1000-digit number: #05’.format(minNum))```
Output:
```Largest 5-digit number extracted from random 1000-digit number: 99902
Smallest 5-digit number extracted from random 1000-digit number: 00043
```

```with Ada.Text_Io;

subtype Digit is Character range '0' .. '9';
package Random_Digits

Gen   : Random_Digits.Generator;
Line  : String (1 .. 1000);
Large : Natural := Natural'First;
Small : Natural := Natural'Last;
begin
Random_Digits.Reset (Gen);
Line := (others => Random_Digits.Random (Gen));

for I in Line'First .. Line'Last - Adjacent_Length + 1 loop
declare
Window : String renames Line (I .. I + Adjacent_Length - 1);
begin
Large := Natural'Max (Large, Natural'Value (Window));
Small := Natural'Min (Small, Natural'Value (Window));
end;
end loop;
Ada.Text_Io.Put_Line ("The largest number : " & Natural'Image (Large));
Ada.Text_Io.Put_Line ("The smallest number: " & Natural'Image (Small));
```
Output:
```The largest number :  99625
The smallest number:  102
```

## ALGOL 68

Adding the minimum number for good measure...

```BEGIN # generate 1000 random digits and find the largest/smallest numbers formed from 5 consecutive digits #
[ 1 : 1000 ]CHAR digits;
FOR i TO UPB digits DO digits[ i ] := REPR ( ENTIER ( next random * 10 ) + ABS "0" ) OD;
STRING max number := digits[ 1 : 5 ];
STRING min number := digits[ 1 : 5 ];
FOR i FROM 2 TO UPB digits - 4 DO
STRING next number = digits[ i : i + 4 ];
IF next number > max number
THEN
# found a new higher number #
max number := next number
FI;
IF next number < min number
THEN
# found a new lower number #
min number := next number
FI
OD;
print( ( "Largest  5 consecutive digits from 1000 random digits: ", max number, newline ) );
print( ( "Smallest 5 consecutive digits from 1000 random digits: ", min number, newline ) )
END```
Output:
```Largest  5 consecutive digits from 1000 random digits: 99987
Smallest 5 consecutive digits from 1000 random digits: 00119
```

## APL

Works with: Dyalog APL
```⌈/((⊣+10×⊢)/(⌽↓))⌺5⊢(-⎕IO)+?1000/10
```
Output:

(example)

`99994`

## Arturo

```N: join to [:string] map 1..100 'x -> random 1000000000 9999999999

i: 0
maxNum: 0
minNum: ∞
while [i < (size N)-5][
num: to :integer join @[N\[i] N\[i+1] N\[i+2] N\[i+3] N\[i+4]]
if num > maxNum -> maxNum: num
if num < minNum -> minNum: num
i: i + 1
]

print "Our random 1000-digit number is:"
print N

print ""

print ["Max 5-adjacent number found:" maxNum]
print ["Min 5-adjacent number found:" (repeat "0" 5-(size to :string minNum)) ++ (to :string minNum)]
```
Output:
```Our random 1000-digit number is:
2540956677308157418624519953263471599696918276171651168484519407031160813613006352660058588944602704848634276542837184618726044674117357036813240557325769932073351534364289297094415941273117151277729576200542643185699525405079189015204192029912043004161916366921458912887890652627268028071729897387395041640352395354106991129061548748712499227024213135531365974620993813773921850969630855401781344832397898392812417729744785629765286216304456806870691502938136795922685099816652448188701308354551593078486609811394420601431484916913833634669083737749230355341380266781803894385432741405633278873213701238310761908151961510643290964548205746238459266137202173265468217401777681775761126374654289733873900330799576500024068191362342162163615972164105625935627483920193464168192083262176697432155066174175594837721476581087940310642712981291006889657297350894628612724944063786324456854104801432247483498384207351647946918119868105898645178074174003550762101547842674605061792172905254724197215648686667

