Find the sum of the divisors of N

Question

Write a program that displays on the screen the sum of the divisors of a number (1 ≤ N ≤ 100) entered by the user in the range of 1 to N.

This is OEIS A000203.

Examples:

Input: 7

Output: 8

Input: 15

15 / 1 = 15
15 / 3 = 5
15 / 5 = 3
15 / 15 = 1

15 + 5 + 3 + 1 = 24

Output: 24

Input: 20

20 / 1 = 20
20 / 2 = 10
20 / 4 = 5
20 / 5 = 4
20 / 10 = 2
20 / 20 = 1

20 + 10 + 5 + 4 + 2 + 1 = 42

Output: 42

Input: 1

1 / 1 = 1

Output: 1

Input: 5

Output: 6

\$\begingroup\$ @H.PWiz I think he means "the divisors of a number N" \$\endgroup\$
– benzene
Commented Sep 8, 2017 at 0:18 — benzene, Commented Sep 8, 2017 at 0:18
\$\begingroup\$ I think you mean sum of divisors, aka, the sigma function? \$\endgroup\$
– Stephen
Commented Sep 8, 2017 at 0:18 — Stephen, Commented Sep 8, 2017 at 0:18
\$\begingroup\$ Sorry, i mean "The sum of the multiple of N". \$\endgroup\$
– Kevin Halley
Commented Sep 8, 2017 at 0:19 — Kevin Halley, Commented Sep 8, 2017 at 0:19
\$\begingroup\$ @H.PWiz this is the sum of those, so I dunno \$\endgroup\$
– Stephen
Commented Sep 8, 2017 at 0:21 — Stephen, Commented Sep 8, 2017 at 0:21
\$\begingroup\$ @Stephen That seems like a trivial change to me \$\endgroup\$
– H.PWiz
Commented Sep 8, 2017 at 0:21 — H.PWiz, Commented Sep 8, 2017 at 0:21

caird coinheringaahin g · Accepted Answer · 2017-09-08 06:37:33Z

1

Add++, 9 bytes

D,f,@,dFs

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I clearly got here too late. This defines a function that gets the factors, then sums them.

answered Sep 8, 2017 at 6:37

caird coinheringaahin g♦

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Mr. Xcoder · Accepted Answer · 2017-09-08 07:37:36Z

1

Ohm v2, 2 bytes

VΣ

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This is pretty straight-forwad:

V   - Divisors.
 Σ  - Sum.

answered Sep 8, 2017 at 7:37

Mr. Xcoder

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\$\begingroup\$ Damnit, you beat me too it! \$\endgroup\$
– ThePlasmaRailgun
Commented Dec 11, 2017 at 3:52

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Husk, 5 bytes

ṁΠuṖp

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How?

ṁΠuṖp  - Full program, implicit input.

     p  - Prime factors.
    Ṗ   - Powerset.
   u    - Remove duplicates.
ṁΠ     - Get the product of each list, sum and implicitly output.

Thanks to Zgarb for the suggestions in chat!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 8, 2017 at 7:43

Mr. Xcoder

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Mr. Xcoder · Accepted Answer · 2017-09-08 09:28:41Z

1

Gaia, 2 bytes

dΣ

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Pretty straight-forward:

dΣ   - Full program.

d    - Divisors.
 Σ   - Sum.

answered Sep 8, 2017 at 9:28

Mr. Xcoder

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Okx · Accepted Answer · 2017-09-08 10:17:50Z

1

Neim, 2 bytes

𝐅𝐬

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answered Sep 8, 2017 at 10:17

Okx

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Luis Mendo · Accepted Answer · 2017-09-08 10:30:54Z

1

Octave, 20 bytes

@(n)~mod(n,t=1:n)*t'

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answered Sep 8, 2017 at 10:30

Luis Mendo

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AdmBorkBork · Accepted Answer · 2017-09-08 13:43:12Z

PowerShell, 36 bytes

param($a)1..$a|%{$j+=$_*!($a%$_)};$j

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Explanation

param($a)1..$a|%{$j+=$_*!($a%$_)};$j
param($a)                            # Takes input $a
         1..$a|%{               };   # For-loop from 1 up to $a
                          $a%$_      # Modulo, if this is zero we've hit a divisor
                        !(     )     # Take the Boolean-not of that. If a divisor, it's 1
                     $_*             # Multiply the current number by that Boolean
                                     # Only if it's a divisor will this be non-zero
                 $j+=                # Add it into our accumulator
                                  $j # Output our accumulator

user58988user58988 · Accepted Answer · 2017-09-09 04:57:21Z

1

Axiom, 42 bytes

f(x:PI):PI==(x=1=>1;reduce(+,divisors(x)))

results

(19) -> [[i,f(i)] for i in [7,15,20,42,1,5] ]
   (19)  [[7,8],[15,24],[20,42],[42,96],[1,1],[5,6]]
                                          Type: List List PositiveInteger

answered Sep 9, 2017 at 4:57

user58988

\$\begingroup\$ Why is not possible upvote the answer of who write it? I have the suspect here people upvote their answer logging with some other login... I say that because in some question there are more star than the people answer... (so why give one star if you not participate to competition?) \$\endgroup\$
– user58988
Commented Sep 9, 2017 at 9:52

