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

7 / 1 = 7
7 / 7 = 1

7 + 1 = 8

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

5 / 1 = 5
5 / 5 = 1

5 + 1 = 6

Output: 6

Answer 1

05AB1E, 2 bytes

ÑO

Try it online!

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

Try it online!

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;}

Try it online!

Answer 4

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;}

C++ optimization by c and c-- : 53 52 bytes

int f(int n,int i=0){return++i<n?f(n,i)+i*!(n%i):n;}

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 5

Japt, 3 bytes

â)x

Try it online!

Answer 6

Brachylog, 2 bytes

f+

Try it online!

Explanation

f       Factors
 +      Sum

Answer 7

Mathematica, 14 bytes

Tr@Divisors@#&   

or an answer by @Loki

Mathematica, 17 bytes

DivisorSum[#,#&]&

Answer 8

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).

Try it online!

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
    }

---

Noncompeting, 19 bytes

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

$n=>sum(factors(n))

Try it online!

Answer 9

Risky, 3 bytes

+/?+??

Try it online!

Answer 10

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 11

Pari/GP, 5 bytes

sigma

Try it online!

Answer 12

J, 23 bytes

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

Try it online!

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

Pyth, 6 bytes

s*M{yP

Try it here!

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 14

Java (OpenJDK 8), 53 51 bytes

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

Try it online!

Answer 15

Jelly, 2 bytes

Æs

Try it online!

Built-in that does exactly as wanted.

Answer 16

Haskell, 30 bytes

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

Try it online!

Answer 17

MATL, 6 bytes

t:\~fs

Try it online!

-4 bytes thanks to @LuisMendo

10 bytes

My previous solution using a loop

:"G@\~@*vs

Try it online!

3 bytes

Using built-in

Z\s

Try it online!

Answer 18

Javascript, 54 44 bytes

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

Saved 10 bytes thanks to Shaggy

Try it online!

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 19

Brain-Flak, 96 bytes

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

Try it online!

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 20

R, 31 26 bytes

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

Try it online!

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

Try it online!

Answer 21

TI-Basic, 16 bytes

sum(seq(Inot(fPart(Ans/I)),I,1,Ans

Takes input in Ans. Output is stored in Ans and displayed.

Answer 22

Vyxal s, 1 byte

K

Try it Online!

K∑ flagless. K just gets the divisors of a number.

Answer 23

Husk, 2 bytes

ΣḊ

Try it online!

Answer 24

Raku, 25 23 22 bytes

thanks ovs for 2 byte improvement
also JoKing for 1 byte improvement

{sum grep $_%%*,1..$_}

declare anonymous block ($_ implicitly declared)
filter (grep) numbers from 1 to $_ inclusive using the whatever variable (*) that are divisible by $_
get the sum of that list

Try it online!

Answer 25

Excel VBA, 41 Bytes

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

For i=1To[A1]:s=s-i*([A1]mod i=0):Next:?s

Excel, 41 Bytes

Worksheet formula that takes input from [A1] and outputs to the caller

Requires MS Excel Version 16.0 or later for access to Let(...) function

=LET(a,SEQUENCE(A1),SUM((MOD(A1,a)=0)*a))

Answer 26

tinylisp, 74 73 bytes

(load library
(d D(q((K N)(i K(a(i(mod N K)0 K)(D(s K 1)N))0
(q((N)(D N N

Try it online!

-1 byte thanks to DLosc.

And without library just for fun:

tinylisp, 97 90 bytes

(d D(q((N K)(i(l N K)N(D(s N K)K
(d F(q((K N)(i K(a(i(D N K)0 K)(F(s K 1)N))0
(q((N)(F N N

Try it online!

-7 bytes thanks to DLosc.

Answer 27

JavaScript, 31 bytes

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

Answer 28

Bash + GNU utilities, 36

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

Try it online.

---

Pure Bash, 41

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

Try it online.

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

Python 2, 41 bytes

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

Try it online!

Answer 30

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