Primes whose sum of digits is 25: Difference between revisions

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===Stretch goal===
===Stretch goal===
{{libheader|mpfr}}
<lang Phix>include mpfr.e
<lang Phix>include mpfr.e
atom t0 = time(), t1 = time()+1
atom t0 = time(), t1 = time()+1

Revision as of 12:49, 21 March 2021

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 digits is 25
Let 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 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

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

Library: mpfr

<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"

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

Ring

<lang ring> load "stdlib.ring"

row = 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

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