Primes whose sum of digits is 25
- Task
Show primes which sum of its decimal digits is 25
Find primes n such that 0 < n < 5000
- Stretch goal
Show the total number of all such primes that do not contain any zeroes (997 <= n <= 1,111,111,111,111,111,111,111,111).
ALGOL 68
<lang algol68>BEGIN # find primes whose digits sum to 25 #
# returns a sieve of primes up to n #
PROC eratosthenes = ( INT n )[]BOOL:
BEGIN
[ 1 : n ]BOOL p;
p[ 1 ] := FALSE; p[ 2 ] := TRUE;
FOR i FROM 3 BY 2 TO n DO p[ i ] := TRUE OD;
FOR i FROM 4 BY 2 TO n DO p[ i ] := FALSE OD;
FOR i FROM 3 BY 2 TO ENTIER sqrt( n ) DO
IF p[ i ] THEN FOR c FROM i * i BY i + i TO n DO p[ c ] := FALSE OD FI
OD;
p
END # eratosthenes # ;
# show all sum25 primes below 5000 #
INT max show sum25 = 4999;
[]BOOL prime = eratosthenes( max show sum25 );
INT p25 count := 0;
FOR n TO max show sum25 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( max show sum25 + 1, 0 ), newline ) )
END</lang>
- 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
Uses the candidate generating algorithm used by Phix, Go
Uses the Miller Rabin primality test and the pow mod procedure from prelude/pow_mod
<lang algol68>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</lang>
- 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 may be faster.
There are 1525141 sum25 primes that contain no zeroes
ALGOL W
<lang algolw>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.</lang>
- Output:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993 Found 17 sum25 primes below 5000
C++
Stretch goal solved the same way as Phix and Go. <lang cpp>#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(), 25) != 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;
}</lang>
- Output:
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: 13.7191s
Delphi
<lang Delphi> 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.</lang>
- Output:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993
Factor
<lang factor>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</lang>
- Output:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993
Forth
<lang forth>: prime? ( n -- ? ) here + c@ 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</lang>
- 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
Go
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. <lang go>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))
}</lang>
- 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
<lang haskell>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 <>)</lang>
- 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
Julia
<lang julia>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: $pcount")
end
</lang>
- Output:
Primes with digits summing to 25 between 0 and 5000: 997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993 Total found: 17
Stretch goal
<lang julia>using Primes, Formatting
function sum25(p::String, rm, res)
if rm == 0
if p[end] in "1379" && isprime(parse(Int128, p))
res += 1
end
else
for i in 1:min(rm, 9)
res = sum25(p * string(i), rm - i, res)
end
end
return res
end
@time println("There are ", format(sum25("", 25, 0), commas=true),
" primes whose digits sum to 25 without any zero digits.")
</lang>
- Output:
There are 1,525,141 primes whose digits sum to 25 without any zero digits. 29.377893 seconds (100.61 M allocations: 4.052 GiB, 0.55% gc time)
Nanoquery
<lang Nanoquery>// find primes using the sieve of eratosthenes // https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes#Pseudocode def find_primes(upper_bound)
a = {true} * (upper_bound + 1)
for i in range(2, int(sqrt(upper_bound)))
if a[i]
for j in range(i ^ 2, upper_bound, i)
a[j] = false
end for
end if
end for
primes = {}
for i in range(2, len(a) - 1)
if a[i]
primes.append(i)
end if
end for
return primes
end find_primes
def sum_digits(num)
digits = str(num) digit_sum = 0
for i in range(0, len(digits) - 1)
digit_sum += int(digits[i])
end for
return digit_sum
end sum_digits
primes_to_check = find_primes(5000) for prime in primes_to_check
if sum_digits(prime) = 25
print prime + " "
end if
end for println</lang>
- Output:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993
Pascal
added only strechted goal.Generating the numbers by permutation.Same result as go.runtime reduced from 5.8s to 0.4s
I don't know why the gmp test is so much faster and why there is one more probably prime.
