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Primes whose sum of digits is 25

From Rosetta Code
Primes whose sum of digits is 25 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.
Task

Show primes which sum of its decimal digits is   25


Find primes     n     such that     n  <  5000


Stretch goal

Show the count of all such primes that do not contain any zeroes in the range:  

(997   ≤   n   ≤   1,111,111,111,111,111,111,111,111).



ALGOL 68[edit]

BEGIN # find primes whose digits sum to 25 #
# show all sum25 primes below 5000 #
PR read "primes.incl.a68" PR
[]BOOL prime = PRIMESIEVE 4999;
INT p25 count := 0;
FOR n TO UPB prime DO
IF prime[ n ] THEN
# have a prime, check for a sum25 prime #
INT digit sum := 0;
INT v := n;
WHILE v > 0 DO
INT digit = v MOD 10;
digit sum +:= digit;
v OVERAB 10
OD;
IF digit sum = 25 THEN
print( ( " ", whole( n, 0 ) ) );
p25 count +:= 1
FI
FI
OD;
print( ( newline, "Found ", whole( p25 count, 0 ), " sum25 primes below ", whole( UPB prime + 1, 0 ), newline ) )
END
Output:
 997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993
Found 17 sum25 primes below 5000

Stretch Goal[edit]

Uses the candidate generating algorithm used by Phix, Go
Uses the Miller Rabin primality test and the pow mod procedure from prelude/pow_mod

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32
BEGIN
PROC pow mod = (LONG LONG INT b,in e, modulus)LONG LONG INT: (
LONG LONG INT sq := b, e := in e;
LONG LONG INT out:= IF ODD e THEN b ELSE 1 FI;
e OVERAB 2;
WHILE e /= 0 DO
sq := sq * sq MOD modulus;
IF ODD e THEN out := out * sq MOD modulus FI ;
e OVERAB 2
OD;
out
);
INT p25 count := 0;
PROC sum25 = ( LONG LONG INT p, INT rem )VOID:
FOR i TO IF rem > 9 THEN 9 ELSE rem FI DO
IF rem > i THEN
sum25( ( p * 10 ) + i, rem - i )
ELIF ODD i AND i /= 5 THEN
LONG LONG INT n = ( p * 10 ) + i;
IF n MOD 3 /= 0 THEN
BOOL is prime := TRUE;
# miller rabin primality test #
INT k = 10;
LONG LONG INT d := n - 1;
INT s := 0;
WHILE NOT ODD d DO
d OVERAB 2;
s +:= 1
OD;
TO k WHILE is prime DO
LONG LONG INT a := 2 + ENTIER (random*(n-3));
LONG LONG INT x := pow mod(a, d, n);
IF x /= 1 THEN
BOOL done := FALSE;
TO s WHILE NOT done DO
IF x = n-1
THEN done := TRUE
ELSE x := x * x MOD n
FI
OD;
IF NOT done THEN IF x /= n-1 THEN is prime := FALSE FI FI
FI
OD;
# END miller rabin primality test #
IF is prime THEN
# IF ( p25 count + 1 ) MOD 100 = 0 THEN print( ( whole( p25 count + 1, -8 ), whole( n, -30 ), newline ) ) FI; #
p25 count +:= 1
FI
FI
FI
OD;
sum25( 0, 25 );
print( ( "There are ", whole( p25 count, 0 ), " sum25 primes that contain no zeroes", newline ) )
END
Output:

Note that ALGOL 68G under Windows is fully interpreted so runtime is not of the same order as the Phix and Go samples. Under Linux with optimisation and compilation, it should be faster than under Windows.

There are 1525141 sum25 primes that contain no zeroes

ALGOL W[edit]

begin % find some primes whose digits sum to 25 %
 % sets p( 1 :: n ) to a sieve of primes up to n %
procedure Eratosthenes ( logical array p( * ) ; integer value n ) ;
begin
p( 1 ) := false; p( 2 ) := true;
for i := 3 step 2 until n do p( i ) := true;
for i := 4 step 2 until n do p( i ) := false;
for i := 3 step 2 until truncate( sqrt( n ) ) do begin
integer ii; ii := i + i;
if p( i ) then for pr := i * i step ii until n do p( pr ) := false
end for_i ;
end Eratosthenes ;
integer MAX_NUMBER;
MAX_NUMBER := 4999;
begin
logical array prime( 1 :: MAX_NUMBER );
integer pCount;
 % sieve the primes to MAX_NUMBER %
Eratosthenes( prime, MAX_NUMBER );
 % find the primes whose digits sum to 25 %
pCount := 0;
for i := 1 until MAX_NUMBER do begin
if prime( i ) then begin
integer dSum, v;
v  := i;
dSum := 0;
while v > 0 do begin
dSum := dSum + ( v rem 10 );
v  := v div 10
end while_v_gt_0 ;
if dSum = 25 then begin
writeon( i_w := 4, s_w := 0, " ", i );
pCount := pCount + 1;
if pCount rem 20 = 0 then write()
end if_prime_pReversed
end if_prime_i
end for_i ;
write( i_w := 1, s_w := 0, "Found ", pCount, " sum25 primes below ", MAX_NUMBER + 1 )
end
end.
Output:
  997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993
Found 17 sum25 primes below 5000

AppleScript[edit]

Functional[edit]

Not fast. This approach takes over 20 seconds here.

use AppleScript version "2.4"
use framework "Foundation"
use scripting additions
 
 
--------- PRIMES WITH DECIMAL DIGITS SUMMING TO 25 -------
 
-- primes :: [Int]
on primes()
-- A non-finite list of primes.
 
set ca to current application
script
property dict : ca's NSMutableDictionary's alloc's init()
property n : 2
on |λ|()
set xs to dict's objectForKey:(n as string)
repeat until missing value = xs
repeat with x in (xs as list)
set m to x as number
set k to (n + m) as string
 
set ys to (dict's objectForKey:(k))
if missing value ≠ ys then
set zs to ys
else
set zs to ca's NSMutableArray's alloc's init()
end if
 
(zs's addObject:(m))
 
(dict's setValue:(zs) forKey:(k))
(dict's removeObjectForKey:(n as string))
end repeat
 
set n to 1 + n
set xs to (dict's objectForKey:(n as string))
end repeat
 
set p to n
dict's setValue:({n}) forKey:((n * n) as string)
set n to 1 + n
set xs to missing value
return p
end |λ|
end script
end primes
 
-- digitSum :: Int -> Int
on digitSum(n)
-- Sum of the decimal digits of n.
 
