Prime a*b+c of N

Question

Given an integer \$N\$, print or return integers \$a\$, \$b\$, and \$c\$ that satisfy all of the following conditions, if such integers exist:

\$a \times b + c = N\$
\$a\$, \$b\$, and \$c\$ are all prime
\$a > b > c\$

If no valid combination of integers exist, you should return nothing, 0, None, an empty list, or raise an error.

If multiple valid combinations of integers exists, you can print or return any of them or all of them in a data type of your choosing.

A list of multiple solutions does not need to be sorted, and since we know that \$a > b > c\$, you can return them in any order.

Examples:

Input: 17
Output: 5 3 2  

Input: 20
Output: None

Input: 37
Output: 7 5 2

Input: 48
Output: None

Input: 208 
Output: [(41, 5, 3), (29, 7, 5)]

This is code golf, so the code with the lowest byte count wins.

Inspired by this Redditor's neat dream.

theorist · Accepted Answer · 2022-02-01 02:47:09Z

11

Wolfram Mathematica, 29 bytes

 Solve[a*b+c==#>a>b>c,Primes]&

Test cases:

Solve[a*b+c==#>a>b>c,Primes]&@17   
Solve[a*b+c==#>a>b>c,Primes]&@20
Solve[a*b+c==#>a>b>c,Primes]&@37
Solve[a*b+c==#>a>b>c,Primes]&@48
Solve[a*b+c==#>a>b>c,Primes]&@208

{{a -> 5, b -> 3, c -> 2}}

{}

{{a -> 7, b -> 5, c -> 2}}

{}

{{a -> 29, b -> 7, c -> 5}, {a -> 41, b -> 5, c -> 3}}

edited Feb 1, 2022 at 2:47

answered Jan 31, 2022 at 4:05

theorist

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ophact · Accepted Answer · 2022-02-03 08:23:37Z

6

+50

tinylisp, 210 bytes

(load library
(d p prime?
(d f(q((n k)(i(e n k)0(i(g n k 2)(g n k 2)(f n(+ k 1
(d g(q((n k j)(i(l(/ n k)j)0(i(*(l(- n(* k j))j)(l j k)(p(- n(* k j)))(p k)(p j))(list k(- n(* k j))j)(g n k(+ j 1
(d F(q((n)(f n 2

Try it online!

Gotta love golfing in tinylisp...

Thanks to DLosc for saving some bytes... 32 to be exact.

edited Feb 3, 2022 at 8:23

answered Feb 2, 2022 at 8:30

ophact

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chunes · Accepted Answer · 2022-01-31 18:14:18Z

4

Factor + `math.combinatorics math.primes`, 64 bytes

[ dup nprimes swap 3 [ first3 * + = ] with filter-combinations ]

Try it online!

Explanation

Get all the combinations of three primes that satisfy a * b + c = N. a > b > c is inherent by virtue of the fact we're filtering combinations without repetition.

edited Jan 31, 2022 at 18:14

answered Jan 31, 2022 at 3:57

chunes

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Pro Odermonicon · Accepted Answer · 2022-02-02 07:04:56Z

Pyth, 37 bytes

FGQIP_GFHGIP_HFNHIP_NIq+*GHNQ
G
H
N.q

hacky, but works

Try it online!

explanation:

                              Q = eval(input())
F                             for
 G                            G
  Q                           in range(Q):
   I                            if
    P_                          prime(
      G                          G
                                ):
       F                          for
        H                         H
         G                        in range(G):
          I                         if
           P_                       prime(
             H                       H
                                    ):
               F                      for
                N                     H
                 H                    in range(H):
                  I                     if
                   P_                   prime(
                     N                   N
                                        ):
                      I                   if
                          G               G
                         *                *
                           H              H
                        +                 +
                            N             N
                       q                  ==
                             Q            Q:
                              (newline)     print(
G                                             G
                                            )
 (newline)                                  print(
H                                             H
                                            )
 (newline)                                  print(
N                                             N
                                            )
 .q                                         exit()