<lang pascal>
program Perm5aus8;
{$IFDEF FPC}
{$mode Delphi}
{$Optimization ON,All}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF} uses
sysutils, gmp;
const
cTotalSum = 25;
cMaxCardsOnDeck = cTotalSum; CMaxCardsUsed = cTotalSum;
type
tDeckIndex = 0..cMaxCardsOnDeck-1; tSequenceIndex = 0..CMaxCardsUsed-1;
tDiffCardCount = 0..9;
tMengenElem = record
Elemcount : tDeckIndex;
Elem : tDiffCardCount;
end;
tMenge = record
RestMenge : array [low(tDiffCardCount)..High(tDiffCardCount)] of tMengenElem;
MaxUsedIdx,
TotElemCnt : word;
end;
tRestmengen = array [low(tSequenceIndex)..High(tSequenceIndex)+1] of tMenge;
tCardSequence = array [low(tSequenceIndex)..High(tSequenceIndex)] of tDiffCardCount;
tPermRec = packed record
permTiefe : byte;// Stelle der Änderung gegenüber Vorgängerpermutation
permCardSequence :tCardSequence ;
end;
var
// globale Variable, um Stelle der Änderung gegenüber der // Vorgänger permutation festzuhalten Restmengen : tRestmengen; TotalUsedDigits : array[tDeckIndex] of Int32; ManifoldOfDigit : array[tDiffCardCount] of Int32; CardSequence : tCardSequence; CardString : string[cMaxCardsOnDeck+3];
Count: NativeInt; PermCount : integer; mindepth : integer;
//***************************************************************************** procedure MengeInit(var ioMenge:tMenge); var
i : integer;
begin
with ioMenge do
begin
MaxUsedIdx := 0;
For i := Low(tDiffCardCount) to High(tDiffCardCount) do
with RestMenge[i] do
begin
ElemCount := 0;
Elem := 0;
end;
end;
end;
procedure CheckPrime(LastDigit:NativeInt); var
z : mpz_t;
begin
if Lastdigit in [1,3,7,9] then
Begin
mpz_init_set_str(z,pChar(@CardString[1]),10);
if mpz_probab_prime_p(z,1)>0 then
inc(count);
mpz_clear(z);
end;
end;
procedure Permute(depth,MaxCardsUsed:integer); var
i : NativeInt; pMengeElem : ^tMengenElem;
begin
i := 0;
pMengeElem := @RestMengen[depth].RestMenge[i];
repeat
if pMengeElem^.Elemcount <> 0 then begin
//take element
dec(pMengeElem^.ElemCount);
//insert in result here string too
CardSequence[depth] := pMengeElem^.Elem;
CardString[depth+1] := chr(pMengeElem^.Elem+Ord('0'));
//done one permutation
IF depth = MaxCardsUsed then begin
inc(permCount);
// ###########
setlength(CardString,depth+1);
CardString[ depth+2] := #0;
CheckPrime(CardSequence[depth]);
// ###########
mindepth := depth;
end
else begin
RestMengen[depth+1]:= RestMengen[depth];
Permute(depth+1,MaxCardsUsed);
//insert element
inc(pMengeElem^.ElemCount);
dec(minDepth);
end;
end;
inc(pMengeElem);
inc(i);
until i >=RestMengen[depth].MaxUsedIdx;
end;
procedure Check(n:nativeInt); var
i,dgtCnt,cnt,dgtIdx : NativeInt;
Begin
MengeInit(Restmengen[0]);
dgtCnt := 0;
dgtIdx := 0;
with Restmengen[0] do
Begin
For i in tDiffCardCount do
Begin
cnt := ManifoldOfDigit[i];
if cnt > 0 then
Begin
with RestMenge[dgtIdx] do
Begin
Elemcount := cnt;
Elem := i;
end;
inc(dgtCnt,cnt);
inc(dgtIdx);
end;
end;
MaxUsedIdx := dgtIdx;
end;
permute(0,dgtCnt-1);
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);
inc(TotalUsedDigits[n]);
end;
end;
dec(ManifoldOfDigit[dgt]);
end;
var
i :NativeInt;
BEGIN
permcount:=0; fillchar(ManifoldOfDigit[0],SizeOf(ManifoldOfDigit),#0);
count := 0;
For i := 1 to 9 do
AppendToSum(0,i,cTotalSum-i);
writeln('Count of generated numbers with digits sum of ',cTotalSum,' are ',permcount);
Writeln('Propably primes ',count);
//For i := 0 to cTotalSum-1 do writeln(i+1:3,TotalUsedDigits[i]:10); writeln; END.</lang>
