# 10001th prime

10001th prime 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.

Find and show on this page the 10001st prime number.

## 11l

```V k = 10001
V n = k * 17
V primes = [1B] * n
primes[0] = primes[1] = 0B

L(i) 2 .. Int(sqrt(n))
I !primes[i]
L.continue
L(j) (i * i .< n).step(i)
primes[j] = 0B

L(i) 0 .< n
I primes[i]
I k == 1
print(i)
L.break
k--```
Output:
```104743
```

## ALGOL 68

Chebyshev showed that the limit of pi( n ) / ( n / ln(n) ) as n approaches infinity is 1 ( pi( n ) is the number of primes up to n ).
The APPROXIMATESIEVESIZEFOR operator uses this to find a rough value for size of sieve needed to contain the required number of primes.

```BEGIN # find the 10001st prime #
PR read "primes.incl.a68" PR
# construct a sieve of primes that should be large enough to contain 10001 primes #
INT required prime = 10 001;
[]BOOL prime = PRIMESIEVE APPROXIMATESIEVESIZEFOR required prime;
INT p count := 1;
FOR n FROM 3 BY 2 WHILE p count < required prime DO
IF prime[ n ] THEN
p count +:= 1;
IF p count = required prime THEN
print( ( "Prime ", whole( required prime, 0 ), " is ", whole( n, 0 ) ) )
FI
FI
OD
END```
Output:
```Prime 10001 is 104743
```

## Arturo

```primes: select 2..110000 => prime?
print primes\[10000]
```
Output:
`104743`

## AWK

#### Converted from FreeBASIC

```# syntax: GAWK -f 10001TH_PRIME.AWK

BEGIN {
printf("%s\n",main(10001))
exit(0)
}
function main(n,  p,pn) {
if (n == 1) { return(2) }
p = 3
pn = 1
while (pn < n) {
if (is_prime(p)) {
pn++
}
p += 2
}
return(p-2)
}
function is_prime(n,  d) {
d = 5
if (n < 2) { return(0) }
if (n % 2 == 0) { return(n == 2) }
if (n % 3 == 0) { return(n == 3) }
while (d*d <= n) {
if (n % d == 0) { return(0) }
d += 2
if (n % d == 0) { return(0) }
d += 4
}
return(1)
}
```

#### Idiomatic

```# awk -f 10001prime.awk

# n: odd number and n>8
function IsOddPrime(n,	i,limit) {
limit = int(sqrt(n))
for (i=3;i <= limit;i+=2)
if (n%i==0) return 0
return 1
}

# pos: positive
function PrimePosition(pos,	prime){
pos -= ( (pos==1) ? prime=2 : prime=3 ) - 1
while (pos)
if (IsOddPrime(prime+=2)) pos--
return prime
}

BEGIN { print PrimePosition(10001) }
```
Output:
```104743
```

## BASIC

### BASIC256

```function isPrime(v)
if v < 2 then return False
if v mod 2 = 0 then return v = 2
if v mod 3 = 0 then return v = 3
d = 5
while d * d <= v
if v mod d = 0 then return False else d += 2
end while
return True
end function

function prime(n)
if n=1 then return 2
p = 3
pn = 1
while pn < n
if isPrime(p) then pn += 1
p += 2
end while
return p-2
end function

print prime(10001)
end```

### Chipmunk Basic

Works with: Chipmunk Basic version 3.6.4

The GW-BASIC solution works without any changes.

### FreeBASIC

```#include "isprime.bas"
function prime( n as uinteger ) as ulongint
if n=1 then return 2
dim as integer p=3, pn=1
while pn<n
if isprime(p) then pn + = 1
p += 2
wend
return p-2
end function

print prime(10001)```
Output:
`104743`

### Gambas

```'Use "isprime.bas"

Public Sub Main()

Print prime(10001)

End

Function prime(n As Integer) As Long

If n = 1 Then Return 2
Dim p As Integer = 3, pn As Integer = 1
While pn < n
If isPrime(p) Then pn += 1
p += 2
Wend
Return p - 2

End Function
```
Output:
`104743`