Axuary · Accepted Answer · 2022-02-02 12:28:16Z

Excel, 164 bytes

=TEXTJOIN(",",,LET(q,SEQUENCE(A4),a,FILTER(q,MMULT((MOD(q,TRANSPOSE(q))=0)*1,q^0)=2),b,TRANSPOSE(a),c,(a>b)*(A4-a*b),IF(XLOOKUP(c,a,a^0,0)*(b>c),a&" "&b&" "&c,"")))

Link to Spreadsheet

Explantion

=TEXTJOIN(",",, ~ )                 # Separates multiple answers with a comma
  LET(q,SEQUENCE(A4),               # q = (1..A4) listed vertically
  a,FILTER(q, ~ ),                  # a = primes in q
    MMULT( ~,q^0)=2                 # a number is prime if the sum of the row is 2 
      MOD(q,TRANSPOSE(q))=0)*1      # creates a qxq array where if row# mod column = 0 then 1 else 0
  b,TRANSPOSE(a),                   # b primes listed horizontally
  c,(a>b)*(A4-a*b),                 # 2-dim array where if a>b then c = A4 - a*b 
  IF(XLOOKUP(c,a,a^0,0)*(b>c),~,"") # if c is prime and b>c then return value else ""
    a&" "&b&" "&c                   # a b c

Insider Beta Version using LAMBDA, 157 bytes

=TEXTJOIN(",",,MAKEARRAY(A1,A1,LAMBDA(a,b,LET(c,MAX(A1-a*b,1),p,LAMBDA(x,SUM((MOD(x,SEQUENCE(x))=0)*1)=2),IF((a>b)*(b>c)*p(a)*p(b)*p(c),a&" "&b&" "&c,"")))))

RGS · Accepted Answer · 2022-02-02 15:38:35Z

APL(Dyalog Unicode) + dfns, ⁶³43 bytes ^SBCS

{⍵⌿⍨(∧/1pco¨⍵)∧⍺=(0∘⌷+1∘⌷×2∘⌷)⍉⍵}∘(3cmat⊢)⍨

Hats off to @ovs for a 2+18 bytes shave!

Tacit function with an anonymous function baked in. Returns a 3-col matrix, with as many rows as possible results.

dfns is like a library of useful functions:

the pco function I'm using, which checks for primality, can be found there; and
the cmat function generates the combinations of possible a, b, and c values.

Try it on online! The TIO link has 2 more bytes because of f←, which I'm using to make it easier to test the function. It also takes a long time because I'm being wasteful with my primality checks; here is the output on my machine:

      f¨ 17 20 37 48 208
┌─────┬┬─────┬┬──────┐
│5 3 2││7 5 2││29 7 5│
│     ││     ││41 5 3│
└─────┴┴─────┴┴──────┘

Explanation:

(3cmat⊢) builds a 3-column matrix with all possible values of a, b, and c, already satisfying the restriction that a > b > c:

Then, {⍵⌿⍨(∧/1pco¨⍵)∧⍺=(0∘⌷+1∘⌷×2∘⌷)⍉⍵} takes the original number on the left and that 3-col matrix on the right, and takes all rows that satisfy two criteria simultaneously:

the rows only have prime numbers (checked with (∧/1pco¨⍵)); and
a×b+c equals the original input (checked with ⍺=(0∘⌷+1∘⌷×2∘⌷)⍉⍵).

You can save 2 bytes by not assigning b and use ⍵⌿⍨(∧/1pco¨⍵)∧⍺=... instead. And if you want to go full library cmat should be able to replace most of the right part — ovs, Commented Feb 2, 2022 at 15:19

Sundar R · Accepted Answer · 2022-02-02 03:17:52Z

2

MATL, 28 25 24 23 bytes

Zq"@tG-YfhXHd0>AHn3=*?H

Try it online!

Zq"   % for each prime less than (or equal to) the input
@t    % duplicate it on the stack
G-    % subtract it from the input
Yf    % get the prime factors of the result
h     % concatenate that with the current iterating prime
XH    % save that in the H clipboard
d0>A  % check that it's strictly increasing
Hn3=  % and has 3 elements 
*?    % if both conditions are true
H     % push that result list back on the stack

Prints all valid results.

edited Feb 2, 2022 at 3:17

answered Feb 2, 2022 at 2:55

Sundar R

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emanresu A · Accepted Answer · 2022-02-05 20:10:02Z

2

Vyxal, 10 bytes

~æ3ḋ'÷*+?=

Try it Online!

