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 =わ 1
15 +たす 5 +たす 3 +たす 1 =わ 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 =わ 1
20 +たす 10 +たす 5 +たす 4 +たす 2 +たす 1 =わ 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

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

Answer 1

05AB1E, 2 bytes

ÑO

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How?

Ñ Divisors
 O Sum

• Mr. Xcoder: Polyglot with 2sable
• Lord Farquaad: ÑO - Rejecting the challenge and winning at the same time. That's pretty badass.

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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• CaydendW: If you're OK with undefined behaviour, you can save a few bytes by replacing that return s; with s=s;. It comes out to 40 bytes.
• CaydendW: Turns out you can shave 1 more byte by making it recursive. 39 bytes

Answer 4

C, C++, C#, D, Java, (削除) 65 (削除ここまで) 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 : (削除) 62 (削除ここまで) 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-- : (削除) 53 (削除ここまで) 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)));
 }
}

• Adalynn: 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.
• c--: 52 bytes, C++

Answer 5

Japt, 3 bytes

â)x

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• Mr. Xcoder: Alternative: â x
• Shaggy: @Mr.Xcoder: not really an alternative; it's doing the exact same thing - only difference is the choice of parenthesising.
• Gymhgy: Or with the flag -x, it could be one byte

Answer 6

Brachylog, 2 bytes

f+

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Explanation

f Factors
 + Sum

Answer 7

Mathematica, 14 bytes

Tr@Divisors@#& 

or an answer by @Loki

Mathematica, 17 bytes

DivisorSum[#,#&]&

• Hans Olo: @Jennymathy Very nice, thanks! An equivalent and funny way to write it is also: DivisorSum[#, # &] &
• Hans Olo: @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 :)
• ZaMoC: @Loki and Tr@Divisors@#& even better ;-)
• ZaMoC: @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
• wchargin: You can use Tr@*Divisors to shave off a byte.

Answer 8

Shnap, (削除) 44 (削除ここまで) 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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• Mr. Xcoder: ==0 is <1 (43 bytes)
• Socratic Phoenix: @Mr. Xcoder thanks... I was outgolfed... In my own language... Which isn't even esoteric xD

Answer 9

Risky, 3 bytes

+/?+??

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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.

• Deera Wijesundara: You could also write it as lambda k:sum(-~i*(k%-~i<1)for i in range(k))
• 97.100.97.109: 41 bytes.
• tsh: @97.100.97.109 Maybe that is different enough to post as another answer.
• tsh: @97.100.97.109 And looks like it is similar to this answer. codegolf.stackexchange.com/a/142108/44718 so you can go up vote that one.

Answer 11

Pari/GP, 5 bytes

sigma

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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.

• miles: This for 13 bytes should be equivalent: +1#.i.*0=i.|]
• Jonah: @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?
• miles: 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.

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), (削除) 53 (削除ここまで) 51 bytes

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

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• shooqie: @corsiKa Only class fields. In a local scope they're unitialized.

Answer 15

Jelly, 2 bytes

Æs

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Built-in that does exactly as wanted.

• Kevin Halley: Jelly seems so beautiful! Where can i find some stuff to study about it?
• hyperneutrino: @KevinHalley You can check out the Tutorial page on Github, or you can visit the official Jelly chatroom once you have 20 reputation, or the Jelly training chatroom, also at 20 reputation, which you'll need to request permission to train in
• Cody Gray: I desperately want to upvote this because you suggested not upvoting trivial solutions.
• hyperneutrino: @CodyGray Yeah, maybe there's no point putting that there :P

Answer 16

Haskell, 30 bytes

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

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Answer 17

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 18

Javascript, (削除) 54 (削除ここまで) 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 19

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

• Wheat Wizard: Do you have an explanation?
• MegaTom: @FunkyComputerMan I got one now!

Answer 20

R, (削除) 31 (削除ここまで) 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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• gstats: For the first one you can save 2 bytes with N=scan();
• Giuseppe: @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.
• Rui Barradas: Shouldn't the second be numbers::Sigma(N)? Like this it outputs the source code of function Sigma.
• Giuseppe: @RuiBarradas a function is a perfectly good submission. to test it, you obviously have to call it as I do in the first submission.
• Anders Ellern Bilgrau: Note: that anonymous functions in the later versions of R now can now use the syntax \(x) instead of function(x). So you can save quite a few characters there.

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

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

Answer 23

Husk, 2 bytes

ΣḊ

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• scpchicken: wow that's pretty good

Answer 24

Raku, (削除) 25 (削除ここまで) (削除) 23 (削除ここまで) 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

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• ovs: {grep($_%%*,1..$_).sum} works for 23
• scpchicken: that that is very smart
• Jo King: you can move sum to the start to save one on the brackets, 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))

• Axuary: For i=1To[A1]:s=s-([A1]mod i=0)*i:Next:?s is 41 bytes. The minus is because VBA seems to convert True to -1.
• Taylor Raine: @Axuary - thanks for having me take a look at this one again - It was fun to make the worksheet formula solution for this

Answer 26

tinylisp, (削除) 74 (削除ここまで) 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

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-1 byte thanks to DLosc.

And without library just for fun:

tinylisp, (削除) 97 (削除ここまで) 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

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-7 bytes thanks to DLosc.

• Giuseppe: @DLosc thanks! I should probably rename D to M since now it's just a mod function.
• DLosc: Huh--I didn't notice that, lol.

Answer 27

APL(Dyalog Unicode), ^{(削除) (削除ここまで)}9 bytes ^SBCS

+/∘⍸0=⍳|⊢

Try it on APLgolf!

Explanation

 ⊢ right argument (let it be N)
 | modulo
 ⍳ numbers from 1 to N 
 = 
 0
 0=⍳|⊢ boolean mask for divisors of N
 ⍸ vector of indices (from 1 to N) of 1s in the mask
 ∘ function composition
 / reduction
+
+/ sum
+/∘⍸0=⍳|⊢ solution

Answer 28

JavaScript, 31 bytes

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

Answer 29

Bash + GNU utilities, 36

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

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Pure Bash, 41

for((;++i<=1ドル;a+=1ドル%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ドル},1ドル%n?0:n\\\))]

Answer 30

Python 2, 41 bytes

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

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