## AutoHotkey

```maxNum := 0, str := ""
loop, 1000
{
Random, rnd, 0, 9
str .= rnd
output .= rnd . (Mod(A_Index, 148) ? "" : "`n")
if A_Index < 5
continue
num := SubStr(str, A_Index-4, 5)
maxNum := maxNum > num ? maxNum : num
minNum := A_Index = 5 ? num : minNum < num ? minNum : num
}
MsgBox % result := output "`n`nLargest five adjacent digits = " maxNum
.  "`n`nSmallest five adjacent digits = " minNum
```
Output:
```3893212622395522104846091986776081862634026945871752892124324578621089065097043281907406149009719673003318226562809101957181871693776164191416491334
2509291361707848297387923254298547833186351133036771338719578505791529263806019240009497155124458943732581184022226943392528107498748575424217651885
3083736872582691290721469942482918430078673685947447234032602113276631102983248999047362916320523840282929255314468323644427797630259187509914424396
1523615571637081320270791095221484894567420630155741441396012393172769867922862248399483054652921274863786220527596050784952102267710198517665662903
6335615800351254988779849447078262460051794249274045128158246939351902901862546960248213286880570086476859341012102414828750098051948784732121573660
9618754338433412518619240496583375235634416473003920360759949694724646721954909867058588446320222792637823988375313876167705092153587245148819122980
2777308429997046827297505483667631338885207838402941712216614732232703459440770039141898763110002290662921501156

Largest five adjacent digits = 99970

Smallest five adjacent digits = 00022```

## AWK

```# syntax: GAWK -f LARGEST_FIVE_ADJACENT_NUMBER.AWK
BEGIN {
limit = 1000
width = 5
max_n = 0
for (i=1; i<=width; i++) {
min_n = min_n "9"
}
srand()
for (i=1; i<=limit; i++) {
digits = digits int(rand() * 10)
}
for (i=1; i<=limit-width+1; i++) {
n = substr(digits,i,width)
if (n > max_n) {
max_n = n
max_pos = i
}
if (n < min_n) {
min_n = n
min_pos = i
}
}
printf("look for %d digit number using %d digits\n",width,limit)
printf("largest  %0*d in positions %d-%d\n",width,max_n,max_pos,max_pos+width-1)
printf("smallest %0*d in positions %d-%d\n",width,min_n,min_pos,min_pos+width-1)
exit(0)
}
```
Output:
```look for 5 digit number using 1000 digits
largest  99873 in positions 300-304
smallest 00099 in positions 697-701
```

## BASIC

### BASIC256

Translation of: FreeBASIC
```dim number(1000)
highest = 0
lowest = 100000
for i = 0 to 999
number[i] = int(rand*10)
if i >= 4 then
tmp = number[i] + 10*number[i-1] + 100*number[i-2] + 1000*number[i-3] + 10000*number[i-4]
if tmp < lowest then lowest = tmp
if tmp > highest then highest = tmp
end if
next i
print highest, lowest
```

### Chipmunk Basic

Works with: Chipmunk Basic version 3.6.4
Translation of: FreeBASIC
```100 randomize timer
110 dim number(999)
120 highest = 0
130 lowest = 100000
140 for i = 0 to 999
150   number(i) = int(rnd(10))
160   if i >= 4 then
170     tmp = number(i)+10*number(i-1)+100*number(i-2)+1000*number(i-3)+10000*number(i-4)
180     if tmp < lowest then lowest = tmp
190     if tmp > highest then highest = tmp
200   endif
210 next i
220 print highest,lowest
230 end
```

### Gambas

Translation of: FreeBASIC
```Public number[1000] As Byte

Public Sub Main()

Randomize
Dim tmp As Integer, highest As Integer = 0, lowest As Integer = 100000
For i As Integer = 0 To 999
number[i] = Int(Rnd(10))
If i >= 4 Then
tmp = number[i] + 10 * number[i - 1] + 100 * number[i - 2] + 1000 * number[i - 3] + 10000 * number[i - 4]
If tmp < lowest Then lowest = tmp
If tmp > highest Then highest = tmp
End If
Next
Print highest, lowest