### GW-BASIC

Works with: BASICA
```10 PN = 1
20 P = 3
30 WHILE PN < 10001
40 GOSUB 100
50 IF Q = 1 THEN PN = PN + 1
60 P = P + 2
70 WEND
80 PRINT P - 2
90 END
100 REM Tests if a number is prime
110 Q = 0
120 IF P = 2 THEN Q = 1: RETURN
130 IF P = 3 THEN Q = 1: RETURN
140 I = 1
150 I = I + 2
160 IF INT(P / I) * I = P THEN RETURN
170 IF I * I <= P THEN GOTO 150
180 Q = 1
190 RETURN```
Output:
`104743`

### PureBasic

```Procedure isPrime(v.i)
If     v <= 1    : ProcedureReturn #False
ElseIf v < 4     : ProcedureReturn #True
ElseIf v % 2 = 0 : ProcedureReturn #False
ElseIf v < 9     : ProcedureReturn #True
ElseIf v % 3 = 0 : ProcedureReturn #False
Else
Protected r = Round(Sqr(v), #PB_Round_Down)
Protected f = 5
While f <= r
If v % f = 0 Or v % (f + 2) = 0
ProcedureReturn #False
EndIf
f + 6
Wend
EndIf
ProcedureReturn #True
EndProcedure

Procedure prime(n.i)
If n = 1
ProcedureReturn 2
EndIf
p.i = 3
pn.i = 1
While pn < n
If isprime(p)
pn + 1
EndIf
p + 2
Wend
ProcedureReturn p-2
EndProcedure

OpenConsole()
n = prime(10001)
PrintN(Str(n))
CloseConsole()```

### QB64

#### Trial Division Method

```max=10001: n=1: p=0: t=Timer ' PRIMES.bas russian DANILIN
While n <= max ' 10001 104743 0.35 seconds
f=0: j=2
While f < 1
If j >= p ^ 0.5 Then f=2
If p Mod j = 0 Then f=1
j=j+1
Wend
If f <> 1 Then n=n+1: ' Print n, p
p=p+1
Wend
Print n-1, p-1, Timer-t
```
Output:
`10001 104743 0.35 seconds`

#### More Efficient TD Method

```'JRace's results:
'Original version: 10001   104743  .21875
'This version:     10001   104743  .109375
max = 10001: n=1: p=0: t = Timer ' PRIMES.bas
While n <= max ' 10001 104743 0.35 seconds
f = 0: j = 2
pp = p^0.5 'NEW VAR: move exponentiation here, to outer loop
While f < 1
' If j >= p ^ 0.5 Then f=2 'original line
' NOTE: p does not change in this loop,
'      therefore p^0.5 doesn't change;
'      moving p^0.5 to the outer loop will eliminate a lot of FP math)
If j >= pp Then f=2 'new version with exponentiation relocated
If p Mod j = 0 Then f=1
j=j+1
Wend
If f <> 1 Then n=n+1: ' Print n, p
p=p+1
Wend
Print n-1, p-1, Timer - t
```
Output:
`10001 104743 0.11 seconds`

### True BASIC

#### Trial Division Method

Translation of: QB64
```LET max = 10001
LET n = 1
LET p = 0
DO WHILE n <= max
LET f = 0
LET j = 2
DO WHILE f < 1
IF j >= p^0.5 THEN LET f = 2
IF REMAINDER(p, j) = 0 THEN LET f = 1
LET j = j+1
LOOP
IF f <> 1 THEN LET n = n+1
LET p = p+1
LOOP
PRINT p-1
END
```
Output:
`104743`

### Yabasic

```sub isPrime(v)
if v < 2 then return False : fi
if mod(v, 2) = 0 then return v = 2 : fi
if mod(v, 3) = 0 then return v = 3 : fi
d = 5
while d * d <= v
if mod(v, d) = 0 then return False else d = d + 2 : fi
wend
return True
end sub

sub prime(n)
if n = 1 then return 2 : fi
p = 3
pn = 1
while pn<n
if isPrime(p) then pn = pn + 1 : fi
p = p + 2
wend
return p-2
end sub

print prime(10001)
end```

## bc

