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.

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Examples:

Input: 7

7 / 1 = 7
7 / 7 = 1

7 + 1 = 8

Output: 8

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Input: 15

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

15 + 5 + 3 + 1 = 24

Output: 24

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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

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Input: 1

1 / 1 = 1

Output: 1

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Input: 5

5 / 1 = 5
5 / 5 = 1

5 + 1 = 6

Output: 6

Answer 1

05AB1E, 2 bytes

ÑO

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How?

Ñ    Divisors
 O   Sum

Answer 2

x86-64 Machine Code, 23 bytes

89 F9 89 FE EB 0D 89 F8 99 F7 F1 85 D2 99 0F 44 D1 01 D6 E2 F1 96 C3

The above bytes of code define a function that accepts a single integer, N, and returns the sum of its multiples as a result.

The single parameter is passed in the EDI register, consistent with the System V AMD64 ABI (as used on *nix-style systems). The result is returned in the EAX register, as with all x86 calling conventions.

The algorithm is a very straightforward one, similar to many of the other submissions in other languages. We loop N times, each time computing the modulo and adding that to our running total.

Ungolfed assembly mnemonics:

; unsigned SumOfMultiples(unsigned N  /* (EDI) */)
    mov     ecx, edi      ; make copy of input N, to be used as our loop counter
    mov     esi, edi      ; make copy of input N, to be used as our accumulator
    jmp     CheckEnd      ; jump directly to 'CheckEnd'
AddModulo:
    mov     eax, edi      ; make copy of input N, to be used as input to DIV instruction
    cdq                   ; short way of setting EDX to 0, based on EAX
    div     ecx           ; divide EDX:EAX by ECX, placing remainder in EDX
    test    edx, edx      ; test remainder, and set ZF if it is zero
    cdq                   ; again, set EDX to 0, without clobbering flags
    cmovz   edx, ecx      ; set EDX to ECX only if remainder was zero (EDX = ZF ? 0 : ECX)
    add     esi, edx      ; add EDX to accumulator
CheckEnd:
    loop    AddModulo     ; decrement loop counter (ECX), and keep looping if it != 0
    xchg    eax, esi      ; move result from accumulator (ESI) into EAX
    ret                   ; return, with result in EAX

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It sure seems like there should be a way to make this shorter, but I can't see it. Computing modulo on x86 takes quite a bit of code, since you do it using the DIV (or IDIV) instruction, and both of those use fixed input registers (EDX and EAX), the values of which get clobbered (because they receive the results, the remainder and quotient, respectively).

The only real tricks here are pretty standard golfing ones:

• I've structured the code in a somewhat unusual way so that I can use the CISC-style LOOP instruction, which is basically just a combination of DEC+JNZ with the ECX register as the implicit operand.
• I'm using XCHG at the end instead of MOV because the former has a special 1-byte encoding when EAX is one of the operands.
• I use CDQ to zero out EDX in preparation for the division, even though for unsigned division you would ordinarily just zero it using a XOR. However, XOR is always 2 bytes, while CDQ is only 1 byte. I use CDQ again a second time inside of the loop to zero EDX, before the CMOVZ instruction. This works because I can be guaranteed that the quotient of the division (in EAX) is always unsigned, so a sign-extension into EDX will set EDX equal to 0.

Answer 3

C (gcc), 45 bytes

i,s;f(n){for(s=i=n;--i;)s+=n%i?0:i;return s;}

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Answer 4

Mathematica, 14 bytes

Tr@Divisors@#&   

or an answer by @Loki

Mathematica, 17 bytes

DivisorSum[#,#&]&

Answer 5

Japt, 3 bytes

â)x

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Answer 6

Shnap, 44 43 bytes

-1 bye thanks to Mr. Xcoder (lol I was outgolfed in my own language)

 $n return:{s=0for d:range(n+1)if n%d<1s+=d}

This is a function ($ starts a function in Shnap).

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Explanation:

$ n                        //Start function with parameter n
    return: {              //Technically, we are returning a scope-block, which evaluates to the last statement run
        s = 0              //Our result
        for d : range(n+1) //For each value in the iterator range(n+1)
            if n % d < 1  // If n is divisible by d
                s += d     // Add d to the sum
                           // Since (s += d) returns (s + d), and a scope-block returns the last run statement, this will be the last statement and equal to our result
    }

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Noncompeting, 19 bytes

After many language updates, this can now be reduced to a measly 19 bytes:

$n=>sum(factors(n))

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Answer 7

Brachylog, 2 bytes

f+

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Explanation

f       Factors
 +      Sum

Answer 8

C, C++, C#, D, Java, 65 62 bytes

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

This works in all theses 5 programming languages because of similarities.