End
```

### PureBasic

Translation of: FreeBASIC
```OpenConsole()
Dim number.i(999)
highest.i = 0
lowest.i = 100000
For i.i = 0 To 999
number(i) = Random(10)
If i >= 4:
tmp = number(i) + 10*number(i-1) + 100*number(i-2) + 1000*number(i-3) + 10000*number(i-4)
If tmp < lowest: lowest = tmp: EndIf
If tmp > highest: highest = tmp: EndIf
EndIf
Next i
PrintN(Str(highest) + #TAB\$ + Str(lowest))
```

### QBasic

Works with: QBasic version 1.1
Works with: QuickBasic version 4.5
Translation of: FreeBASIC
```RANDOMIZE TIMER
DIM number(0 TO 999)
highest = 0
lowest = 100000
FOR i = 0 TO 999
number(i) = INT(RND * 10)
IF i >= 4 THEN
tmp = number(i) + 10 * number(i - 1) + 100 * number(i - 2) + 1000 * number(i - 3) + 10000 * number(i - 4)
IF tmp < lowest THEN lowest = tmp
IF tmp > highest THEN highest = tmp
END IF
NEXT i
PRINT highest, lowest
END
```

### True BASIC

Translation of: QBasic
```RANDOMIZE
DIM number(0 TO 999)
LET highest = 0
LET lowest = 100000
FOR i = 0 TO 999
LET number(i) = INT(RND*10)
IF i >= 4 THEN
LET tmp = number(i)+10*number(i-1)+100*number(i-2)+1000*number(i-3)+10000*number(i-4)
IF tmp < lowest THEN LET lowest = tmp
IF tmp > highest THEN LET highest = tmp
END IF
NEXT i
PRINT highest, lowest
END
```

### Yabasic

Translation of: FreeBASIC
```dim number(999)
highest = 0
lowest = 100000
for i = 0 to 999
number(i) = int(ran(10))
if i >= 4 then
tmp = number(i) + 10*number(i-1) + 100*number(i-2) + 1000*number(i-3) + 10000*number(i-4)
if tmp < lowest  lowest = tmp
if tmp > highest  highest = tmp
fi
next i
print highest, lowest
```

## BQN

```⌈´(⊣+10×⊢)˝⌽⍉5↕1000 •rand.Range 10
```
Output:

(example)

`99991`

## C

```#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <time.h>

#define DIGITS 1000
#define NUMSIZE 5

uint8_t randomDigit() {
uint8_t d;
do {d = rand() & 0xF;} while (d >= 10);
return d;
}

int numberAt(uint8_t *d, int size) {
int acc = 0;
while (size--) acc = 10*acc + *d++;
return acc;
}

int main() {
uint8_t digits[DIGITS];
int i, largest = 0;

srand(time(NULL));

for (i=0; i<DIGITS; i++) digits[i] = randomDigit();
for (i=0; i<DIGITS-NUMSIZE; i++) {
int here = numberAt(&digits[i], NUMSIZE);
if (here > largest) largest = here;
}

printf("%d\n", largest);
return 0;
}
```
Output:

(example)

`99931`

## CLU

```% Generate a number of N random digits
random_digits = proc (n: int) returns (sequence[int])
digits: array[int] := array[int]\$predict(1,n)

% A number never starts with a zero
for i: int in int\$from_to(1,n-1) do
end
return(sequence[int]\$a2s(digits))
end random_digits

% Find the largest and smallest N-adjacent number in the digits
find_min_max = proc (n: int, digits: sequence[int]) returns (int,int)
min: int := 10**n  % Guaranteed to be bigger than any N-adjacent number
max: int := 0

for i: int in int\$from_to(1, sequence[int]\$size(digits)-n) do
cur: int := 0
for j: int in int\$from_to(0, n-1) do
cur := 10*cur + digits[i+j]
end
if cur<min then min:=cur end
if cur>max then max:=cur end
end
return(min, max)
end find_min_max

start_up = proc ()
% Seed the RNG with the current time
d: date := now()
random\$seed(d.second + 60*(d.minute + 60*d.hour))

% Find the minimum and maximum 5-adjacent numbers in a 1000-digit number
min, max: int := find_min_max(5, random_digits(1000))

po: stream := stream\$primary_output()
stream\$putl(po, "Smallest: " || int\$unparse(min))
stream\$putl(po, "Largest: " || int\$unparse(max))
end start_up```
Output:
```Smallest: 144
Largest: 99951```

## Delphi

Works with: Delphi version 6.0