set m to 0
set cs to characters of (n as string)
repeat with c in cs
set m to m + ((id of c) - 48)
end repeat
end digitSum
 
--------------------------- TEST -------------------------
on run
script q
on |λ|(x)
5000 > x
end |λ|
end script
 
script p
on |λ|(n)
25 = digitSum(n)
end |λ|
end script
 
 
set startTime to current date
set xs to takeWhile(q, filterGen(p, primes()))
set elapsedSeconds to ((current date) - startTime) as string
 
showList(xs)
end run
 
------------------------- GENERIC ------------------------
 
-- filterGen :: (a -> Bool) -> Gen [a] -> Gen [a]
on filterGen(p, gen)
-- Non-finite stream of values which are
-- drawn from gen, and satisfy p
script
property mp : mReturn(p)'s |λ|
on |λ|()
set v to gen's |λ|()
repeat until mp(v)
set v to gen's |λ|()
end repeat
return v
end |λ|
end script
end filterGen
 
 
-- intercalateS :: String -> [String] -> String
on intercalate(delim, xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, delim}
set s to xs as text
set my text item delimiters to dlm
s
end intercalate
 
 
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
-- The list obtained by applying f
-- to each element of xs.
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
 
 
-- mReturn :: First-class m => (a -> b) -> m (a -> b)
on mReturn(f)
-- 2nd class handler function lifted into 1st class script wrapper.
if script is class of f then
f
else
script
property |λ| : f
end script
end if
end mReturn
 
 
-- showList :: [a] -> String
on showList(xs)
"[" & intercalate(",", map(my str, xs)) & "]"
end showList
 
 
-- str :: a -> String
on str(x)
x as string
end str
 
 
-- takeWhile :: (a -> Bool) -> Gen [a] -> [a]
on takeWhile(p, xs)
set ys to {}
set v to |λ|() of xs
tell mReturn(p)
repeat while (its |λ|(v))
set end of ys to v
set v to xs's |λ|()
end repeat
end tell
return ys
end takeWhile
Output:
[997,1699,1789,1879,1987,2689,2797,2887,3499,3697,3769,3877,3967,4597,4759,4957,4993]

Idiomatic[edit]

Primes with silly properties are getting a bit tedious. But hey. This takes just under 0.02 seconds.

on sieveOfEratosthenes(limit)
script o
property numberList : {missing value}
end script
 
repeat with n from 2 to limit
set end of o's numberList to n
end repeat
 
repeat with n from 2 to (limit ^ 0.5) div 1
if (item n of o's numberList is n) then
repeat with multiple from n * n to limit by n
set item multiple of o's numberList to missing value
end repeat
end if
end repeat
 
return o's numberList's numbers
end sieveOfEratosthenes
 
on sumOfDigits(n) -- n assumed to be a positive decimal integer.
set sum to n mod 10
set n to n div 10
repeat until (n = 0)
set sum to sum + n mod 10
set n to n div 10
end repeat
 
return sum
end sumOfDigits
 
on numbersWhoseDigitsSumTo(numList, targetSum)
script o
property numberList : numList
property output : {}
end script
 
repeat with n in o's numberList
if (sumOfDigits(n) = targetSum) then set end of o's output to n's contents
end repeat
 
return o's output
end numbersWhoseDigitsSumTo
 
-- Task code:
return numbersWhoseDigitsSumTo(sieveOfEratosthenes(4999), 25)
Output:
{997, 1699, 1789, 1879, 1987, 2689, 2797, 2887, 3499, 3697, 3769, 3877, 3967, 4597, 4759, 4957, 4993}

Arturo[edit]

primes: select 1..5000 => prime?
 
loop split.every: 3 select primes 'p [25 = sum digits p] 'a ->
print map a => [pad to :string & 5]
Output:
  997  1699  1789 
 1879  1987  2689 
 2797  2887  3499 
 3697  3769  3877 
 3967  4597  4759 
 4957  4993

AWK[edit]

 
# syntax: GAWK -f PRIMES_WHICH_SUM_OF_DIGITS_IS_25.AWK
BEGIN {
start = 1
stop = 5000
for (i=start; i<=stop; i++) {
if (is_prime(i)) {
sum = 0
for (j=1; j<=length(i); j++) {
sum += substr(i,j,1)
}
if (sum == 25) {
printf("%d ",i)
count++
}
}
}
printf("\nPrime numbers %d-%d whose digits sum to 25: %d\n",start,stop,count)
exit(0)
}
function is_prime(x, i) {
if (x <= 1) {
return(0)
}
for (i=2; i<=int(sqrt(x)); i++) {
if (x % i == 0) {
return(0)
}
}
return(1)
}
 
Output:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993
Prime numbers 1-5000 whose digits sum to 25: 17

BASIC[edit]

BASIC256[edit]

Translation of: FreeBASIC
 
function isprime(num)
for i = 2 to int(sqr(num))
if (num mod i = 0) then return False
next i
return True
end function
 
function digit_sum(num)
sum25 = 0
for j = 1 to length(num)
sum25 += int(mid(string(num),j,1))
next j
return sum25
end function
 
inicio = 1: final = 5000
total = 0
for i = inicio to final
if isprime(i) and (digit_sum(i) = 25) then
total += 1
print i; " ";
end if
next i
print chr(13) + chr(13)
print "Se encontraron "; total; " primos sum25 por debajo de "; final
end
 
Output:
Igual que la entrada de FreeBASIC.


C[edit]

#include <stdbool.h>
#include <stdio.h>
 
bool is_prime(int n) {
int i = 5;
 
if (n < 2) {
return false;
}
if (n % 2 == 0) {
return n == 2;
}
if (n % 3 == 0) {
return n == 3;
}
 
while (i * i <= n) {
if (n % i == 0) {
return false;
}
i += 2;
 
if (n % i == 0) {
return false;
}
i += 4;
}
 
return true;
}
 
int digit_sum(int n) {
int sum = 0;
while (n > 0) {
int rem = n % 10;
n /= 10;
sum += rem;
}
return sum;
}
 
int main() {
int n;
 
for (n = 2; n < 5000; n++) {
if (is_prime(n) && digit_sum(n) == 25) {
printf("%d ", n);
}
}
 
return 0;
}
Output:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993

C++[edit]

Library: GMP

Stretch goal solved the same way as Phix and Go.