C, C++ and D optimization : 62 60 bytes

In C++ and D, integers convert implicitly to booleans ( Zero => false, Not Zero => true ), so you don't need to have the !=0

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

D optimization : golfy template system, 55 bytes

T d(T)(T n){T s,i=1;for(;i<=n;++i)s+=n%i?0:i;return s;}

Code to test :

C :

printf("%d %d %d %d %d", d(7), d(15), d(20), d(1), d(5));

C++ :

std::cout << d(7) << ' ' << d(15) << ' ' << d(20) << ' ' << d(1) << ' ' << d(5);

C# :

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

    static void Main(string[] args)
    {
        var f = new FindSum();
        Console.WriteLine(string.Format("{0}, {1}, {2}, {3}, {4}", f.d(7), f.d(15), f.d(20), f.d(1), f.d(5)));
    }
}

D :

writeln(d(7));
writeln(d(15));
writeln(d(20));
writeln(d(1));
writeln(d(5));

Java :

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

    public static void main(String[] args) {
        FindSum f = new FindSum();
        System.out.println(String.format("%d, %d, %d, %d, %d", f.d(7), f.d(15), f.d(20), f.d(1), f.d(5)));
    }
}

Answer 9

R, 31 26 bytes

function(N)(x=1:N)%*%!N%%x

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Returns a 1x1 matrix.

Computes !N%%x maps elements d of 1:N by: d->(1 if d divides N, 0 otherwise)

Then x%*%x!N%%x is the matrix product of 1:N which results in the sum of x where !N%%x is 1. Neat! Technically a port of Luis Mendo's Octave answer but I only saw that after I thought of this.

R+ numbers, 14 bytes

numbers::Sigma

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Answer 10

Javascript, 54 44 bytes

n=>[...Array(x=n)].reduce(y=>y+!(n%x)*x--,0)

Saved 10 bytes thanks to Shaggy

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const f = n=>[...Array(x=n)].reduce(y=>y+!(n%x)*x--,0)

console.log(f(7))
console.log(f(15))
console.log(f(20))
console.log(f(1))
console.log(f(5))

Answer 11

Python, 44 bytes

lambda k:sum(i*(k%i<1)for i in range(1,1+k))

• Thanks to Stephen, save 1 byte by removing whitespace.
• Thanks to Jonathan Frech, save another 1 byte by changing if to multiply.

Answer 12

J, 23 bytes

[:+/](([:=&0]|[)#])1+i.

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For J fans, there is a clever 13 byte solution: >:@#.~/.~&.q: but since it wasn't my invention I'm not posting it as my official answer.

My own solution simply filters 1..n, finding divisors, then sums them. The crux of it is the dyadic fork

](([:=&0]|[)#])

Note that in this context ] is 1..n, and [ is n itself. Hence ]|[ are the remainders when dividing each element of 1..n into n, and =&0 tells you if they're equal to 0.

Answer 13

Java (OpenJDK 8), 53 51 bytes

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

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Answer 14

MATL, 6 bytes

t:\~fs

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-4 bytes thanks to @LuisMendo

10 bytes

My previous solution using a loop

:"G@\~@*vs

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3 bytes

Using built-in

Z\s

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Answer 15

Haskell, 30 bytes

f n=sum[i|i<-[1..n],n`mod`i<1]

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Answer 16

Brain-Flak, 96 bytes

((({})<>){<(([()]{})){<>(({})(<()>))<>{(({})){({}[()])<>}{}}{}<>([{}()]({})){((<{}{}>))}}{}>{}})

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Explanation:

_{Now outdated by improvements.}

The heart of the algorithm is this:

({}(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]({})) turns |N, M...| into |N mod M, M...|
{((<{}{}>))} if the top of stack is not zero, replace it and the second with zero

That is a modification on mod that will give us M if it is a factor of N and 0 otherwise. Full code is below.