```a[0] = 2
a[o = 1] = 3
for (n = 5; o < 10000; n += 2) {
for (i = 1; n % (p = a[i]) != 0; ++i) {
if (p * p > n) {
a[++o] = n
break
}
}
}
a[o]
```
Output:
`104743`

## C

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

int isprime( int p ) {
int i;
if(p==2) return 1;
if(!(p%2)) return 0;
for(i=3; i*i<=p; i+=2) {
if(!(p%i)) return 0;
}
return 1;
}

int prime( int n ) {
int p, pn=1;
if(n==1) return 2;
for(p=3;pn<n;p+=2) {
if(isprime(p)) pn++;
}
return p-2;
}

int main(void) {
printf( "%d\n", prime(10001) );
return 0;
}
```
Output:
`104743`

## C#

### Sieve vs Trial Division

Comparing performance of the one-at-a-time trial division method vs the sieve of Eratosthenes method. About ten times faster for the sieve. It may appear that the sieve may be off by one, `pr[10000]` but since the array is zero based, it's the 10001st value.

```using System; class Program {

static bool isprime(uint p ) { if ((p & 1) == 0) return p == 2;
if ((p % 3) == 0) return p == 3;
for (uint i = 5, d = 4; i * i <= p; i += (d = 6 - d))
if (p % i == 0) return false; return true; }

static uint prime(uint n) { uint p, d, pn;
for (p = 5, d = 4, pn = 2; pn < n; p += (d = 6 - d))
if (isprime(p)) pn++; return p - d; }

static void Main(string[] args) {
Console.WriteLine("One-at-a-time trial division vs sieve of Eratosthenes");
var sw = System.Diagnostics.Stopwatch.StartNew();
var t = prime(10001); sw.Stop(); double e1, e2;
Console.Write("{0:n0} {1} ms", prime(10001),
e1 = sw.Elapsed.TotalMilliseconds);
sw.Restart(); uint n = 105000, i, j; var pr = new uint[10100];
pr[0] = 2; pr[1] = 3; uint pc = 2, r, d, ii;
var pl = new bool[n + 1]; r = (uint)Math.Sqrt(n);
for (i = 9; i < n; i += 6) pl[i] = true;
for (i = 5, d = 4; i <= r; i += (d = 6 - d)) if (!pl[i]) {
pr[pc++] = i; for (j = i * i, ii = i << 1; j <= n; j += ii)
pl[j] = true; }
for ( ;i <= n; i += (d = 6 - d)) if (!pl[i]) pr[pc++] = i;
t = pr[10000]; sw.Stop();
Console.Write("  {0:n0} {1} μs  {2:0.000} times faster", t,
(e2 = sw.Elapsed.TotalMilliseconds) * 1000.0, e1 / e2); } }
```
Output @ Tio.run:
```One-at-a-time trial division vs sieve of Eratosthenes
104,743 3.8943 ms  104,743 357.9 μs  10.881 times faster```

### Alternative Trial Division Method

```using System; using System.Text; // PRIME_numb.cs russian DANILIN
namespace p10001 // 1 second  10001  104743
{ class Program // rextester.com/ZBEPGE34760
{ static void Main(string[] args)
{ int max=10001; int n=1; int p=1; int f; int j; long s;
while (n <= max)
{ f=0; j=2; s=Convert.ToInt32(Math.Pow(p,0.5));
while (f < 1)
{ if (j >= s)
{ f=2; }
if (p % j == 0) { f=1; }
j++;
}
if (f != 1) { n++; } // Console.WriteLine("{0} {1}", n, p);
p++;
}
Console.Write("{0} {1}", n-1, p-1);
}}}
```
Output:
`104743`

## C++

Library: Primesieve
```#include <iostream>
#include <locale>