```function Get5DigitNumber(S: string; Off: integer): integer;
{Extract 5 digit number from string at Off}
var I: integer;
var NS: string;
begin
NS:=Copy(S,Off,5);
Result:=StrToIntDef(NS,-1);
end;

function BreakupString(S: string): string;
{Breakup thousand digit number for easy display}
var I: integer;
begin
for I:=1 to Length(S) do
begin
Result:=Result+S[I];
if (I mod 55)=0 then Result:=Result+#\$0D#\$0A;
end;
end;

procedure FiveDigitNumber(Memo: TMemo);
{Find the largest and small 5 digit sequence}
{in 1000 digit number}
var S: string;
var N,I: integer;
var Largest,Smallest: integer;
begin
Smallest:=High(Integer);
Largest:=0;
for I:=1 to 1000 do
S:=S+Char(Random(10)+\$30);
for I:=1 to Length(S)-5 do
begin
N:=Get5DigitNumber(S,I);
if N>Largest then Largest:=N;
if N<Smallest then Smallest:=N;
end;
end;
```
Output:
```0082263134040802937368731342824182794880115050767752659
6926207485596307977119758620628125911215421677000178364
7438810001625238336693427757455861441056098692774612931
9301856160395349334087194184285169534216966507128749101
0333045468523586265833674268791722749102838792380205401
7335212073765802860114410575280403628540910018912794058
9569778977033072890894634763659190635686944921467068416
0978402580498879216810854417805724457730620420683349740
8203884243646784563247619038458645194136841413688117232
0960606571886477139587251334596793042923055521495533796
5592094928040937883628134090110628164451939278452734493
5741344340195488542852682604882967292438604245256357719
4755578568409079269700382959730067457921191314413220282
3502307407547002586284406642530858066838890257743184196
5040611036453640792847940715686736822030381083124941163
3588177613294220880152655471721880286144478485085399563
1095924640071825166992021998152653370680394470682198029
3879102724160697653653330275506532525946257246355415772
4978409544
Largest: 99815
Smallest: 16

```

## F#

```// Largest five adjacent number. Nigel Galloway: September 28th., 2021
let n()=let n()=System.Random().Next(10) in Seq.unfold(fun g->Some(g,(g%10000)*10+n()))(n()*10000+n()*1000+n()*100+n()*10+n())
printfn \$"Largest 5 adjacent digits are %d{(n()|>Seq.take 995|>Seq.max)}"
```
Output:
```Largest 5 adjacent digits are 99914
```

## EasyLang

```for i to 1000
n\$ &= random 10 - 1
.
min = 99999
for i = 1 to 995
n = number substr n\$ i 5
min = lower min n
max = higher max n
.
print min & " " & max```
Output:
```21 99768
```

## Factor

Works with: Factor version 0.99 2021-06-02
```USING: grouping io math.functions prettyprint random sequences ;

1000 10^ random unparse 5 <clumps> supremum print
```
Output:
```99987
```

## FreeBASIC

Generate the number digit by digit, and test as we go. If the task didn't specifically ask to generate the whole 1,000 digit number I wouldn't bother storing more than five of its digits at a time.

```randomize timer
dim as ubyte number(0 to 999)
dim as uinteger seg, highest = 0, lowest = 100000
for i as uinteger = 0 to 999
number(i) = int(rnd*10)
if i >= 4 then
seg = number(i) + 10*number(i-1) + 100*number(i-2) +_
1000*number(i-3) + 10000*number(i-4)
if seg < lowest then lowest = seg
if seg > highest then highest = seg
end if
next i
print highest, lowest
```
Output:
```99748         31
```

## Go

Translation of: Wren
Library: Go-rcu
```package main

import (
"fmt"
"math/rand"
"rcu"
"strings"
"time"
)

func main() {
rand.Seed(time.Now().UnixNano())
var sb strings.Builder
for i := 0; i < 1000; i++ {
sb.WriteByte(byte(rand.Intn(10) + 48))
}
number := sb.String()
for i := 99999; i >= 0; i-- {
quintet := fmt.Sprintf("%05d", i)
if strings.Contains(number, quintet) {
ci := rcu.Commatize(i)
fmt.Printf("The largest  number formed from 5 adjacent digits (%s) is: %6s\n", quintet, ci)
break
}
}
for i := 0; i <= 99999; i++ {
quintet := fmt.Sprintf("%05d", i)
if strings.Contains(number, quintet) {
ci := rcu.Commatize(i)
fmt.Printf("The smallest number formed from 5 adjacent digits (%s) is: %6s\n", quintet, ci)
return
}
}
}
```
Output:

Sample run:

```The largest  number formed from 5 adjacent digits (99928) is: 99,928
The smallest number formed from 5 adjacent digits (00120) is:    120
```

## J

```>./5([+10*])/@|:\?1000#10
```
Output:

(example)

`99929`

## jq

Works with: jq

Also works with gojq, the Go implementation of jq.

First, a direct solution using only jq's standard library and a line for generating the PRN:

```< /dev/random tr -cd '0-9' | head -c 1000 | jq -R '
length as \$n
| . as \$s
| (\$s[0:5] | tonumber) as \$m
| reduce range(1; \$n - 5) as \$i ( {min: \$m, max: \$m};
(\$s[\$i: \$i+5] | tonumber) as \$x
| if   \$x < .min then .min = \$x
elif \$x > .max then .max = \$x
else . end)
'
```
Output:
```{
"min": 224,
"max": 99772
}
```

Next, a "one-line solution" apart from generic helper functions and the line for generating the PRN:

```< /dev/random tr -cd '0-9' | head -c 1000 | jq -R '
# Input: an array
# Output: a stream of the width-long subarrays
def windows(width):
range(0; 1 + length - width)  as \$i | .[\$i:\$i+width];

def minmax(s):
reduce s as \$x ( {};
if .min == null then {min: \$x, max: \$x}
elif \$x < .min then .min = \$x
elif \$x > .max then .max = \$x else . end);

explode | minmax(windows(5) | implode | tonumber)
```

## Julia

```dig = rand(0:9, 1000)
@show maximum(evalpoly(10, dig[i:i+4]) for i in 1:length(dig)-4)
```
Output:
```maximum((evalpoly(10, dig[i:i + 4]) for i = 1:length(dig) - 4)) = 99993
```

Or, using strings, and see Nigel's comment in the discussion:

```julia> setprecision(3324)
3324

julia> s = string(BigFloat(pi))[3:end]
"141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609433057270365" ⋯ 180 bytes ⋯ "66940513200056812714526356082778577134275778960917363717872146844090122495343014654958537105079227968925892354201995611212902196086403441815981362977477130996051870721134999999837297804995105973173281609631859502445945534690830264252230825334468503526193118817101000313783875288658753320838142061717766914730359825349042875546873115956286388235378759375195778185778053217122680661300192787661119590921642019898"

julia> m, pos = maximum((s[i:i+4], i) for i in 1:length(s)-4)
("99999", 763)

julia> println("Maximum is \$m at position \$pos.")
Maximum is 99999 at position 763.
```

## Mathematica / Wolfram Language

```MinMax[FromDigits /@ Partition[RandomInteger[{0, 9}, 1000], 5, 1]]
```
Output:
`{104,99984}`

## Nim

```import random, strutils

randomize()

const N = 1000
type Digit = 0..9

# Build a 1000-digit number.
var number: array[1..N, Digit]
number[1] = rand(1..9)  # Make sure the first digit is not 0.
for i in 1..N: number[i] = rand(9)

func `>`(s1, s2: seq[Digit]): bool =
## Compare two digit sequences.
## Defining `<` rather than `>` would work too.
assert s1.len == s2.len
for i in 0..s1.high:
let comp = cmp(s1[i], s2[i])
if comp != 0: return comp == 1
result = false

var max = @[Digit 0, 0, 0, 0, 0]
for i in 5..N:
let n = number[i-4..i]
if n > max: max = n

echo "Largest 5-digit number extracted from random 1000-digit number: ", max.join()
```
Output:
`Largest 5-digit number extracted from random 1000-digit number: 99855`