~          # Filter (implicit) range 1...n by...
 æ         # Is prime?
  3ḋ       # Combinations of length 3
    '      # Filtered by...
     ÷*+   # a*b+c
        ?= # Equals input?

Outputs a list of all valid lists.

edited Feb 5, 2022 at 20:10

answered Jan 31, 2022 at 3:06

emanresu A

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3

\$\begingroup\$ I don't think this is a valid submission. The algorithm here is a semi-decision algorithm, meaning it goes to an infinite loop when it encounters a number that doesn't have a solution, but OP asked for a deciding algorithm. Usually things like memory exhaustion are ignored in code golf. It's fine to say that the memory exhaustion is part of the algorithm, but now this algorithm can only calculate a finite amount of N:s, since if N is very large, the memory exhaustion will happen, even if N has a solution. \$\endgroup\$
– AnttiP
Commented Jan 31, 2022 at 20:39
1

\$\begingroup\$ @AnttiP The memory exhaustion will always eventually happen on any machine if impossible. But this can calculate any value on a machine with enough memory to do so. \$\endgroup\$
– emanresu A
Commented Jan 31, 2022 at 20:55
\$\begingroup\$ Hmm, I still disagree, but I think this would better be discussed in meta. \$\endgroup\$
– AnttiP
Commented Jan 31, 2022 at 20:59
\$\begingroup\$ @AnttiP Agreed. Also see l4m2's second solution. \$\endgroup\$
– emanresu A
Commented Jan 31, 2022 at 21:00
\$\begingroup\$ It seems that concensus has gone in favor of this being invalid (codegolf.meta.stackexchange.com/questions/24411/…). \$\endgroup\$
– hyper-neutrino ♦
Commented Feb 5, 2022 at 5:51

| Show 1 more comment

l4m2 · Accepted Answer · 2022-02-06 02:04:26Z

2

JavaScript (Node.js), 89 bytes

f=(n,b)=>p(b)&p(a=n/b|0)&p(c=n%b)&a>b?[a,b,c]:b!=n&&f(n,-~b);p=(n,i=n)=>n%--i?p(n,i):i==1

Try it online!

If stackoverflow allowed as an error, then

JavaScript (Node.js), 83 bytes

f=(n,b)=>p(b)&p(a=n/b|0)&p(c=n%b)&a>b?[a,b,c]:f(n,-~b);p=(n,i=n)=>n%--i?p(n,i):i==1

Try it online!

edited Feb 6, 2022 at 2:04

answered Jan 31, 2022 at 8:47

l4m2

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\$\begingroup\$ It seems that concensus has gone in favor of this being invalid (codegolf.meta.stackexchange.com/questions/24411/…). \$\endgroup\$
– hyper-neutrino ♦
Commented Feb 6, 2022 at 0:10

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Deera Wijesundara · Accepted Answer · 2022-02-15 06:42:41Z

2

Python 3, 137 bytes

lambda n:[(a,b,c)for a in R(n)for b in R(n)for c in R(n)if all([a>b>c,a*b+c==n,all(all(n%m for m in R(2,n))for n in (a,b,c)),c])]
R=range

Try it online!

Thanks to @attiP for R=range -8 bytes

Python 3, 139 bytes

eval("lambda n:[(a,b,c)for aRfor bRfor cRif all([a>b>c,a*b+c==n,all(all(n%m for mR[2:])for n in (a,b,c)),c])]".replace("R"," in range(n)"))

Try it online!

edited Feb 15, 2022 at 6:42

answered Feb 13, 2022 at 13:31

Deera Wijesundara

3492 silver badges4 bronze badges

\$\begingroup\$ You can do just R=range for 137 bytes: Try it online! \$\endgroup\$
– AnttiP
Commented Feb 13, 2022 at 15:59

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ovs · Accepted Answer · 2022-01-31 13:01:31Z

1

Python 2, 112 bytes

P=k=1
p=r=[]
n=input()
while k<n:
 if P%k:r=r+[(k,b,n-k*b)for b in p if b>n-k*b in p];p+=k,
 P*=k*k;k+=1
print r

Try it online!