#include <algorithm>
#include <chrono>
#include <iomanip>
#include <iostream>
#include <string>
 
#include <gmpxx.h>
 
bool is_probably_prime(const mpz_class& n) {
return mpz_probab_prime_p(n.get_mpz_t(), 3) != 0;
}
 
bool is_prime(int n) {
if (n < 2)
return false;
if (n % 2 == 0)
return n == 2;
if (n % 3 == 0)
return n == 3;
for (int p = 5; p * p <= n; p += 4) {
if (n % p == 0)
return false;
p += 2;
if (n % p == 0)
return false;
}
return true;
}
 
int digit_sum(int n) {
int sum = 0;
for (; n > 0; n /= 10)
sum += n % 10;
return sum;
}
 
int count_all(const std::string& str, int rem) {
int count = 0;
if (rem == 0) {
switch (str.back()) {
case '1':
case '3':
case '7':
case '9':
if (is_probably_prime(mpz_class(str)))
++count;
break;
default:
break;
}
} else {
for (int i = 1; i <= std::min(9, rem); ++i)
count += count_all(str + char('0' + i), rem - i);
}
return count;
}
 
int main() {
std::cout.imbue(std::locale(""));
const int limit = 5000;
std::cout << "Primes < " << limit << " whose digits sum to 25:\n";
int count = 0;
for (int p = 1; p < limit; ++p) {
if (digit_sum(p) == 25 && is_prime(p)) {
++count;
std::cout << std::setw(6) << p << (count % 10 == 0 ? '\n' : ' ');
}
}
std::cout << '\n';
 
auto start = std::chrono::steady_clock::now();
count = count_all("", 25);
auto end = std::chrono::steady_clock::now();
std::cout << "\nThere are " << count
<< " primes whose digits sum to 25 and include no zeros.\n";
std::cout << "Time taken: "
<< std::chrono::duration<double>(end - start).count() << "s\n";
return 0;
}
Output:
//https://tio.run/#cpp-gcc -lgmp -O3
Primes < 5,000 whose digits sum to 25:
   997  1,699  1,789  1,879  1,987  2,689  2,797  2,887  3,499  3,697
 3,769  3,877  3,967  4,597  4,759  4,957  4,993 

There are 1,525,141 primes whose digits sum to 25 and include no zeros.
Time taken: 10.6088s
.....
Real time: 11.214 s
User time: 11.075 s
Sys. time: 0.082 s
CPU share: 99.50 %
Exit code: 0

D[edit]

Translation of: C
import std.bigint;
import std.stdio;
 
bool isPrime(BigInt n) {
if (n < 2) {
return false;
}
 
if (n % 2 == 0) {
return n == 2;
}
if (n % 3 == 0) {
return n == 3;
}
 
auto i = BigInt(5);
while (i * i <= n) {
if (n % i == 0){
return false;
}
i += 2;
 
if (n % i == 0){
return false;
}
i += 4;
}
 
return true;
}
 
int digitSum(BigInt n) {
int result;
while (n > 0) {
result += n % 10;
n /= 10;
}
return result;
}
 
void main() {
for (auto n = BigInt(2); n < 5_000; n++) {
if (n.isPrime && n.digitSum == 25) {
write(n, ' ');
}
}
writeln;
}
Output:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993 

Delphi[edit]

Library: PrimTrial
Translation of: Ring
 
program Primes_which_sum_of_digits_is_25;
 
{$APPTYPE CONSOLE}
 
uses
System.SysUtils,
PrimTrial;
 
var
row: Integer = 0;
limit1: Integer = 25;
limit2: Integer = 5000;
 
function Sum25(n: Integer): boolean;
var
sum: Integer;
str: string;
c: char;
begin
sum := 0;
str := n.ToString;
for c in str do
inc(sum, strToInt(c));
Result := sum = limit1;
end;
 
begin
for var n := 1 to limit2-1 do
begin
if isPrime(n) and sum25(n) then
begin
inc(row);
write(n: 4, ' ');
if (row mod 5) = 0 then
writeln;
end;
end;
readln;
end.
Output:
 997 1699 1789 1879 1987
2689 2797 2887 3499 3697
3769 3877 3967 4597 4759
4957 4993

F#[edit]

 
// Primes to 5000 who's sum of digits is 25. Nigel Galloway: April 1st., 2021
let rec fN g=function n when n<10->n+g=25 |n->fN(g+n%10)(n/10)
primes32()|>Seq.takeWhile((>)5000)|>Seq.filter fN|>Seq.iter(printf "%d "); printfn ""
 
Output:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993

Factor[edit]

Works with: Factor version 0.99 2021-02-05
USING: kernel lists lists.lazy math math.primes.lists prettyprint ;
 
: digit-sum ( n -- sum )
0 swap [ 10 /mod rot + swap ] until-zero ;
 
: lprimes25 ( -- list ) lprimes [ digit-sum 25 = ] lfilter ;
 
lprimes25 [ 5,000 < ] lwhile [ . ] leach
Output:
997
1699
1789
1879
1987
2689
2797
2887
3499
3697
3769
3877
3967
4597
4759
4957
4993

Forth[edit]

Works with: Gforth
: prime? ( n -- ? ) here + [email protected] 0= ;
: notprime! ( n -- ) here + 1 swap c! ;
 
: prime_sieve { n -- }
here n erase
0 notprime!
1 notprime!
n 4 > if
n 4 do i notprime! 2 +loop
then
3
begin
dup dup * n <
while
dup prime? if
n over dup * do
i notprime!
dup 2* +loop
then
2 +
repeat
drop ;
 
: digit_sum ( u -- u )
dup 10 < if exit then
10 /mod recurse + ;
 
: prime25? { p -- ? }
p prime? if
p digit_sum 25 =
else
false
then ;
 
: .prime25 { n -- }
." Primes < " n . ." whose digits sum to 25:" cr
n prime_sieve
0
n 0 do
i prime25? if
i 5 .r
1+ dup 10 mod 0= if cr then
then
loop
cr ." Count: " . cr ;
 
5000 .prime25
bye
Output:
Primes < 5000 whose digits sum to 25:
  997 1699 1789 1879 1987 2689 2797 2887 3499 3697
 3769 3877 3967 4597 4759 4957 4993
Count: 17 


FreeBASIC[edit]

Translation of: AWK
 
Function isprime(num As Ulongint) As Boolean
For i As Integer = 2 To Sqr(num)
If (num Mod i = 0) Then Return False
Next i
Return True
End Function
 
Function digit_sum(num As Integer) As Integer
Dim As Integer sum25 = 0
For j As Integer = 1 To Len(num)
sum25 += Val(Mid(Str(num),j,1))
Next j
Return sum25
End Function
 