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

APL, 9 bytes

+/⍳×0=⍳|⊢

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How? (input n)

⍳|⊢, [n mod 1, ..., n mod n]
0=, [0 == n mod 1, ... 0 == n mod n]
⍳×, [1*(0 == n mod 1), ..., n*(0 == n mod n)]
+/, sum([1*(0 == n mod 1), ..., n*(0 == n mod n)])

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 10, 2017 at 21:41

Adalynn

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Shaggy · Accepted Answer · 2020-09-14 18:43:58Z

1

Rockstar, 96 bytes

listen to N
X's0
T's0
while N-X
let X be+1
let D be N/X
turn up D
let T be+D is N/X and X

say T

Try it here (Code will need to be pasted in)

edited Sep 14, 2020 at 18:43

answered Sep 14, 2020 at 18:20

Shaggy

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2014MELO03 · Accepted Answer · 2020-09-15 16:39:34Z

GolfScript, 15 bytes

~.,{.3$\%!*+}*+

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~                # Parses the input to integer
 .,              # Makes an array with numbers from 0 to (N-1)
   {.3$\%!*+}    # This block goes to the top of the stack without being executed
             *   # Folds
              +  # Adds the input and the result

What the block does:

    .            # Duplicates the number of the array
     3$          # Makes a copy of N
       \%!       # Tests if the number of the array is a divisor of N
          *      # Multiplies, resulting in the divisor if it was divisible or 0 if it wasn't
           +     # Adds to the acumulator

640KB · Accepted Answer · 2020-09-15 20:02:57Z

x86-16 machine code, 18 bytes

00000000: 89c1 9988 0e09 0150 d40a 7502 02d4 58e2  .......P..u...X.
00000010: f2c3                                     ..

Listing:

89 C1           MOV  CX, AX                 ; input number as loop counter 
99              CWD                         ; zero DX as running sum 
            DIVLOOP:
88 0E 0109      MOV  BYTE PTR[AAM1+1], CL   ; move loop counter to divisor 
50              PUSH AX                     ; save input number 
            AAM1:
D4 0A           AAM                         ; ZF if (AL % CL == 0), AH = quotient
75 02           JNZ  END_LOOP               ; if not ZF, continue loop 
02 D4           ADD  DL, AH                 ; otherwise add to running sum 
            END_LOOP:
58              POP  AX                     ; restore input number 
E2 F2           LOOP DIVLOOP                ; loop until CL == 0 
C3              RET                         ; return to caller

Callable function, input N in AX, result in DX.

Inspired by @CodyGray's excellent answer, and the comment "It sure seems like there should be a way to make this shorter" I just had to try!

As Cody mentioned, Using DIV/IDIV is inconvenient because it clobbers two registers including the dividend. Another is that the ZF flag is set when the quotient is 0, however in this case we're only interested in when the remainder is 0. Enter AAM, a trusty and underrated byte-sized division/modulo instruction that can be golfy. Even though it does clobber the dividend just like DIV, it will set ZF when the remainder is 0 which is what we want here. It's downside is that the divisor is encoded in the instruction opcode, which can be modified at runtime at a cost of 4 bytes (in real mode at least).

this is for MS-DOS ? what assembler and debugger you using ? TASM ? — Febriyanto Nugroho, Commented Sep 19, 2020 at 16:22
@NotStar I use MASM 1.1 for DOS asm and then just DOS DEBUG.EXE for the x86-16 stuff. Quick and dirty! — 640KB, Commented Sep 19, 2020 at 21:06

Larry Bagel · Accepted Answer · 2021-12-07 18:32:43Z

1

Python 3, 45 bytes

lambda n:sum(i for i in range(1,n+1)if n%i<1)

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answered Dec 7, 2021 at 18:32

Larry Bagel

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random person · Accepted Answer · 2022-01-29 11:10:39Z

1

Kotlin, 42 bytes

{var s=0;for(i in 1..it)if(it%i==0)s+=i;s}

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answered Jan 29, 2022 at 11:10

random person

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coltim · Accepted Answer · 2022-01-30 16:33:44Z