A port of l4m2's Javascript answer with a bit of their help comes in at ~~101~~ 99 bytes:

f=lambda n,b=1:all(p%k for p in(n/b,b,n%b)for k in range((n%b>1)+(n/b>b),p))*(n/b,b,n%b)or f(n,b+1)

Try it online!

edited Jan 31, 2022 at 13:01

answered Jan 31, 2022 at 10:23

ovs

60.2k3 gold badges47 silver badges160 bronze badges

1

\$\begingroup\$ Do this work? \$\endgroup\$
– l4m2
Commented Jan 31, 2022 at 10:49
\$\begingroup\$ That's a great python 2 answer. \$\endgroup\$
– solid.py
Commented Jan 31, 2022 at 18:06

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caird coinheringaahin g · Accepted Answer · 2022-02-01 03:54:06Z

Jelly (fork), 10 bytes

ÆRUŒƈ×+ƭ/Ƙ

Try it online! (the Jelly equivalent)

Outputs all possible solutions, or [] if there are none. And, the equivalent vanilla Jelly answer:

Jelly, 13 bytes

ÆRUœc3×+ƭ/=¥Ƈ

Try it online!

How they work

ÆRUŒƈ×+ƭ/Ƙ - Main link. Takes an integer N on the left
ÆR         - Primes from 2 to N
  U        - Reverse
   Œƈ      - Combinations of 3, no replacement
         Ƙ - Keep those that equal N under:
        /  -   Reduce [a, b, c] by:
       ƭ   -     Tie two dyads f, g:
     ×     -       Product
      +    -       Addition
                This maps [a, b, c] -> [a×b, c] -> a×b+c

The normal Jelly one is identical, except that œc3 is special-cased as Œƈ and =¥Ƈ is equivalent to Ƙ

G B · Accepted Answer · 2022-02-01 07:35:32Z

1

Ruby, 83 78 bytes

->n{(2..n).select{|x|/^(!!+)\1+$/!~?!*x}.combination(3).find{|c,b,a|a*b+c==n}}

Try it online!

edited Feb 1, 2022 at 7:35

answered Jan 31, 2022 at 8:28

G B

22.5k1 gold badge20 silver badges53 bronze badges

\$\begingroup\$ I gather 1 isn't a prime? You can fix this by using Prime.take(n), and either including require'prime' or -rprime as a flag. \$\endgroup\$
– dingledooper
Commented Jan 31, 2022 at 21:57

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Dominic van Essen · Accepted Answer · 2022-02-02 23:52:40Z

Husk, 14 bytes

ḟo=⁰§+←oΠtṖ3İp

Try it online!

Outputs c<b<a if a solution exists, otherwise quits after 1 minute with a 'the request exceeded the 60 second time limit and was terminated' warning from the 'try it online' interpreter.
Arguably this isn't a real 'error' (and Husk itself would happily run forever looking for a solution among higher and higher entries in the infinite list of primes), so for 1 more byte we can have:
ḟo=¹§+←oΠtṖ3fṗḣ (which only searches among primes less than the input).

ḟo              # return the first...
          Ṗ3    # ...list of 3 items selected from
            İp  # ...the infinite list of primes
                # that satisfies:           
    §+          #   the sum of...
      ←         #   ...the first item
       oΠ       #   ...and the product of 
         t      #   ...the tail (all except first item)
  =⁰            #   is equal to the input

Julian · Accepted Answer · 2022-02-03 02:02:06Z

1

PowerShell Core, 176 bytes

param($n)for($c=1;++$c-lt$n){for($b=$c;++$b-lt$n){for($a=$b;++$a-lt$n){if($a*$b+$c-eq$n-and(&($p={param($t)!($t-2)-!(2..($t-1)|?{!($t%$_)})})$a)*(&$p $b)*(&$p $c)){$a,$b,$c}}}}

Try it online!

