# Find the sum of the divisors of N

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

• @H.PWiz I think he means "the divisors of a number N" – benzene Sep 8 '17 at 0:18
• I think you mean sum of divisors, aka, the sigma function? – Stephen Sep 8 '17 at 0:18
• Sorry, i mean "The sum of the multiple of N". – Kevin Halley Sep 8 '17 at 0:19
• @H.PWiz this is the sum of those, so I dunno – Stephen Sep 8 '17 at 0:21
• @Stephen That seems like a trivial change to me – H.PWiz Sep 8 '17 at 0:21

# 05AB1E, 2 bytes

ÑO

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### How?

Ñ    Divisors
O   Sum
• Polyglot with 2sable – Mr. Xcoder Sep 8 '17 at 9:31
• ÑO - Rejecting the challenge and winning at the same time. That's pretty badass. – Lord Farquaad Sep 8 '17 at 20:26

# 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'
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)
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.

# C (gcc), 45 bytes

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

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# Japt, 3 bytes

â)x

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• Alternative: â x – Mr. Xcoder Sep 8 '17 at 22:11
• @Mr.Xcoder: not really an alternative; it's doing the exact same thing - only difference is the choice of parenthesising. – Shaggy Oct 10 '17 at 16:57
• Or with the flag -x, it could be one byte – Embodiment of Ignorance Apr 20 '19 at 19:30

f+

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f       Factors
+      Sum

# Mathematica, 14 bytes

Tr@Divisors@#&

# Mathematica, 17 bytes

DivisorSum[#,#&]&
• @Jennymathy Very nice, thanks! An equivalent and funny way to write it is also: DivisorSum[#, # &] & – Rebel-Scum Sep 8 '17 at 18:15
• @Jennymathy Hmm, this is even better: Total@Divisors@ is only 15 characters long! And it works: eg Total@Divisors@15 gives 24 as expected. Mathematica FTW :) – Rebel-Scum Sep 8 '17 at 19:27
• @Loki and Tr@Divisors@#& even better ;-) – J42161217 Sep 8 '17 at 19:29
• @Loki the program must be a function f= that takes an input f[x] that's why I present it in this way.Welcome to PPCG – J42161217 Sep 8 '17 at 19:36
• You can use Tr@*Divisors to shave off a byte. – wchargin Sep 9 '17 at 1:52

## 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)));
}
}
• A few things: First, I don't think you need parentheses around the n%i/n%i!=0 in any of the languages. Second, your first solution should be able to have n%i>0 instead of n%i!=0. Third, D's solution can be T d(T)(T n){T s,i=1;for(;i<=n;++i)s+=n%i?0:i;return s;} by abusing the template system and default values. – Zacharý Sep 9 '17 at 13:51

# 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:

$n //Start function with parameter n return: { //Technically, we are returning a scope-block, which evaluates to the last statement run s = 0 //Our result for d : range(n+1) //For each value in the iterator range(n+1) if n % d < 1 // If n is divisible by d s += d // Add d to the sum // Since (s += d) returns (s + d), and a scope-block returns the last run statement, this will be the last statement and equal to our result } # Noncompeting, 19 bytes After many language updates, this can now be reduced to a measly 19 bytes:$n=>sum(factors(n))

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• ==0 is <1 (43 bytes) – Mr. Xcoder Sep 8 '17 at 9:22
• @Mr. Xcoder thanks... I was outgolfed... In my own language... Which isn't even esoteric xD – Socratic Phoenix Sep 8 '17 at 10:51

# 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.

# 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.

• This for 13 bytes should be equivalent: +1#.i.*0=i.|] – miles Sep 8 '17 at 6:22
• @miles, that is really nice. This part is i.|] is a great improvement on my approach. I don't fully understand this part though: +1#.i. -- could you explain it? – Jonah Sep 8 '17 at 6:27
• 1#. is base 1 conversion, which is equivalent to +/"1. First i.|] to get the remainders, then 0= to find the ones equal to 0 (the divisors), then i.* to zero out the non-divisors in the range, then sum using 1#., then add + itself since i. is an exclusive range. – miles Sep 8 '17 at 6:31