## Pascal

Works with: Free Pascal

inspired by Wren
Assumes that there at least is a "1" 4 digits before end of all digits.Else I have to include sysutils and s := Format('%.5d',[i]); for leading zeros.

```var
digits,
s : AnsiString;
i : LongInt;
begin
randomize;
setlength(digits,1000);
for i := 1 to 1000 do
digits[i] := chr(random(10)+ord('0'));
for i := 99999 downto 0 do
begin
str(i:5,s);
if Pos(s,digits) > 0 then
break;
end;
writeln(s, ' found as largest 5 digit number ')
end.
```
Output:
`99889 found as largest 5 digit number `

## Perl

```#!/usr/bin/perl

use warnings;

\$_ = join '', map int rand 10, 1 .. 1e3;
my @n;
@n[ /(?=(\d{5}))/g ] = ();
print "\$#n\n";
```
Output:
```99899
```

## Phix

```with javascript_semantics
procedure shlong(string s)
string hi5 = s[1..5], lo5 = hi5
for i=2 to length(s)-4 do
string s5 = s[i..i+4]
hi5 = max(hi5,s5)
lo5 = min(lo5,s5)
end for
printf(1,"String %s: hi5:%s, lo5:%s\n",{shorten(s),hi5,lo5})
end procedure

string s = repeat(' ',1000)
for i=1 to length(s) do
s[i] = rand(10)-1+'0'
end for
shlong(s)

include mpfr.e
mpfr pi = mpfr_init(0,-1001) -- (set precision to 1,000 dp, plus the "3.")
mpfr_const_pi(pi)
s = mpfr_get_fixed(pi,1000)
s = s[3..\$]
shlong(s)
```
Output:
```String 35369847249221789712...55814915156742014134 (1,000 digits): hi5:99969, lo5:00013
String 14159265358979323846...66111959092164201989 (1,000 digits): hi5:99999, lo5:00031
```

## Python

Seeding the random number generator directly with the datetime stamp produces a warning that it will be deprecated in Python 3.9, hence the "hack" of creating a string out of the timestamp and then seeding with it.

```#Aamrun, 5th October 2021

from random import seed,randint
from datetime import datetime

seed(str(datetime.now()))

largeNum = [randint(1,9)]

for i in range(1,1000):
largeNum.append(randint(0,9))

maxNum,minNum = 0,99999

for i in range(0,994):
num = int("".join(map(str,largeNum[i:i+5])))
if num > maxNum:
maxNum = num
elif num < minNum:
minNum = num

print("Largest 5-adjacent number found ", maxNum)
print("Smallest 5-adjacent number found ", minNum)
```

Results from multiple runs :

Output:
```

```

## Quackery

```  9 random 1+
999 times [ 10 * 10 random + ]
dup 10 995 ** / swap
say "1000 digit number"
cr cr
dup echo
cr cr
[] swap 1000 times
[ 10 /mod
rot join swap ]
drop dup
0 0 rot witheach
[ swap 10 *
100000 mod +
tuck max swap ]
drop
say "largest 5 adjacent digits " echo
cr cr
5 split nip
dip dup witheach
[ swap 10 *
100000 mod +
tuck min swap ]
drop
say "smallest 5 adjacent digits "
number\$
char 0 over size 5 swap - of
swap join echo\$```
Output:
```1000 digit number

6840907174710710253578773992410923828010161316527489025709598588564725782830158923520744533291356763925463645174705745049218864529135157750600471363289558510223445011025163844199130052941524405130793922050669143532883592357897096269697903780509770222546659289832999639637730759831717125055857319129937934353617386529810429642261048827016148476352187592939822910964334104828550764225596939965675519243696921514153715258715961987394884393797714002723369560598384723111928648279375269590756880538160907807290640466592734345970439851284217252141914792365031610947925633607292897379320456985054219371373707477609843617810620097343420379245258762479642377134776965386535533204636182773979582543243782455626021964121509778973939346873293400502531060571761381532229278485105166678017234489439222625767334040651185482277484204647473910364297105035077787620562600454016296114868335345408156093266755340971022669397814048919735693462065796634326535292979494128432997646841467835174156471055078228524511787150409

## Raku

Show minimum too because... why not?