Dim As Integer inicio = 1, final = 5000, total = 0
For i As Integer = inicio To final
If (isprime(i)) And (digit_sum(i) = 25) Then
total += 1
Print Using " ####"; i;
If (total Mod 9) = 0 Then Print
End If
Next i
Print !"\n\nSe encontraron"; total; " primos sum25 por debajo de"; finalSleep
 
Output:
  997 1699 1789 1879 1987 2689 2797 2887 3499
 3697 3769 3877 3967 4597 4759 4957 4993

Se encontraron 17 primos sum25 por debajo de 5000


Go[edit]

This uses the Phix routine for the stretch goal though I've had to plug in a GMP wrapper to better the Phix time. Using Go's native big.Int, the time was slightly slower than Phix at 1 minute 28 seconds.

package main
 
import (
"fmt"
big "github.com/ncw/gmp"
"time"
)
 
// for small numbers
func sieve(limit int) []bool {
limit++
// True denotes composite, false denotes prime.
c := make([]bool, limit) // all false by default
c[0] = true
c[1] = true
// no need to bother with even numbers over 2 for this task
p := 3 // Start from 3.
for {
p2 := p * p
if p2 >= limit {
break
}
for i := p2; i < limit; i += 2 * p {
c[i] = true
}
for {
p += 2
if !c[p] {
break
}
}
}
return c
}
 
func sumDigits(n int) int {
sum := 0
for n > 0 {
sum += n % 10
n /= 10
}
return sum
}
 
func min(a, b int) int {
if a < b {
return a
}
return b
}
 
// for big numbers
func countAll(p string, rem, res int) int {
if rem == 0 {
b := p[len(p)-1]
if b == '1' || b == '3' || b == '7' || b == '9' {
z := new(big.Int)
z.SetString(p, 10)
if z.ProbablyPrime(1) {
res++
}
}
} else {
for i := 1; i <= min(9, rem); i++ {
res = countAll(p+fmt.Sprintf("%d", i), rem-i, res)
}
}
return res
}
 
func commatize(n int) string {
s := fmt.Sprintf("%d", n)
if n < 0 {
s = s[1:]
}
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
if n >= 0 {
return s
}
return "-" + s
}
 
func main() {
start := time.Now()
c := sieve(4999)
var primes25 []int
for i := 997; i < 5000; i += 2 {
if !c[i] && sumDigits(i) == 25 {
primes25 = append(primes25, i)
}
}
fmt.Println("The", len(primes25), "primes under 5,000 whose digits sum to 25 are:")
fmt.Println(primes25)
n := countAll("", 25, 0)
fmt.Println("\nThere are", commatize(n), "primes whose digits sum to 25 and include no zeros.")
fmt.Printf("\nTook %s\n", time.Since(start))
}
Output:
The 17 primes under 5,000 whose digits sum to 25 are:
[997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993]

There are 1,525,141 primes whose digits sum to 25 and include no zeros.

Took 25.300758564s

Haskell[edit]

import Data.Bifunctor (second)
import Data.List (replicate)
import Data.List.Split (chunksOf)
import Data.Numbers.Primes (primes)
 
--------- PRIMES WITH DECIMAL DIGITS SUMMING TO 25 -------
 
matchingPrimes :: [Int]
matchingPrimes =
takeWhile
(< 5000)
[n | n <- primes, 25 == decimalDigitSum n]
 
decimalDigitSum :: Int -> Int
decimalDigitSum n =
snd $
until
((0 ==) . fst)
(\(n, x) -> second (+ x) $ quotRem n 10)
(n, 0)
 
--------------------------- TEST -------------------------
main :: IO ()
main = do
let w = length (show (last matchingPrimes))
mapM_ putStrLn $
( show (length matchingPrimes)
<> " primes (< 5000) with decimal digits totalling 25:\n"
) :
( unwords
<$> chunksOf
4
(justifyRight w ' ' . show <$> matchingPrimes)
)
 
justifyRight :: Int -> Char -> String -> String
justifyRight n c = (drop . length) <*> (replicate n c <>)
Output:
17 primes (< 5000) with decimal digits totalling 25:

 997 1699 1789 1879
1987 2689 2797 2887
3499 3697 3769 3877
3967 4597 4759 4957
4993

Java[edit]

Translation of: Kotlin
import java.math.BigInteger;
 
public class PrimeSum {
private static int digitSum(BigInteger bi) {
int sum = 0;
while (bi.compareTo(BigInteger.ZERO) > 0) {
BigInteger[] dr = bi.divideAndRemainder(BigInteger.TEN);
sum += dr[1].intValue();
bi = dr[0];
}
return sum;
}
 
public static void main(String[] args) {
BigInteger fiveK = BigInteger.valueOf(5_000);
BigInteger bi = BigInteger.valueOf(2);
while (bi.compareTo(fiveK) < 0) {
if (digitSum(bi) == 25) {
System.out.print(bi);
System.out.print(" ");
}
bi = bi.nextProbablePrime();
}
System.out.println();
}
}
Output:
997  1699  1789  1879  1987  2689  2797  2887  3499  3697  3769  3877  3967  4597  4759  4957  4993  

JavaScript[edit]

(() => {
"use strict";
 
// ---- PRIMES WITH DECIMAL DIGITS SUMMING TO 25 -----
 
// digitSum :: Int -> Int
const digitSum = n =>
`${n}`.split("").reduce(
(a, c) => a + (c.codePointAt(0) - 48),
0
);
 
 
// primes :: [Int]
const primes = function* () {
// Non finite sequence of prime numbers.
const dct = {};
let n = 2;
 
while (true) {
if (n in dct) {
dct[n].forEach(p => {
const np = n + p;
 
dct[np] = (dct[np] || []).concat(p);
delete dct[n];
});
} else {
yield n;
dct[n * n] = [n];
}
n = 1 + n;
}
};
 
 
// ---------------------- TEST -----------------------
// main :: IO ()
const main = () =>
unlines(
chunksOf(5)(
takeWhileGen(n => 5000 > n)(
filterGen(n => 25 === digitSum(n))(
primes()
)
).map(str)
).map(unwords)
);
 
 
// --------------------- GENERIC ---------------------
 
// chunksOf :: Int -> [a] -> [[a]]
const chunksOf = n => {
// xs split into sublists of length n.
// The last sublist will be short if n
// does not evenly divide the length of xs .
const go = xs => {
const chunk = xs.slice(0, n);
 
return 0 < chunk.length ? (
[chunk].concat(
go(xs.slice(n))
)
) : [];
};
 
return go;
};
 