I tried using ranges, but it gets really slow, probably because it is allocating too many of them, so for for now :)

Slow draft with ranges, 175 bytes

edited Feb 3, 2022 at 2:02

answered Feb 2, 2022 at 1:00

Julian

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ceilingcat · Accepted Answer · 2022-02-12 22:11:38Z

1

C (gcc), 139 130 122 118 115 bytes

a,b,c,z,y;p(x){for(y=z=1;++z<x;)y&=x%z>0;y=!y;}f(n){for(a=n;a&&p(c=(c<3?b<3?b=--a-1:--b:c)-1)|p(b)|p(a)|a*b+c-n;);}

Try it online!

Slightly golfed less.

a,b,c,z,y;
p(x){
  y=z=1;
  for(;++z<x;)
    y&=x%z>0;
  y=!y;
}
f(n){
  a=n;
  while(a&&p(c=(c<3?b<3?b=--a-1:--b:c)-1)|p(b)|p(a)|a*b+c-n);
}

edited Feb 12, 2022 at 22:11

answered Feb 3, 2022 at 17:15

ceilingcat

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xiver77 · Accepted Answer · 2022-03-03 12:03:07Z

80386 Machine Code, 110 107 99 bytes

Try it online!

When there is no valid answer, the program crashes, and I hope this belongs to "raise an error" category.

Thanks for the wonderful tip for skipping instructions. I'm still trying to get rid of the calls which take 5 bytes each, but it's not clear for now.

ecx is the input number, and edx is the pointer to the output array

00000000 <f>:
0:  89 cf                   mov    edi,ecx
2:  31 db                   xor    ebx,ebx
4:  89 d8                   mov    eax,ebx
00000006 <L0>:
6:  e3 40                   jecxz  48 <err>
8:  83 f8 02                cmp    eax,0x2
b:  7f 0b                   jg     18 <_1>
d:  83 fb 02                cmp    ebx,0x2
10: 7f 03                   jg     15 <_0>
12: 49                      dec    ecx
13: 89 cb                   mov    ebx,ecx
00000015 <_0>:
15: 4b                      dec    ebx
16: 89 d8                   mov    eax,ebx
00000018 <_1>:
18: 48                      dec    eax
19: 89 c5                   mov    ebp,eax
1b: e8 29 00 00 00          call   49 <p>
20: 73 e4                   jae    6 <L0>
22: 89 dd                   mov    ebp,ebx
24: e8 20 00 00 00          call   49 <p>
29: 73 db                   jae    6 <L0>
2b: 89 cd                   mov    ebp,ecx
2d: e8 17 00 00 00          call   49 <p>
32: 73 d2                   jae    6 <L0>
34: 89 ce                   mov    esi,ecx
36: 0f af f3                imul   esi,ebx
39: 01 c6                   add    esi,eax
3b: 39 fe                   cmp    esi,edi
3d: 75 c7                   jne    6 <L0>
3f: 89 0a                   mov    DWORD PTR [edx],ecx
41: 89 5a 04                mov    DWORD PTR [edx+0x4],ebx
44: 89 42 08                mov    DWORD PTR [edx+0x8],eax
47: c3                      ret
00000048 <err>:
48: f4                      hlt
00000049 <p>:
49: 50                      push   eax
4a: 51                      push   ecx
4b: 52                      push   edx
4c: 89 e9                   mov    ecx,ebp
0000004e <L1>:
4e: 49                      dec    ecx
4f: 83 f9 02                cmp    ecx,0x2
52: 7c 0a                   jl     5e <end>
54: 89 e8                   mov    eax,ebp
56: 99                      cdq
57: f7 f1                   div    ecx
59: 85 d2                   test   edx,edx
5b: 75 f1                   jne    4e <L1>
5d: a8                      .byte 0xa8
0000005e <end>:
5e: f9                      stc
5f: 5a                      pop    edx
60: 59                      pop    ecx
61: 58                      pop    eax
62: c3                      ret

output

the original assembly code

\$\begingroup\$ Nice! And yes, a crashed program is a fair result. \$\endgroup\$
– drmosley
Commented Mar 2, 2022 at 22:30 — drmosley, Commented Mar 2, 2022 at 22:30

Neil · Accepted Answer · 2022-01-31 10:55:47Z

Retina 0.8.2, 120 bytes

.+
$*
^(11(1)+)(?<!^\3+(11+))((?<9-2>\1))+(?!1?$|(11+)\5+$)((?<-4>1)+)$(?<=(?!(11+)\7+$)((?>1(?<-9>1)+)))|.+
$.1 $.8 $.6

Try it online! Link includes test cases. Explanation:

.+
$*

Convert to unary.