Use some Tamil Unicode numbers for brevity, and for amusement purposes.

```   ௰ - Tamil number ten
௲ - Tamil number one thousand
```

Do it 5 times for variety, it's random after all.

```(^௰).roll(௲).rotor(5 => -4)».join.minmax.bounds.put xx 5
```
Sample output:
```00371 99975
00012 99982
00008 99995
00012 99945
00127 99972```

## Ring

```digit = ""
max = 0
min = 99999
limit = 1000

for n = 1 to limit
rand = random(9)
randStr = string(rand)
digit += randStr
next

for n = 1 to len(digit)-5
res = substr(digit,n,5)
resNum = number(res)
if resNum > max
max = resNum
ok
if resNum < min
min = res
ok
next

see "The largest number is:" + nl
see max + nl
see "The smallest number is:" + nl
see min + nl```
Output:
```The largest number is:
99638
The smallest number is:
00118
```

## RPL

RPL can not handle 1000-digit numbers, so we use a 1000-digit string.

```≪ ""
1 1000 START RAND 9 * 0 RND →STR + NEXT
{ -99999 0 }
1 3 PICK SIZE 4 - FOR j
OVER j DUP 4 + SUB
STR→ NEG LASTARG 2 →LIST MAX
NEXT ABS
```
Output:
```2: "46725324552811522…
1: { 198 99886 }
```

## Ruby

```digits = %w(0 1 2 3 4 5 6 7 8 9)
arr = Array.new(1000){ digits.sample }
puts "minimum sequence %s, maximum sequence %s." % arr.each_cons(5).minmax_by{|slice| slice.join.to_i}.map(&:join)
```
Output:
```minimum sequence 00096, maximum sequence 99508.
```

## Sidef

```var k = 5
var n = 1e1000.irand

say "length(n) = #{n.len}"

var c = n.digits.cons(k)

say ("Min #{k}-digit sub-number: ", c.min_by { .digits2num }.flip.join)
say ("Max #{k}-digit sub-number: ", c.max_by { .digits2num }.flip.join)
```
Output:
```length(n) = 1000
Min 5-digit sub-number: 00072
Max 5-digit sub-number: 99861
```

## V (Vlang)

Translation of: go
```import rand
import rand.seed
import strings

fn main() {
rand.seed(seed.time_seed_array(2))
mut sb := strings.new_builder(128)
for _ in 0..1000 {
sb.write_byte(u8(rand.intn(10) or {0} + 48))
}
number := sb.str()
println('>> \$number')
for i := 99999; i >= 0; i-- {
quintet := "\${i:05}"
if number.contains(quintet) {
println("The largest  number formed from 5 adjacent digits (\$quintet) is: \${i:6}")
break
}
}
for i := 0; i <= 99999; i++ {
quintet := "\${i:05}"
if number.contains(quintet) {
println("The smallest number formed from 5 adjacent digits (\$quintet) is: \${i:6}")
return
}
}
}```
Output:
```The largest  number formed from 5 adjacent digits (99928) is: 99,928
The smallest number formed from 5 adjacent digits (00120) is:    120```

## Wren

Library: Wren-fmt

Very simple approach as there's little need for speed here.

```import "random" for Random
import "./fmt" for Fmt

var rand = Random.new()
var digits = List.filled(1000, 0)
for (i in 0...999) digits[i] = rand.int(10)
var number = digits.join()
for (r in [99999..0, 0..99999]) {
var target = (r.from == 0) ? "smallest" : "largest "
for (i in r) {
var quintet = Fmt.swrite("\$05d", i)
if (number.contains(quintet)) {
Fmt.print("The \$s number formed from 5 adjacent digits (\$s) is: \$,6d", target, quintet, i)
break
}
}
}
```
Output:

Sample output:

```The largest  number formed from 5 adjacent digits (99830) is: 99,830
The smallest number formed from 5 adjacent digits (00154) is:    154
```

## XPL0

```char    Number(1000);
int     Num, Max, I, J;
[for I:= 0 to 1000-1 do         \generate 1000-digit number
Number(I):= Ran(10);
Max:= 0;                        \find its largest 5-digit number
for I:= 0 to 1000-5 do
[Num:= 0;
for J:= 0 to 5-1 do
Num:= Num*10 + Number(I+J);
if Num > Max then
Max:= Num;
];
IntOut(0, Max);
]```
Output:
```99930
```