 
// filterGen :: (a -> Bool) -> Gen [a] -> Gen [a]
const filterGen = p => xs => {
// Non-finite stream of values which are
// drawn from gen, and satisfy p
const go = function* () {
let x = xs.next();
 
while (!x.done) {
const v = x.value;
 
if (p(v)) {
yield v;
}
x = xs.next();
}
};
 
return go(xs);
};
 
 
// str :: a -> String
const str = x =>
x.toString();
 
 
// takeWhileGen :: (a -> Bool) -> Gen [a] -> [a]
const takeWhileGen = p =>
// Values drawn from xs until p matches.
xs => {
const ys = [];
let
nxt = xs.next(),
v = nxt.value;
 
while (!nxt.done && p(v)) {
ys.push(v);
nxt = xs.next();
v = nxt.value;
}
 
return ys;
};
 
 
// 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");
 
 
// unwords :: [String] -> String
const unwords = xs =>
// A space-separated string derived
// from a list of words.
xs.join(" ");
 
return main();
})();
997 1699 1789 1879 1987
2689 2797 2887 3499 3697
3769 3877 3967 4597 4759
4957 4993

jq[edit]

Works with jq
Works with gojq, the Go implementation of jq

The stretch goal is currently beyond the practical capabilities of both the C and Go-based implementations of jq, so only a simple solution to the primary task is shown here.

A suitable definition of `is_prime` may be found at Erdős-primes#jq and is therefore not repeated here.

Preliminaries

def digits: tostring | explode | map( [.]|implode|tonumber);
 
def emit_until(cond; stream): label $out | stream | if cond then break $out else . end;

The Task

# Output: primes whose decimal representation has no 0s and whose sum of digits is $sum > 2
def task($sum):
# Input: array of digits
def nozeros: select(all(.[]; . != 0));
range(3;infinite;2)
| select(digits | (.[-1] != 5 and nozeros and (add == $sum)) )
| select(is_prime);
 
emit_until(. >= 5000; task(25) )
Output:
997
1699
1789
1879
1987
2689
2797
2887
3499
3697
3769
3877
3967
4597
4759
4957
4993

Julia[edit]

using Primes
 
let
pmask, pcount = primesmask(1, 5000), 0
issum25prime(n) = pmask[n] && sum(digits(n)) == 25
 
println("Primes with digits summing to 25 between 0 and 5000:")
for n in 1:4999
if issum25prime(n)
pcount += 1
print(rpad(n, 5))
end
end
println("\nTotal found: $pcount")
end
 
Output:
Primes with digits summing to 25 between 0 and 5000:
997  1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993 
Total found: 17

Stretch goal[edit]

Translation of: Phix
using Primes, Formatting
 
function sum25(p::String, rm, res)
if rm == 0
if p[end] in "1379" && isprime(parse(Int128, p))
res += 1
end
else
for i in 1:min(rm, 9)
res = sum25(p * string(i), rm - i, res)
end
end
return res
end
 
@time println("There are ", format(sum25("", 25, 0), commas=true),
" primes whose digits sum to 25 without any zero digits.")
 
Output:
There are 1,525,141 primes whose digits sum to 25 without any zero digits.
 29.377893 seconds (100.61 M allocations: 4.052 GiB, 0.55% gc time)

Kotlin[edit]

import java.math.BigInteger
 
fun digitSum(bi: BigInteger): Int {
var bi2 = bi
var sum = 0
while (bi2 > BigInteger.ZERO) {
val dr = bi2.divideAndRemainder(BigInteger.TEN)
sum += dr[1].toInt()
bi2 = dr[0]
}
return sum
}
 
fun main() {
val fiveK = BigInteger.valueOf(5_000)
 
var bi = BigInteger.valueOf(2)
while (bi < fiveK) {
if (digitSum(bi) == 25) {
print(bi)
print(" ")
}
 
bi = bi.nextProbablePrime()
}
println()
}
Output:
997  1699  1789  1879  1987  2689  2797  2887  3499  3697  3769  3877  3967  4597  4759  4957  4993  

Mathematica/Wolfram Language[edit]

Select[Prime[[email protected][4999]], IntegerDigits /* Total /* EqualTo[25]]
Output:
{997, 1699, 1789, 1879, 1987, 2689, 2797, 2887, 3499, 3697, 3769, 3877, 3967, 4597, 4759, 4957, 4993}

Nanoquery[edit]

// find primes using the sieve of eratosthenes
// https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes#Pseudocode
def find_primes(upper_bound)
a = {true} * (upper_bound + 1)
for i in range(2, int(sqrt(upper_bound)))
if a[i]
for j in range(i ^ 2, upper_bound, i)
a[j] = false
end for
end if
end for
 
primes = {}
for i in range(2, len(a) - 1)
if a[i]
primes.append(i)
end if
end for
return primes
end find_primes
 
def sum_digits(num)
digits = str(num)
digit_sum = 0
 
for i in range(0, len(digits) - 1)
digit_sum += int(digits[i])
end for
 
return digit_sum
end sum_digits
 
primes_to_check = find_primes(5000)
for prime in primes_to_check
if sum_digits(prime) = 25
print prime + " "
end if
end for
println
Output:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993 

Nim[edit]

Task[edit]

import strutils, sugar
 
func isPrime(n: Natural): bool =
if n < 2: return false
if n mod 2 == 0: return n == 2
if n mod 3 == 0: return n == 3
var d = 5
while d * d <= n:
if n mod d == 0: return false
inc d, 2
if n mod d == 0: return false
inc d, 4
result = true
 
func digitSum(n: Natural): int =
var n = n
while n != 0:
result += n mod 10
n = n div 10
 
let result = collect(newSeq):
for n in countup(3, 5000, 2):
if digitSum(n) == 25 and n.isPrime: n
 
for i, n in result:
stdout.write ($n).align(4), if (i + 1) mod 6 == 0: '\n' else: ' '
echo()
Output:
 997 1699 1789 1879 1987 2689
2797 2887 3499 3697 3769 3877
3967 4597 4759 4957 4993 

Stretch goal[edit]

Translation of: Julia
Library: bignum
import std/monotimes, strformat, strutils
import bignum
 
func sum25(p: string; rm, res: Natural): Natural =
result = res
if rm == 0:
if p[^1] in "1379" and probablyPrime(newInt(p), 25) != 0:
inc result
else:
for i in 1..min(rm, 9):
result = sum25(p & chr(i + ord('0')), rm - i, result)
 
let t0 = getMonoTime()
let count = $sum25("", 25, 0)
echo &"There are {count.insertSep()} primes whose digits sum to 25 without any zero digits."
echo "\nExecution time: ", getMonoTime() - t0
Output:
There are 1_525_141 primes whose digits sum to 25 without any zero digits.