^(11(1)+)(?<!^\3+(11+))

Search for a prime number a ($.1 or 2+$#2)...

((?<9-2>\1))+

... that when repeated b (1+$#4 or 1+$#9) times ($#4/$#9 is less than or equal to $#2 so b is less than a)...

(?!1?$|(11+)\5+$)((?<-4>1)+)$

... leaves a remainder c ($.6) that is prime ($.6 is less than or equal to $#4 so c is less than b)...

(?<=(?!(11+)\7+$)((?>1(?<-9>1)+)))

... and b ($.8) is prime.

|.+

But if a match couldn't be found, replace the input anyway.

$.1 $.8 $.6

Replace the input with the desired values.

solid.py · Accepted Answer · 2022-01-31 18:11:55Z

0

Python 3, 174 bytes

from itertools import*
f=lambda n:[i for i in [(a,b,c)for a,b,c in combinations([*range(n-1,-1,-1)],3)if a*b+c==n]if all(map(lambda n:all(n%i for i in range(2,n))==(n>1),i))]

Try it online!

edited Jan 31, 2022 at 18:11

answered Jan 31, 2022 at 9:59

solid.py

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ophact · Accepted Answer · 2022-02-01 06:17:07Z

0

Python 3.8 (pre-release), 138 bytes

lambda i,F=lambda n:all(n%a for a in range(2,n)):[[j,k,l]for j in range(3,i)for k in range(3,i//j+1)if F(j)+F(k)+F(l:=i-j*k)>2<=l<k<j][:1]

Try it online!

Semi-brute-force method. Thanks AnttiP for spotting an obvious superfluous 5-byte segment.

edited Feb 1, 2022 at 6:17

answered Jan 31, 2022 at 10:05

ophact

3,1741 gold badge7 silver badges24 bronze badges

\$\begingroup\$ Isn't 2<=l|0==l the same as 2<=l? \$\endgroup\$
– AnttiP
Commented Jan 31, 2022 at 20:44
\$\begingroup\$ @AnttiP of course, that l|0 thing was just something from my previous code. Will fix. \$\endgroup\$
– ophact
Commented Feb 1, 2022 at 6:15

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Kevin Cruijssen · Accepted Answer · 2022-02-01 07:53:40Z

05AB1E, 10 bytes

ÅP3.Æʒ`*+Q

Outputs the triplets in the order \$[c,b,a]\$.

Try it online or verify all test cases.

Explanation:

ÅP          # Push a list of prime numbers ≤ the (implicit) input-integer
  3.Æ       # Get all unique 3-element combinations of this list
     ʒ      # Filter these [c,b,a]-triplets by:
      `     #  Pop and push c,b,a separated to the stack
       *    #  Multiply the top two: b*a
        +   #  Add it to the third: c+b*a
         Q  #  Check if it's equal to the (implicit) input-integer
            # (after which the filtered list of triplets is output implicitly)

Neil · Accepted Answer · 2022-02-01 17:28:22Z

Charcoal, 39 bytes

ＮθＩΦＥ…²θ⟦ι÷θι﹪θι⟧∧⬤⊞ＯΦιμ¹‹λ§ιμ⬤ι⬤…²λ﹪λν

Try it online! Link is to verbose version of code. Outputs all solutions. Explanation: Test all values a from 2 up to n and checks that the list [a, n/a, n%a, 1] is descending and that the first three elements are prime. Conveniently this optimised version turns out to be golfier than a naïve brute force algorithm.