# 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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• @corsiKa Only class fields. In a local scope they're unitialized. – shooqie Sep 9 '17 at 9:03

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

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# 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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Using built-in

Z\s

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# 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))

# 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
>
}) push the sum
• Do you have an explanation? – Wheat Wizard Sep 19 '17 at 2:33
• @FunkyComputerMan I got one now! – MegaTom Sep 21 '17 at 19:56

# 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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• For the first one you can save 2 bytes with N=scan(); – gstats Sep 8 '17 at 12:07
• @gstats yes, but then I should get +4 bytes per meta discussion. If you have a strong opinion you can weigh in on Jarko's answer but as nobody has suggested an alternative, that stands in my mind. – Giuseppe Sep 8 '17 at 12:18
• Shouldn't the second be numbers::Sigma(N)? Like this it outputs the source code of function Sigma. – Rui Barradas Sep 8 '17 at 15:58
• @RuiBarradas a function is a perfectly good submission. to test it, you obviously have to call it as I do in the first submission. – Giuseppe Sep 8 '17 at 15:59

# JavaScript, 31 bytes

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

sigma

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# Python 2, 41 bytes

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

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# 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
• This answer is invalid as it is a collection of snippets that cannot be run as a single unit as stands. To make this valid you will need to convert this to a subroutine or an anonymous VBE immediate window function. – Taylor Scott Sep 17 '17 at 19:44
• I am not very familiar in what you had said. Can you help me a bit more? – remoel Sep 18 '17 at 4:43
• To make this post valid you would have to convert it into one of the following formats, 1 - Subroutine, 2 - Function, 3 - Anonymous VBE immediate window function (a single line that can be executed in the Immediate window); For your implementation, the simplest implementation of this would be to convert to a subroutine by wrapping with Sub Y...End Sub to get the 85 Byte solution Sub y A=Cells(1,1) x=1 While x<=A If A Mod x=0 Then b=b+x x=x+1 Wend MsgBox b End Sub – Taylor Scott Sep 20 '17 at 0:32
• That however can be optimized quite heavily down to the 72 byte solution Sub y While x<=[A1] x=x+1 If [A1]Mod x=0Then b=b+x Wend Debug.?b End Sub which assumes that it is run in a clean module (x = default int value, 0) and outputs to the VBE immediate window (? autoformats to Print ) – Taylor Scott Sep 20 '17 at 0:34
• Beyond this, and recognizing that your solution does not take input via the subroutine call, this can then be converted to a VBE immediate window function for 50 Bytes While x<=[A1]:x=x+1:b=IIf([A1]Mod x,b,b+x):Wend:?b which assumes that x,b are the default value of 0 and outputs to the VBE immediate window (from the VBE immediate window ? is equivalent to Debug.Print ) – Taylor Scott Sep 20 '17 at 0:38

# 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.

# Ohm v2, 2 bytes

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V   - Divisors.
Σ  - Sum.
• Damnit, you beat me too it! – ThePlasmaRailgun Dec 11 '17 at 3:52

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!

# Octave, 20 bytes

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

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# RProgN 2, 2 bytes

ƒ+

## Explained

ƒ+
ƒ   # Factorize
+  # Sum

Trivial, but felt it needed to be posted.

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

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

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# Bash + GNU utilities, 36

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

# Pure Bash, 41

for((;++i<=$1;a+=$1%i?0:i))
{
:
}
echo $a I first tried a fancy bash expansion answer, but it ended up being longer than the simple loop above: echo$[\$(eval echo +\\$$n={1..1},1%n?0:n\\$$)]

D,f,@,dFs

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

# 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

# Gaia, 2 bytes

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Pretty straight-forward:

dΣ   - Full program.

d    - Divisors.
Σ   - Sum.