Execution time: (seconds: 12, nanosecond: 182051288)

Pascal[edit]

added only strechted goal.Generating the combination of the digits for the numbers and afterwards generating the Permutations with some identical elements
Now seting one digit out of 1,3,7,9 to the end and permute the rest of the digits in front.
So much less numbers have to be tested.10.5e6 instead of 16.4e6.Generating of the numbers is reduced in the same ratio.

program Perm5aus8;
//formerly roborally take 5 cards out of 8
{$IFDEF FPC}
{$mode Delphi}
{$Optimization ON,All}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
uses
sysutils,
gmp;
const
cTotalSum = 31;
 
cMaxCardsOnDeck = cTotalSum;//8
CMaxCardsUsed = cTotalSum;//5
 
type
tDeckIndex = 0..cMaxCardsOnDeck-1;
tSequenceIndex = 0..CMaxCardsUsed-1;
 
tDiffCardCount = 0..9;
 
tSetElem = record
Elem : tDiffCardCount;
Elemcount : tDeckIndex;
end;
 
tSet = record
RemSet : array [low(tDiffCardCount)..High(tDiffCardCount)] of tSetElem;
MaxUsedIdx,
TotElemCnt : byte;
end;
 
tRemainSet = array [low(tSequenceIndex)..High(tSequenceIndex)+1] of tSet;
 
tCardSequence = array [low(tSequenceIndex)..High(tSequenceIndex)] of tDiffCardCount;
 
var
ManifoldOfDigit : array[tDiffCardCount] of Byte;
TotalUsedDigits : array[tDeckIndex] of Byte;
RemainSets : tRemainSet;
 
CardString : AnsiString;
 
PrimeCount : integer;
PermCount : integer;
 
//*****************************************************************************
var
CS : pchar;
z : mpz_t;
 
procedure SetInit(var ioSet:tSet);
var
i : integer;
begin
with ioSet do
begin
MaxUsedIdx := 0;
For i := Low(tDiffCardCount) to High(tDiffCardCount) do
with RemSet[i] do
begin
ElemCount := 0;
Elem := 0;
end;
end;
end;
 
procedure CheckPrime;inline;
begin
mpz_set_str(z,CS,10);
inc(PrimeCount,ORD(mpz_probab_prime_p(z,3)>0));
end;
 
procedure Permute(depth,MaxCardsUsed:NativeInt);
var
pSetElem : ^tSetElem;
i : NativeInt;
begin
i := 0;
pSetElem := @RemainSets[depth].RemSet[i];
repeat
if pSetElem^.Elemcount <> 0 then begin
//take one of the same elements of the stack
//insert in result here string
CS[depth] := chr(pSetElem^.Elem+Ord('0'));
//done one permutation
IF depth = MaxCardsUsed then
begin
inc(permCount);
CheckPrime;
end
else
begin
dec(pSetElem^.ElemCount);
RemainSets[depth+1]:= RemainSets[depth];
Permute(depth+1,MaxCardsUsed);
//re-insert that element
inc(pSetElem^.ElemCount);
end;
end;
//move on to the next digit
inc(pSetElem);
inc(i);
until i >=RemainSets[depth].MaxUsedIdx;
end;
 
procedure Check(n:nativeInt);
var
i,dgtCnt,cnt,dgtIdx : NativeInt;
Begin
SetInit(RemainSets[0]);
dgtCnt := 0;
dgtIdx := 0;
//creating the start set.
with RemainSets[0] do
Begin
For i in tDiffCardCount do
Begin
cnt := ManifoldOfDigit[i];
if cnt > 0 then
Begin
with RemSet[dgtIdx] do
Begin
Elemcount := cnt;
Elem := i;
end;
inc(dgtCnt,cnt);
inc(dgtIdx);
end;
end;
TotElemCnt := dgtCnt;
MaxUsedIdx := dgtIdx;
 
CS := @CardString[1];
//Check only useful end-digits
For i := 0 to dgtIdx-1 do
Begin
if RemSet[i].Elem in[1,3,7,9]then
Begin
CS[dgtCnt-1] := chr(RemSet[i].Elem+Ord('0'));
CS[dgtCnt] := #00;
 
dec(RemSet[i].ElemCount);
permute(0,dgtCnt-2);
inc(RemSet[i].ElemCount);
end;
end;
end;
end;
 
procedure AppendToSum(n,dgt,remsum:NativeInt);
var
i: NativeInt;
begin
inc(ManifoldOfDigit[dgt]);
IF remsum > 0 then
For i := dgt to 9 do
AppendToSum(n+1,i,remsum-i)
else
Begin
if remsum = 0 then
Begin
Check(n);
//n is 0 based PrimeCount combinations of length n
inc(TotalUsedDigits[n+1]);
end;
end;
dec(ManifoldOfDigit[dgt]);
end;
 
procedure CheckAll(SumGoal:NativeInt);
var
i :NativeInt;
begin
setlength(CardString,SumGoal);
IF sumGoal>cTotalSum then
EXIT;
fillchar(ManifoldOfDigit[0],SizeOf(ManifoldOfDigit),#0);
permcount:=0;
PrimeCount := 0;
 
For i := 1 to 9 do
AppendToSum(0,i,SumGoal-i);
 
writeln('PrimeCount of generated numbers with digits sum of ',SumGoal,' are ',permcount);
writeln('Propably primes ',PrimeCount);
writeln;
end;
var
T1,T0 : Int64;
SumGoal: NativeInt;
BEGIN
writeln('GMP-Version ',gmp.version);
mpz_init_set_ui(z,0);
T0 := GetTickCount64;
For SumGoal := 25 to 25 do
Begin
CheckAll(SumGoal);
T1 := GetTickCount64;Writeln((T1-T0)/1000:7:3,' s');
T0 := T1;
end;
mpz_clear(z);
END.
 