- Output:
Count of generated numbers with digits sum of 25 are 16499120
Propably primes 1525142<pre>
real 0m6,875s
=={{header|Perl}}==
{{libheader|ntheory}}
<lang perl>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 < $limit with digital sum 25:\n" . join ' ', @p25;
</lang>
{{out}}
<pre>17 primes < 5000 with digital sum 25:
997 1699 1789 1879 1987 2689 2797 2887 3499 3697 3769 3877 3967 4597 4759 4957 4993
Phix
<lang Phix>function sum25(integer p) return sum(sq_sub(sprint(p),'0'))=25 end function sequence res = filter(get_primes_le(5000),sum25) string r = join(shorten(apply(res,sprint),"",4)) printf(1,"%d sum25 primes less than 5000 found: %s\n",{length(res),r})</lang>
- Output:
17 sum25 primes less than 5000 found: 997 1699 1789 1879 ... 4597 4759 4957 4993
Stretch goal
<lang Phix>include mpfr.e atom t0 = time(), t1 = time()+1 mpz pz = mpz_init(0)
function sum25(string p, integer rem, res=0)
if rem=0 then
if find(p[$],"1379") then -- (saves 13s)
mpz_set_str(pz,p)
if mpz_prime(pz) then
res += 1
if 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)</lang>
- Output:
There are 1,525,141 sum25 primes that contain no zeroes "1 minute and 27s"
Python
<lang python>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()</lang>
- 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
<lang perl6>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 }</lang>
- 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
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)
<lang rexx>/*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. */
@primes= ' primes that are < ' commas(hi) " and whose decimal digits sum to "
if cols>0 then say ' index │'center(@primes commas(target), 1 + cols*(w+1) ) if cols>0 then say '───────┼'center("" , 1 + cols*(w+1), '─') primesT= 0; idx= 1 /*define # target primes found & index.*/ $= /*list of target primes found (so far).*/
do j=1 for #; y= @.j /*examine all the primes generated. */
if sumDigs(y)\==target then iterate /*Is sum≡target sum? No, then skip it.*/
primesT= primesT + 1 /*bump the number of target primes. */
if cols==0 then iterate /*Build the list (to be shown later)? */
c= commas(y) /*maybe add commas to the number. */
$= $ right(c, max(w, length(c) ) ) /*add a prime ──► list, allow big #'s.*/
if primesT//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.*/ say say 'Found ' commas(primesT) @primes commas(target) 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; s.#= @.# **2 /*number of primes so far; prime². */
/* [↓] generate more primes ≤ high.*/
do j=@.#+2 by 2 to hi /*find odd primes from here on. */
parse var j -1 _; if _==5 then iterate /*J divisible by 5? (right dig)*/
if j// 3==0 then iterate /*" " " 3? */
if j// 7==0 then iterate /*" " " 7? */
if j//11==0 then iterate /*" " " 11? */
do k=6 while s.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; s.#= 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</lang>
- 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
<lang ring> 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
</lang>
- 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...
Wren
Although do-able, the stretch goal would take too long in Wren so I haven't bothered. <lang ecmascript>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)</lang>
- 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