#include <primesieve.hpp>

int main() {
std::cout.imbue(std::locale(""));
std::cout << "The 10,001st prime is " << primesieve::nth_prime(10001) << ".\n";
}
```
Output:
```The 10,001st prime is 104,743.
```

## Dart

```import 'dart:math';

bool isPrime(int n) {
if (n <= 1) return false;
if (n == 2) return true;
for (int i = 2; i <= sqrt(n); ++i) {
if (n % i == 0) return false;
}
return true;
}

int prime(int n) {
int p, pn = 1;
if (n == 1) return 2;
for (p = 3; pn < n; p += 2) {
if (isPrime(p)) pn++;
}
return p - 2;
}

void main() {
print(prime(10001));
}
```
Output:
`104743`

## Delphi

Works with: Delphi version 6.0
Library: none
```function IsPrime(N: integer): boolean;
{Fast, optimised prime test}
var I,Stop: integer;
begin
if (N = 2) or (N=3) then Result:=true
else if (n <= 1) or ((n mod 2) = 0) or ((n mod 3) = 0) then Result:= false
else
begin
I:=5;
Stop:=Trunc(sqrt(N));
Result:=False;
while I<=Stop do
begin
if ((N mod I) = 0) or ((N mod (I + 2)) = 0) then exit;
Inc(I,6);
end;
Result:=True;
end;
end;

function Find10001thPrime: integer;
{Return the 10,001th prime number}
var Count: integer;
begin
Count:=1;
Result:= 3;
while true do
begin
if IsPrime(Result) then Count:= Count+1;
if Count=10001 then exit;
Inc(Result,2);
end;
end;
```
Output:
```10,001th Prime= 104,743
```

## D

Translation of: C
```import std.stdio;

int isprime( int p ) {
int i;

if(p==2) return 1;

if(!(p%2)) return 0;

for(i=3; i*i<=p; i+=2) {
if(!(p%i)) return 0;
}

return 1;
}

int prime( int n ) {

if(n==1) return 2;

int p, pn=1;

for(p=3;pn<n;p+=2) {
if(isprime(p)) pn++;
}

return p-2;
}

void main()
{
writeln(prime(10_001));
}
```
Output:
`104743`

## Erlang

```-module(task_10001th_prime).

-export([main/0]).

% 2 and 3 are both primes, so we start searching at 5
main() -> search(5, 2, 2).

search(N, Inc, Count) when Count < 10_001 ->
case is_prime(N) of
true -> search(N + Inc, 6 - Inc, Count + 1);
false -> search(N + Inc, 6 - Inc, Count)
end;
search(N, Inc, _) -> N - 6 + Inc.

is_prime(P) -> is_prime(P, 5, 2).

is_prime(N, P, _) when P * P > N -> true;
is_prime(N, P, _) when N rem P =:= 0 -> false;
is_prime(N, P, Inc) -> is_prime(N, P + Inc, 6 - Inc).
```
Output:
``` timer:tc(task_10001th_prime, main, []).
{68615,104743}
```

## EasyLang

```fastfunc isprim num .
if num mod 2 = 0 and num > 2
return 0
.
i = 3
while i <= sqrt num
if num mod i = 0
return 0
.
i += 2
.
return 1
.
i = 2
repeat
if isprim i = 1
nprim += 1
.
until nprim = 10001
i += 1
.
print i```
Output:
`104743`

## F#

This task uses Extensible Prime Generator (F#)

```// 10001st prime. Nigel Galloway: November 22nd., 2021
printfn \$"%d{Seq.item 10000 (primes32())}"
```
Output:
```104743
```