Output:
//Runnning on TIO.RUN
GMP-Version 6.1.2
PrimeCount of generated numbers with digits sum of 25 are 10488498
Propably primes 1525141

  9.932 s
....
Free Pascal Compiler version 3.0.4 [2018/07/13] for x86_64
Copyright (c) 1993-2017 by Florian Klaempfl and others
Target OS: Linux for x86-64
Compiling .code.tio.pp
Linking .bin.tio
/usr/bin/ld: warning: link.res contains output sections; did you forget -T?
204 lines compiled, 0.2 sec

Real time: 10.135 s
User time: 10.027 s
Sys. time: 0.052 s
CPU share: 99.45 %
Exit code: 0

Perl[edit]

Library: ntheory
use strict;
use warnings;
use feature 'say';
use List::Util 'sum';
use ntheory 'is_prime';
 
my($limit, @p25) = 5000;
is_prime($_) and 25 == sum(split '', $_) and push @p25, $_ for 1..$limit;
say @p25 . " primes < $limit with digital sum 25:\n" . join ' ', @p25;
 
Output:
17 primes < 5000 with digital sum 25:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993

Phix[edit]

function sum25(integer p) return sum(sq_sub(sprint(p),'0'))=25 end function
sequence res = filter(get_primes_le(5000),sum25)
string r = join(shorten(apply(res,sprint),"",4))
printf(1,"%d sum25 primes less than 5000 found: %s\n",{length(res),r})
Output:
17 sum25 primes less than 5000 found: 997 1699 1789 1879 ... 4597 4759 4957 4993

Stretch goal[edit]

Library: Phix/mpfr
without js
include mpfr.e
atom t0 = time(), t1 = time()+1
mpz pz = mpz_init(0)

function sum25(string p, integer rem, res=0)
    if rem=0 then
        if find(p[$],"1379") then -- (saves 13s)
            mpz_set_str(pz,p)
            if mpz_prime(pz) then
                res += 1
                if platform()!=JS and time()>t1 then
                    progress("%d, %s...",{res,p})
                    t1 = time()+1
                end if
            end if
        end if
    else
        for i=1 to min(rem,9) do
            res = sum25(p&'0'+i,rem-i,res)
        end for
    end if
    return res
end function
 
printf(1,"There are %,d sum25 primes that contain no zeroes\n",sum25("",25))
?elapsed(time()-t0)
Output:
There are 1,525,141 sum25 primes that contain no zeroes
"1 minute and 27s"

Note this works under pwa/p2js but you would get to stare at a blank screen for 8½ minutes with 100% cpu, hence it has been marked "without js".

Python[edit]

'''Primes with a decimal digit sum of 25'''
 
from itertools import takewhile
 
 
# primesWithGivenDigitSum :: Int -> Int -> [Int]
def primesWithGivenDigitSum(below, n):
'''Primes below a given value with
decimal digits sums equal to n.
'''

return list(
takewhile(
lambda x: below > x,
(
x for x in primes()
if n == sum(int(c) for c in str(x))
)
)
)
 
 
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Test'''
matches = primesWithGivenDigitSum(5000, 25)
print(
str(len(matches)) + (
' primes below 5000 with a decimal digit sum of 25:\n'
)
)
print(
'\n'.join([
' '.join([str(x).rjust(4, ' ') for x in xs])
for xs in chunksOf(4)(matches)
])
)
 
 
# ----------------------- GENERIC ------------------------
 
# chunksOf :: Int -> [a] -> [[a]]
def chunksOf(n):
'''A series of lists of length n, subdividing the
contents of xs. Where the length of xs is not evenly
divible, the final list will be shorter than n.
'''

def go(xs):
return (
xs[i:n + i] for i in range(0, len(xs), n)
) if 0 < n else None
return go
 
 
# primes :: [Int]
def primes():
''' Non-finite sequence of prime numbers.
'''

n = 2
dct = {}
while True:
if n in dct:
for p in dct[n]:
dct.setdefault(n + p, []).append(p)
del dct[n]
else:
yield n
dct[n * n] = [n]
n = 1 + n
 
 
# MAIN ---
if __name__ == '__main__':
main()
Output:
17 primes below 5000 with a decimal digit sum of 25:

 997 1699 1789 1879
1987 2689 2797 2887
3499 3697 3769 3877
3967 4597 4759 4957
4993

Raku[edit]

unit sub MAIN ($limit = 5000);
say "{+$_} primes < $limit with digital sum 25:\n{$_».fmt("%" ~ $limit.chars ~ "d").batch(10).join("\n")}",
with ^$limit .grep: { .is-prime and .comb.sum == 25 }
Output:
17 primes < 5000 with digital sum 25:
 997 1699 1789 1879 1987 2689 2797 2887 3499 3697
3769 3877 3967 4597 4759 4957 4993

REXX[edit]

This REXX version allows the following to be specified on the command line:

  •   the high number   (HI)
  •   the number of columns shown per line   (COLS)
  •   the target sum   (TARGET)
/*REXX pgm finds and displays primes less than  HI  whose decimal digits sum to  TARGET.*/
parse arg hi cols target . /*obtain optional argument from the CL.*/
if hi=='' | hi=="," then hi= 5000 /*Not specified? Then use the default.*/
if cols=='' | cols=="," then cols= 10 /* " " " " " " */
if target=='' | target=="," then target= 25 /* " " " " " " */
call genP /*build array of semaphores for primes.*/
w= 10 /*width of a number in any column. */
title= ' primes that are < ' commas(hi) " and whose decimal digits sum to " ,
commas(target)
if cols>0 then say ' index │'center(title, 1 + cols*(w+1) )
if cols>0 then say '───────┼'center("" , 1 + cols*(w+1), '─')
found= 0; idx= 1 /*define # target primes found and IDX.*/
$= /*list of target primes found (so far).*/
do j=1 for # /*examine all the primes generated. */
if sumDigs(@.j)\==target then iterate /*Is sum≡target sum? No, then skip it.*/
found= found + 1 /*bump the number of target primes. */
if cols<1 then iterate /*Build the list (to be shown later)? */
c= commas(@.j) /*maybe add commas to the number. */
$= $ right(c, max(w, length(c) ) ) /*add a prime ──► list, allow big #'s.*/
if found//cols\==0 then iterate /*have we populated a line of output? */
say center(idx, 7)'│' substr($, 2); $= /*display what we have so far (cols). */
idx= idx + cols /*bump the index count for the output*/
end /*j*/
 