((({})<>) place input, N on both stacks
{ Loop to find factors
 <
  (([()]{})) Decrement and Duplicate; get next factor to check
  { if not zero
   (<>({})<>) Copy N from other stack
   ({}(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]({})){((<{}{}>))} Code explained above
  }
  {} drop the zero
 >
 {} add the factor
}) push the sum

Answer 17

JavaScript, 31 bytes

f=(n,i=n)=>i&&!(n%i)*i+f(n,i-1)

Answer 18

VBA (Excel), 73 bytes

a=Cells(1,1)
x=1
While x<=a
If a Mod x = 0 Then b=b+x
x=x+1
Wend
MsgBox b

Answer 19

Pari/GP, 5 bytes

sigma

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Answer 20

Python 2, 41 bytes

f=lambda n,i=1:i<=n and(n%i<1)*i+f(n,i+1)

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Answer 21

Pyth, 6 bytes

s*M{yP

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Pyth doesn't have a built-in for divisors, so I think this is reasonable.

Explanation

s*M{yP    - Full program with implicit input.

     P    - The prime factors of the input.
    y     - The powerset of its prime factors.
   {      - Deduplicate.
 *M       - Map with multiplication.
s         - Sum.
          - Implicitly display the result.

Given 20, for instance, this is what our program does after each instruction:

• P: [2, 2, 5].

• y: [[], [2], [2], [5], [2, 2], [2, 5], [2, 5], [2, 2, 5]].

• {: [[], [2], [5], [2, 2], [2, 5], [2, 2, 5]].

• *M: [1, 2, 5, 4, 10, 20].

• s: 42.

Answer 22

Ohm v2, 2 bytes

VΣ

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This is pretty straight-forwad:

V   - Divisors.
 Σ  - Sum.

Answer 23

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!

Answer 24

Octave, 20 bytes

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

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Answer 25

Jelly, 2 bytes

Æs

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Built-in that does exactly as wanted.

Answer 26

RProgN 2, 2 bytes

ƒ+

Explained

ƒ+
ƒ   # Factorize
 +  # Sum

Trivial, but felt it needed to be posted.

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Answer 27

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

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

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Answer 28

Bash + GNU utilities, 36

bc<<<`seq -f"n=%g;a+=n*!$1%%n;" $1`a

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Pure Bash, 41

for((;++i<=$1;a+=$1%i?0:i))
{
:
}
echo $a

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I first tried a fancy bash expansion answer, but it ended up being longer than the simple loop above:

echo $[$(eval echo +\\\(n={1..$1},$1%n?0:n\\\))]

Answer 29

Recursiva, 20 18 bytes

smBa++'%'Va'a:0!a'

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Explanation:

smBa++'%'Va'a:0!a' - Input to a; 20
s                    - sum up; 42
 m                   - map with; [1, 0, 3, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 15]
  Ba                 - Range; [1,2..20]
    ++'%'Va'a:0!a'   - function, evaluates to 0 if param cannot divide a i.e. 20

Answer 30

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.

Answer 31

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

Answer 32

Gaia, 2 bytes

dΣ

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Pretty straight-forward:

dΣ   - Full program.

d    - Divisors.
 Σ   - Sum.

Answer 33

Neim, 2 bytes

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Answer 34

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

Answer 35

C#, 56 bytes

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Data

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

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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;
};

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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();
        }
    }
}

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Releases

• v1.0 - 56 bytes - Initial solution.

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Notes

• None

Answer 36

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

Answer 37

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%

Answer 38

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

Answer 39

int divSum(int n)
{
    // Final result of sum of divisors
    int res = 0;

    // find all divisors which divides n
    for (int i=2; i<=sqrt(num); i++)
    {
        // if i is divisor of n
        if (num%i==0)
        {
            // if both divisors are same then add
            // it only once else add both
            if (i==(n/i))
                res += i;
            else
                res += (i + n/i);
        }
    }

    // Add 1 to the res as 1 is also a divisor
    return (res + 1);
}

Answer 40

MY, 4 bytes

ωḊΣ↵

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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.

Answer 41

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)])

Answer 42

Ruby, 29 bytes

->x{(1..x).sum{|r|x%r>0?0:r}}

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Answer 43

Excel VBA, 44 Bytes

Anonymous VBE immediate window function that takes input from [A1] and outputs to the VBE immediate window

For i=1To[A1]:s=s+IIf([A1]mod i,0,i):Next:?s

Answer 44

Clojure, 53 bytes

#(apply +(for[i(range 1(inc %)):when(=(mod % i)0)]i))

Answer 45

Casio-Basic, 42 bytes

sum(seq(piecewise(n/x=int(n/x),x,0),x,1,n

The piecewise acts as a shorter If statement. seq loops values of x from 1 to n, and the piecewise returns x if the number is a factor, otherwise it returns 0. sum adds the whole list together to output the total.

41 bytes for the function, +1 to add n in the parameters box.