## Factor

```USING: math math.primes prettyprint ;

2 10,000 [ next-prime ] times .
```
Output:
```104743
```

## Fermat

```Prime(10001);
```
Output:
`104743`

## Frink

`nth[primes[], 10001-1]`
Output:
```104743
```

## FutureBasic

```local fn IsPrime( n as NSUInteger ) as BOOL
BOOL       isPrime = YES
NSUInteger i

if n < 2        then exit fn = NO
if n = 2        then exit fn = YES
if n mod 2 == 0 then exit fn = NO
for i = 3 to int(n^.5) step 2
if n mod i == 0 then exit fn = NO
next
end fn = isPrime

local fn Prime( n as NSUInteger ) as long
long p = 3, pn = 1

if n = 1 then exit fn = 2
while ( pn < n )
if fn IsPrime( p ) then pn++
p += 2
wend
p -= 2
end fn = p

print fn Prime(10001)

HandleEvents```
Output:
```104743
```

## Go

```package main

import "fmt"

func isPrime(n int) bool {
if n == 1 {
return false
}
i := 2
for i*i <= n {
if n%i == 0 {
return false
}
i++
}
return true
}

func main() {
var final, pNum int

for i := 1; pNum < 10001; i++ {
if isPrime(i) {
pNum++
}
final = i
}
fmt.Println(final)
}
```
Output:
`104743`

## J

```p:10000      NB. the index starts at 0; p:0 = 2
```
Output:
`104743`

## Java

Uses the PrimeGenerator class from Extensible prime generator#Java.

```public class NthPrime {
public static void main(String[] args) {
System.out.printf("The 10,001st prime is %,d.\n", nthPrime(10001));
}

private static int nthPrime(int n) {
assert n > 0;
PrimeGenerator primeGen = new PrimeGenerator(10000, 100000);
int prime = primeGen.nextPrime();
while (--n > 0)
prime = primeGen.nextPrime();
return prime;
}
}
```
Output:
```The 10,001st prime is 104,743.
```

## JavaScript

Prime 10001 from Russia jdoodle.com/h/2V1

```var max = 10001, n=1, p=1; var f,j,s
while (n <= max)
{ f=0; j=2; s = parseInt(Math.pow(p, 0.5))
while (f < 1)
{ if (j >= s) f=2
if ( p % j == 0 ) f=1
j++
}
if (f != 1) n++ // { document.write(n +" "+ p +"<br>") }
p++
}
document.write("<br>"+ (n-1) +" "+ (p-1) +"<br>" )
```
Output:
```10001 104743
```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

A solution without any pre-determined numbers or guesses.

See Erdős-primes#jq for a suitable definition of `is_prime` as used here.

```# Output: a stream of the primes
def primes: 2, (range(3; infinite; 2) | select(is_prime));

# jq uses an index-origin of 1 and so:
nth(10000; primes)```
Output:
```104743
```

## Julia

```julia> using Primes

julia> prime(10001)
104743
```

## Mathematica / Wolfram Language

```Prime[10001]
```
Output:
```
104743

```

## Maxima

```primes_first_n(n) := (
if n<=1 then
[]
else (
L : [2],
for i:2 thru n do (
L : append(L, [next_prime(last(L))])
),
L
)
)\$
last(primes_first_n(10001));
```
Output:
```
104743

```

## MiniScript

```isPrime = function(n)
s = floor(n ^ 0.5)
for i in range(2, s)
if n % i == 0 then return false
end for
return true
end function

tt = time
c = 2
p = 2
lastPrime = 2
while c < 10001
p += 1
if isPrime(p) then
c += 1
end if
end while
print p
```
Output:
`104743`

## Nim

```func isPrime(n: Positive): bool =
## Return true if "n" is prime.
## Assume n >= 5 and n not multiple of 2 and 3.
var d = 5
var step = 2
while d * d <= n:
if n mod d == 0:
return false
inc d, step
step = 6 - step
result = true

iterator primes(): tuple[rank, value: int] =
## Yield the primes preceded by their rank in the sequence.
yield (1, 2)
yield (2, 3)
var idx = 2
var n = 5
var step = 2
while true:
if n.isPrime:
inc idx
yield (idx, n)
inc n, step
step = 6 - step

for (i, n) in primes():
if i == 10001:
echo "10001st prime: ", n
break
```
Output:
```10001st prime: 104743
```