Ｎθ                                      Input as an integer
     …                                  Range from
      ²                                 Literal integer `2`
       θ                                To input integer
    Ｅ                                   Map over values
         ι                              Current value
           θ                            Input integer
          ÷                             Integer divided by
            ι                           Current value
              θ                         Input integer
             ﹪                          Modulo
               ι                        Current value
        ⟦       ⟧                       Make into a list
   Φ                                    Filtered where
                       ι                Current list
                      Φ                 Filtered where
                        μ               Inner index
                   ⊞Ｏ                  Concatenated with
                        ¹               Literal integer `1`
                  ⬤                     All values satisfy
                          λ             Current value
                         ‹              Is less than
                            ι           Current list
                           §            Indexed by
                             μ          Inner index
                 ∧                      Logical And
                               ι        Current list
                              ⬤         All elements satisfy
                                 …      Range from
                                  ²     Literal integer `2`
                                   λ    To current element
                                ⬤       All values satisfy
                                     λ  Current element
                                    ﹪   Is not divisible by
                                      ν Innermost value
  Ｉ                                         Cast to string
                                            Implicitly print

ovs · Accepted Answer · 2022-02-02 14:50:57Z

0

BQN, 37 bytes^SBCS

Main idea taken from l4m2's answer.

(∨⊸≡∧´(1=·+´0=↕⊸|)¨)¨⊸/⊢(⌊÷∾⊢∾|˜)¨1↓↕

Run online!

answered Feb 2, 2022 at 14:50

ovs

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ovs · Accepted Answer · 2022-02-02 17:39:11Z

APL (Dyalog Unicode) + dfns, 29 bytes

I spent a bit too much time trying to golf RGS's answer, so I decided to post this on its own now.

{m⌿⍨⍵=2(⌷+0⌷×⌿)⍉m←pco⌽3cmat⍵}

Try it online!

There are two significant differences compared to the original answer:

When pco is applied to a single argument, it returns the ⍵-th prime number. This can be used to construct triples of primes directly, avoiding a more complicated primality test.
The implementation for \$a\times b+c\$ is a bit shorter using dyadic / to get the product.

Shaggy · Accepted Answer · 2022-02-02 17:48:43Z

0

Japt, 16 bytes

... So very out of practice. Returns all possible solutions, each in reverse order.

o ðj à f@¶XÎ+XÅ×

Try it

edited Feb 2, 2022 at 17:48

answered Feb 1, 2022 at 23:22

Shaggy

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MarcMush · Accepted Answer · 2022-03-03 14:45:30Z

0

Julia 0.4, 62 bytes

N->for a=P=primes(N),b=P,c=P a>b>c&&a*b+c==N&&return a,b,c end

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Julia 0.4 for the primes function (later moved in the Primes package). Returns nothing is there is no solution

answered Mar 3, 2022 at 14:45

MarcMush

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Benji · Accepted Answer · 2022-03-06 14:50:29Z

0

Haskell, 91 bytes

f n=[[a,b,c]|a<-q$n+1,b<-q a,c<-q b,a*b+c==n,and[0<mod y x|y<-[a,b,c],x<-q y]];q z=[2..z-1]

Try it online!

answered Mar 6, 2022 at 14:50

Benji

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Prime a*b+c of N

27 Answers 27

Wolfram Mathematica, 29 bytes

tinylisp, 210 bytes

Factor + math.combinatorics math.primes, 64 bytes

Explanation

Pyth, 37 bytes

Excel, 164 bytes

Explantion

Insider Beta Version using LAMBDA, 157 bytes

APL(Dyalog Unicode) + dfns, 6343 bytes SBCS

MATL, 28 25 24 23 bytes

Vyxal, 10 bytes

JavaScript (Node.js), 89 bytes

JavaScript (Node.js), 83 bytes

Python 3, 137 bytes

Python 3, 139 bytes

Python 2, 112 bytes

Jelly (fork), 10 bytes

Jelly, 13 bytes

How they work

Ruby, 83 78 bytes

Husk, 14 bytes

PowerShell Core, 176 bytes

C (gcc), 139 130 122 118 115 bytes

80386 Machine Code, 110 107 99 bytes

Retina 0.8.2, 120 bytes

Python 3, 174 bytes

Python 3.8 (pre-release), 138 bytes

05AB1E, 10 bytes

Charcoal, 39 bytes

BQN, 37 bytesSBCS

APL (Dyalog Unicode) + dfns, 29 bytes

Japt, 16 bytes

Julia 0.4, 62 bytes

Haskell, 91 bytes

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Factor + `math.combinatorics math.primes`, 64 bytes

APL(Dyalog Unicode) + dfns, ⁶³43 bytes ^SBCS

BQN, 37 bytes^SBCS

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