1

K (ngn/k), 14 bytes

{+/&~1!x%!1+x}

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!1+x generate sequence from 1..x
x% divide the input by each term in this range
&~1! identify values in the range that evenly divide the input
+/ take (and implicitly return) their sum

answered Jan 30, 2022 at 16:33

coltim

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chunes · Accepted Answer · 2022-02-24 22:22:42Z

1

Factor + `math.primes.factors math.unicode`, 15 bytes

[ divisors Σ ]

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answered Feb 24, 2022 at 22:22

chunes

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Seggan · Accepted Answer · 2022-09-26 18:08:05Z

1

Fig, \$3\log_{256}(96)\approx\$ 2.469 bytes

+Sk

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+Sk
 S  # Sum
  k # The divisors
+   # And add the input

answered Sep 26, 2022 at 18:08

Seggan

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Aiden Chow · Accepted Answer · 2022-09-27 16:02:37Z

1

Prolog (SWI), 60 56 bytes

-4 bytes thanks to @Steffan

N+X:-bagof(M,(between(1,N,M),N mod M<1),L),sumlist(L,X).

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edited Sep 27, 2022 at 16:02

answered Sep 26, 2022 at 21:13

Aiden Chow

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\$\begingroup\$ =:=0 can be <1 \$\endgroup\$
– naffetS
Commented Sep 26, 2022 at 23:32
\$\begingroup\$ also you can use bagof instead of findall \$\endgroup\$
– naffetS
Commented Sep 27, 2022 at 3:04

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bigyihsuan · Accepted Answer · 2022-09-27 18:25:12Z

1

Go, 62 bytes

func p(n int)(p int){i:=1
for;i<=n;i++{if n%i<1{p+=i}}
return}

Attempt This Online!

-2 bytes compared to my answer to Product of Divisors via use of Go's 0-initialization for variables.

answered Sep 27, 2022 at 18:25

bigyihsuan

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oeuf · Accepted Answer · 2022-11-13 12:56:33Z

K (ngn/k), 20 bytes

{+/t(*=(t:1+!x)!'x)}

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I honestly didn't know how I manage to make a for-each that find the divisors, lol.

Explanations:

{+/t(*=(t:1+!x)!'x)}  Main function. x is input
    (           'x)   For x amount of times
               !      Modulo it with
       (    !x)       Range, create a list of [0..x-1]
          1+          + 1 to each number in list to turn it into [1..x]
        t:            Assign it to variable t
                      Now, we got the modulo results of each number from [1..x],
                      we now need to
      =               Group them into a dictionary
     *                And get the value of the first key
                      i.e. the occurence positions of the 0s
                      (Implicit at)
   t                  The variable t
 +/                   Sum

The Thonnu · Accepted Answer · 2023-02-14 10:42:00Z

1

Thunno, \$ 2 \log_{256}(96) \approx \$ 1.65 bytes

fS

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f   # Factors
 S  # Sum

answered Feb 14, 2023 at 10:42

The Thonnu

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The Thonnu · Accepted Answer · 2023-06-14 16:34:31Z

1

Thunno 2 `S`, 1 byte

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Or, FS flagless.

answered Jun 14, 2023 at 16:34

The Thonnu

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chunes · Accepted Answer · 2023-06-14 17:22:25Z

1

Arturo, 19 16 bytes

$=>[∑factors&]

Try it

edited Jun 14, 2023 at 17:22

answered Nov 13, 2022 at 11:34

chunes

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ATaco · Accepted Answer · 2017-09-08 02:11:19Z

0

RProgN 2, 2 bytes

ƒ+

Explained

ƒ+
ƒ   # Factorize
 +  # Sum

Trivial, but felt it needed to be posted.

Try it online!

answered Sep 8, 2017 at 2:11

ATaco

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Xcali · Accepted Answer · 2017-09-08 04:05:19Z

0

Perl 5, 35 + 1 (-p) = 36 bytes

$\+=($n%$_==0)&&$_ for 1..($n=$_)}{

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answered Sep 8, 2017 at 4:05

Xcali

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steenbergh · Accepted Answer · 2017-09-08 07:22:09Z

0

QBIC, 17 bytes

[:|~b%a|\p=p+a}?p

Explanation

[:|      FOR a = 1; a <= b (read from cmd line); a++
~b%a|    IF b modulo a has a remainder THEN - empty block - 
\p=p+a   ELSE add divisor 'a' to running total 'p'
}        END IF, NEXT
?p       PRINT p

answered Sep 8, 2017 at 7:22

steenbergh

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

CJam, 16 bytes

ri:X{)_X\%!*}%:+

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Explanation

ri                 e# Read integer, n
  :X               e# Write to variable X
    {       }%     e# Map this block over the array [0 1 ... n-1]
     )             e# Increment current value. Will give 1, 2, ..., n
      _            e# Duplicate
       X           e# Push input
        \          e# Swap
         %         e# Modulus
          !        e# Negate
           *       e# Multiply
              :+   e# Sum of array. Implicitly display