## OCaml

Using the function `seq_primes` from Extensible prime generator#OCaml:

```let () =
seq_primes |> Seq.drop 10000 |> Seq.take 1 |> Seq.iter (Printf.printf "%u\n")
```
Output:
`104743`

## PARI/GP

`prime(10001)`
Output:
`%1 = 104743`

## Pascal

### Free Pascal

```Program nth_prime;

Function nthprime(x : uint32): uint32;
Var primes : array Of uint32;
i,pr,count : uint32;
found : boolean;
Begin
setlength(primes, x);
primes[1] := 2;
count := 1;
i := 3;
Repeat
found := FALSE;
pr := 0;
Repeat
inc(pr);
found := i Mod primes[pr] = 0;
Until (primes[pr]*primes[pr] > i) Or found;
Begin
inc(count);
primes[count] := i;
End;
inc(i,2);
Until count = x;
nthprime := primes[x];
End;

Begin
writeln(nthprime(10001));
End.
```
Output:
```104743
```

## Perl

Library: ntheory
```use strict;
use warnings;
use feature 'say';

# the lengthy wait increases the delight in finally seeing the answer
my(\$n,\$c) = (1,0);
while () {
\$c++ if (1 x ++\$n) !~ /^(11+)\1+\$/;
\$c == 10_001 and say \$n and last;
}

# or if for some reason you want the answer quickly
use ntheory 'nth_prime';
say nth_prime(10_001);
```
Output:
```104743
104743```

## Phix

```with js ?get_prime(10001)
```
Output:
```104743
```

## Picat

```go ?=>
println(nth_prime(10001)),

% faster but probably considered cheating
nth(10001,primes(200000),P),
println(P).

nth_prime(Choosen) = P =>
nth_prime(1,0,Choosen, P).

nth_prime(Num, Choosen, Choosen, Num) :-
prime(Num).
nth_prime(Num, Ix, Choosen, P) :-
Ix < Choosen,
next_prime(Num, P2),
nth_prime(P2, Ix+1, Choosen, P).

next_prime(Num, P) :-
next_prime2(Num+1, P).
next_prime2(Num, Num) :-
prime(Num).
next_prime2(Num, P) :-
next_prime2(Num+1,P).```
Output:
```104743
104743```

## Prolog

for swi-prolog

```isPrime(2).                 % prime generator
isPrime(N):-
between(3, inf, N),
N /\ 1 > 0,             % odd
M is floor(sqrt(N)) - 1, % reverse 2*I+1
Max is M div 2,
forall(between(1, Max, I), N mod (2*I+1) > 0).

do:- Index is 10001,
findnsols(Index, N, isPrime(N), PrimeList),!,
last(PrimeList, PrimeAtIndex),
format('prime(~d) is ~d', [Index, PrimeAtIndex]), nl.
```
Output:
```?- do.
prime(10001) is 104743
true.```

## Python

### Trial Division Method

```#!/usr/bin/python

def isPrime(n):
for i in range(2, int(n**0.5) + 1):
if n % i == 0:
return False
return True

def prime(n: int) -> int:
if n == 1:
return 2
p = 3
pn = 1
while pn < n:
if isPrime(p):
pn += 1
p += 2
return p-2

if __name__ == '__main__':
print(prime(10001))
```
Output:
`104743`

### Alternative Trial Division Method

```import time; max=10001; n=1; p=1; # PRIME_numb.py russian DANILIN
while n<=max: # 78081 994271 45 seconds
f=0; j=2; s = int(p**0.5) # rextester.com/AAOHQ6342
while f < 1:
if j >= s:
f=2
if p % j == 0:
f=1
j+=1
if f != 1:
n+=1;
#print(n,p);
p+=1
print(n-1,p-1)
print(time.perf_counter())
```
Output:
`10001 104743 7 seconds`

## Quackery

The 10001st prime number is the 10000th odd prime number.

`prime` is defined at Miller–Rabin primality test#Quackery.