if $\=='' then say center(idx, 7)"│" substr($, 2) /*possible display residual output.*/
if cols>0 then say '───────┴'center("" , 1 + cols*(w+1), '─')
say
say 'Found ' commas(found) title
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg ?; do jc=length(?)-3 to 1 by -3; ?=insert(',', ?, jc); end; return ?
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP: !.= 0 /*placeholders for primes' semaphores. */
@.1=2; @.2=3; @.3=5; @.4=7; @.5=11; @.6=13 /*define some low primes. */
 !.2=1; !.3=1; !.5=1; !.7=1; !.11=1; @.13=1 /* " " " primes' semaphores. */
#= 6; sq.#= @.# ** 2 /*number of primes so far; prime². */
/* [↓] generate more primes ≤ high.*/
do [email protected].#+2 by 2 to hi /*find odd primes from here on. */
parse var j '' -1 _ /*obtain the last decimal digit of J. */
if _==5 then iterate; if j// 3==0 then iterate /*J ÷ by 5? J ÷ by 3? */
if j//7==0 then iterate; if j//11==0 then iterate /*" " " 7? " " " 11? */
do k=6 while sq.k<=j /* [↓] divide by the known odd primes.*/
if j // @.k == 0 then iterate j /*Is J ÷ X? Then not prime. ___ */
end /*k*/ /* [↑] only process numbers ≤ √ J */
#= #+1; @.#= j; sq.#= j*j;  !.j= 1 /*bump # of Ps; assign next P; P²; P# */
end /*j*/; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
sumDigs: parse arg x 1 s 2 '' -1 z; L= length(x); if L==1 then return s; s= s + z
do m=2 for L-2; s= s + substr(x, m, 1); end; return s
output   when using the default inputs:
 index │                         primes that are  <  5,000  and whose decimal digits sum to  25
───────┼───────────────────────────────────────────────────────────────────────────────────────────────────────────────
   1   │        997      1,699      1,789      1,879      1,987      2,689      2,797      2,887      3,499      3,697
  11   │      3,769      3,877      3,967      4,597      4,759      4,957      4,993
───────┴───────────────────────────────────────────────────────────────────────────────────────────────────────────────

Found  17  primes that are  <  5,000  and whose decimal digits sum to  25
output   when using the input of:     1000000   0
Found  6,198  primes that are  <  1,000,000  and whose decimal digits sum to  25

Ring[edit]

 
load "stdlib.ring"
 
see "working..." + nl
decimals(0)
row = 0
num = 0
nr = 0
numsum25 = 0
limit1 = 25
limit2 = 5000
 
for n = 1 to limit2
if isprime(n)
bool = sum25(n)
if bool = 1
row = row + 1
see "" + n + " "
if (row%5) = 0
see nl
ok
ok
ok
next
 
see nl + "Found " + row + " sum25 primes below 5000" + nl
 
time1 = clock()
see nl
row = 0
 
while true
num = num + 1
str = string(num)
for m = 1 to len(str)
if str[m] = 0
loop
ok
next
if isprime(num)
bool = sum25(num)
if bool = 1
nr = num
numsum25 = numsum25 + 1
ok
ok
time2 = clock()
time3 = (time2-time1)/1000/60
if time3 > 30
exit
ok
end
 
see "There are " + numsum25 + " sum25 primes that contain no zeroes (during 30 mins)" + nl
see "The last sum25 prime found during 30 mins is: " + nr + nl
see "time = " + time3 + " mins" + nl
see "done..." + nl
 
func sum25(n)
sum = 0
str = string(n)
for n = 1 to len(str)
sum = sum + number(str[n])
next
if sum = limit1
return 1
ok
 
Output:
working...
997 1699 1789 1879 1987 
2689 2797 2887 3499 3697 
3769 3877 3967 4597 4759 
4957 4993 
Found 17 sum25 primes below 5000

There are 1753 sum25 primes that contain no zeroes (during 30 mins)
The last sum25 prime found during 30 mins is: 230929
time = 30 mins
done...

Ruby[edit]

require 'prime'
 
def digitSum(n)
sum = 0
while n > 0
sum += n % 10
n /= 10
end
return sum
end
 
for p in Prime.take_while { |p| p < 5000 }
if digitSum(p) == 25 then
print p, " "
end
end
Output:
997  1699  1789  1879  1987  2689  2797  2887  3499  3697  3769  3877  3967  4597  4759  4957  4993  

Sidef[edit]

Simple solution:

5000.primes.grep { .sumdigits == 25 }.say

Generate such primes from digits (asymptotically faster):

func generate_from_prefix(limit, digitsum, p, base, digits, t=p) {
 
var seq = [p]
 
digits.each {|d|
var num = (p*base + d)
num <= limit || return seq
 
var sum = (t + d)
sum <= digitsum || return seq
 
seq << __FUNC__(limit, digitsum, num, base, digits, sum)\
.grep { .is_prime }...
}
 
return seq
}
 
func primes_with_digit_sum(limit, digitsum = 25, base = 10, digits = @(^base)) {
digits.grep { _ > 0 }\
.map { generate_from_prefix(limit, digitsum, _, base, digits)... }\
.grep { .sumdigits(base) == digitsum }\
.sort
}
 
say primes_with_digit_sum(5000)
Output:
[997, 1699, 1789, 1879, 1987, 2689, 2797, 2887, 3499, 3697, 3769, 3877, 3967, 4597, 4759, 4957, 4993]

Tcl[edit]

Library: Tcllib (Package: math::numtheory)

Could be made prettier with the staple helper proc lfilter.

package require Tcl 8.5
package require math::numtheory
namespace path ::tcl::mathop
 
puts [lmap x [math::numtheory::primesLowerThan 5000] {
if {[+ {*}[split $x {}]] == 25} {set x} else continue
}]
Output:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993

Wren[edit]

Library: Wren-math
Library: Wren-fmt
Library: Wren-seq

Although do-able, the stretch goal would take too long in Wren so I haven't bothered.

import "/math" for Int
import "/fmt" for Fmt
import "/seq" for Lst
 
var sumDigits = Fn.new { |n|
var sum = 0
while (n > 0) {
sum = sum + (n % 10)
n = (n/10).floor
}
return sum
}
 
var primes = Int.primeSieve(4999).where { |p| p >= 997 }
var primes25 = []
for (p in primes) {
if (sumDigits.call(p) == 25) primes25.add(p)
}
System.print("The %(primes25.count) primes under 5,000 whose digits sum to 25 are:")
for (chunk in Lst.chunks(primes25, 6)) Fmt.print("$,6d", chunk)
Output:
The 17 primes under 5,000 whose digits sum to 25 are:
   997  1,699  1,789  1,879  1,987  2,689
 2,797  2,887  3,499  3,697  3,769  3,877
 3,967  4,597  4,759  4,957  4,993