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 8, 2017 at 10:39

Luis Mendo

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Community · Accepted Answer · 2020-06-17 09:04:33Z

C#, 56 bytes

Data

Input Int32 i A number
Output Int32 The sum of the divisors of the number

Golfed

i=>{int s=0,d=1;for(;d<=i;d++)s+=i%d<1?i/d:0;return s;};

Ungolfed

i => {
    int
        s = 0,
        d = 1;
        
    for( ; d <= i; d++ )
        s += i % d < 1 ? i / d : 0;
    
    return s;
};

Ungolfed readable

// Takes an int
i => {
    // Initializes the sum 's' and divider 'd' vars
    int
        s = 0,
        d = 1;
        
    // Cycles each number lower than the the number 'i'
    for( ; d <= i; d++ )
    
        // Sums the result of the division between 'i' and 'd' if the modulus is 0
        s += i % d < 1 ? i / d : 0;
    
    // Returns the sum
    return s;
};

Full code

using System;
using System.Collections.Generic;

namespace TestBench {
    public class Program {
        // Methods
        static void Main( string[] args ) {
            Func<Int32, Int32> f = i => {
                int s = 0, d = 1;

                for( ; d <= i; d++ )
                    s += i % d < 1 ? i / d : 0;

                return s;
            };

            List<Int32>
                testCases = new List<Int32>() {
                    7,
                    15,
                    20,
                    42,
                    1,
                    5,
                };

            foreach( Int32 testCase in testCases ) {
                Console.WriteLine( $" INPUT: {testCase}\nOUTPUT: {f( testCase )}" );
            }

            Console.ReadLine();
        }
    }
}

Releases

v1.0 - 56 bytes - Initial solution.

Notes

None

In this statement s+=i%d<1?i/d:0;, you can save 2 bytes by replacing i/d by d, the results will be the same ( mostly because it will add the same numbers by in the reverse order : For example, in the second test case, all the division result are added : 15+5+3+1=24. If you add all the denominators ( i believe it's the word to describe the number used to divide the input, sorry if it's the wrong word, i'm not a native english speaker ), we get : 1+3+5+15=24, which is the same ) — HatsuPointerKun, Commented Sep 8, 2017 at 20:03

Neil · Accepted Answer · 2017-09-08 23:48:08Z

0

Batch, 70 bytes

@set/ai=%2+1,s=%3+i*!(%1%%i)
@if not %i%==%1 %0 %1 %i% %s%
@echo %s%

Alternative solution, also 70 bytes:

@set s=0
@for /l %%i in (1,1,%1)do @set/as+=%%i*!(%1%%%%i)
@echo %s%

answered Sep 8, 2017 at 23:48

Neil

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Adalynn · Accepted Answer · 2017-09-10 21:36:37Z

0

MY, 4 bytes

ωḊΣ↵

Try it online!

How?

It's really simple: ω is the argument, Ḋ is divisors, Σ is sum, ↵ is output. I thought I already answered this challenge, for some reason.

answered Sep 10, 2017 at 21:36

Adalynn

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Find the sum of the divisors of N

67 Answers 67

Add++, 9 bytes

Ohm v2, 2 bytes

Husk, 5 bytes

How?

Gaia, 2 bytes

Neim, 2 bytes

Octave, 20 bytes

PowerShell, 36 bytes

Explanation

Axiom, 42 bytes

APL, 9 bytes

How? (input n)

Rockstar, 96 bytes

GolfScript, 15 bytes

x86-16 machine code, 18 bytes

Python 3, 45 bytes

Kotlin, 42 bytes

K (ngn/k), 14 bytes

Factor + math.primes.factors math.unicode, 15 bytes

Fig, \$3\log_{256}(96)\approx\$ 2.469 bytes

Prolog (SWI), 60 56 bytes

Go, 62 bytes

K (ngn/k), 20 bytes

Thunno, \$ 2 \log_{256}(96) \approx \$ 1.65 bytes

Thunno 2 S, 1 byte

Arturo, 19 16 bytes

RProgN 2, 2 bytes

Explained

Perl 5, 35 + 1 (-p) = 36 bytes

QBIC, 17 bytes

Explanation

CJam, 16 bytes

Explanation

C#, 56 bytes

Data

Golfed

Ungolfed

Ungolfed readable

Full code

Releases

Notes

Batch, 70 bytes

MY, 4 bytes

How?

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