`  1 10000 times [ 2 + dup prime until ] echo`
Output:
`104743`

## R

```library(primes)
nth_prime(10001)
```
Output:
`104743`

## Racket

```#lang racket
(require math/number-theory)
; Index starts at 0, (nth-prime 0) is 2
(nth-prime 10000)
```
Output:
`104743`

## Raku

```say (^∞).grep( &is-prime )[10000]
```
Output:
`104743`

## REXX

```/* REXX */
Parse Version v
Say v
Call Time 'R'
z=1
p.0=3
p.1=2
p.2=3
p.3=5
Do n=5 By 2 Until z=10001
If right(n,1)=5 Then Iterate
Do i=2 To p.0 Until b**2>n
b=p.i
If n//b=0 Then Leave
End
If b**2>n Then Do
z=p.0+1
p.z=n
p.0=z
End
End
Say z n time('E')
```
Output:
```H:\>rexx prime10001
REXX-ooRexx_5.0.0(MT)_64-bit 6.05 8 Sep 2020
10001 104743 0.219000

H:\>regina prime10001
REXX-Regina_3.9.4(MT) 5.00 25 Oct 2021
10001 104743 .615000```

## Ring

```load "stdlib.ring"
see "working..." + nl
num = 0 pr = 0 limit = 10001

while true
num++
if isprime(num) pr++ ok
if pr = limit exit ok
end

see "" + num + nl + "done..." + nl```
Output:
```working...
The 10001th prime is: 104743
done...
```

## RPL

Works with: HP version 49
```« 2
1 10000 START NEXTPRIME NEXT
```
Output:
```1: 104743
```

## Ruby

```require "prime"
puts  Prime.lazy.drop(10_000).next
```
Output:
```104743
```

## Rust

```// [dependencies]
// primal = "0.3"

fn main() {
let p = primal::StreamingSieve::nth_prime(10001);
println!("The 10001st prime is {}.", p);
}
```
Output:
```The 10001st prime is 104743.
```

## Scala

```object Prime10001 extends App {

val oddPrimes: LazyList[Int] = 3 #:: LazyList.from(5, 2)
.filter(p => oddPrimes.takeWhile(_ <= math.sqrt(p)).forall(p % _ > 0))
val primes = 2 #:: oddPrimes

val index = 10_001
println(s"prime(\$index): " + primes.drop(index - 1).take(1).head)
}
```
Output:
```prime(10001): 104743
```

## Sidef

`say 10001.prime`
Output:
```104743
```

## Wren

Library: Wren-math
Library: Wren-fmt
```import "./math" for Int
import "./fmt" for Fmt

var n = 10001
var limit = (n.log * n * 1.2).floor  // should be enough
var primes = Int.primeSieve(limit)
Fmt.print("The \$,r prime is \$,d.", n, primes[n-1])
```
Output:
```The 10,001st prime is 104,743.
```

## XPL0

```func IsPrime(N); \Return 'true' if odd N > 2 is prime
int  N, I;
[for I:= 3 to sqrt(N) do
[if rem(N/I) = 0 then return false;
I:= I+1;
];
return true;
];

int C, N;
[C:= 1;         \count 2 as first prime
N:= 3;
loop    [if IsPrime(N) then
[C:= C+1;
if C = 10001 then quit;
];
N:= N+2;
];
IntOut(0, N);
]```
Output:
```104743
```

## Zig

```const std = @import("std");
const stdout = @import("std").io.getStdOut().writer();

const limit = 10001;

fn isPrime(x: usize) bool {
if (x % 2 == 0) return false;

var i: usize = 3;
while (i * i <= x) : (i += 2) {
if (x % i == 0) return false;
}
return true;
}

pub fn main() !void {
var cnt: usize = 0;
var last: usize = 0;
var n: usize = 1;

while (cnt < limit) : (n += 1) {
if (isPrime(n)) {
cnt += 1;
last = n;
}
}
try stdout.print("{d}st prime number is: {d}\n", .{ limit, last });
}
```
Output:
```10001st prime number is: 104743
```