The current Perfect Numbers challenge is rather flawed and complicated, since it asks you to output in a complex format involving the factors of the number. This is a purely repost of the challenge.


Given a positive integer through any standard input format, distinguish between whether it is perfect or not.

A perfect number is a number that is equal to the sum of all its proper divisors (its positive divisors less than itself). For example, \$6\$ is a perfect number, since its divisors are \$1,2,3\$, which sum up to \$6\$, while \$12\$ is not a perfect number since its divisors ( \$1,2,3,4,6\$ ) sum up to \$16\$, not \$12\$.

Test Cases:




  • Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time.
  • Output can be two distinct and consistent values through any allowed output format. If it isn't immediately obvious what represents Perfect/Imperfect, please make sure to specify in your answer.
  • \$\begingroup\$ Wait, so truthy is for values that aren't perfect, and falsey is for values that are? \$\endgroup\$ – Esolanging Fruit Mar 12 '19 at 2:57
  • 2
    \$\begingroup\$ @Tvde1 Proper divisors have to less than the number, otherwise no number other than 1 would be perfect, since every number is divisible by 1 and itself. The sum of proper divisors of 1 is 0 \$\endgroup\$ – Jo King Mar 12 '19 at 7:40
  • 3
    \$\begingroup\$ @Grimy Only if you can prove so. Good luck! (though I'm wondering how that would save bytes) \$\endgroup\$ – Jo King Mar 12 '19 at 9:15
  • 1
    \$\begingroup\$ So no, too bad. It would cut the size of an ECMA regex answer by a factor of about 3. \$\endgroup\$ – Grimmy Mar 12 '19 at 9:18
  • 3
    \$\begingroup\$ "Output can be two distinct and consistent values" - may we not use "truthy vs falsey" here (e.g. for Python using zero vs non zero; a list with content vs an empty list; and combinations thereof)? \$\endgroup\$ – Jonathan Allan Mar 12 '19 at 10:15

45 Answers 45


Brachylog, 4 bytes


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The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true. or false. if run as a complete program (except on the last test case which takes more than a minute on TIO).

        The input's
f       factors
 k      without the last element
  +     sum to
   ?    the input.
  • 1
    \$\begingroup\$ I like how the code says fk :x \$\endgroup\$ – Ismael Miguel Mar 13 '19 at 1:38

Neim, 3 bytes


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(I don't actually know how to run all of the test cases at once, since I started learning Neim about fifteen minutes ago, but I did check them individually.)

Prints 0 for imperfect, 1 for perfect.

𝐕      Pop an int from the stack and push its proper divisors,
       implicitly reading the int from a line of input as the otherwise absent top of the stack.
 𝐬     Pop a list from the stack and push the sum of the values it contains.
  𝔼    Pop two ints from the stack and push 1 if they are equal, 0 if they are not;
       implicitly reading the same line of input that was already read as the second int, I guess?
       Implicitly print the contents of the stack, or something like that.
  • 2
    \$\begingroup\$ "I guess?"; "or something like that.". When you're not even sure what you've written yourself, haha. ;) But yes, that's indeed how it works. I don't know Neim, but using the input implicitly like that and outputting implicitly at the end implicitly, is similar in 05AB1E. \$\endgroup\$ – Kevin Cruijssen Mar 13 '19 at 12:26
  • \$\begingroup\$ How is 𝔼 1 byte? Does Neim use only 128 such non-standart characters? \$\endgroup\$ – kajacx Mar 13 '19 at 13:39
  • 3
    \$\begingroup\$ @kajacx Neim has its own code page. Therefore, each of the 256 characters present in the codepage can be encoded using 1 byte. \$\endgroup\$ – Mr. Xcoder Mar 13 '19 at 17:23

R, 33 29 bytes


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Returns TRUE for perfect numbers and FALSE for imperfect ones.

  • \$\begingroup\$ What do the 2 !s in a row get you? \$\endgroup\$ – CT Hall Mar 12 '19 at 3:16
  • \$\begingroup\$ @CTHall I misread the spec; they originally mapped 0 (perfect) to FALSE and nonzero to TRUE but I removed one of them to reverse the mapping. It's a useful golfing trick to cast from numeric to logical, often in conjunction with which or [. \$\endgroup\$ – Giuseppe Mar 12 '19 at 3:37

Jelly, 3 bytes


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Japt -!, 4 bytes

            Implicit Input U
¥           Equal to
   x        Sum of
 â          Factors of U
  ¬         Without itself

For some reason ¦ doesnt work on tio so I need to use the -! flag and ¥ instead

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  • \$\begingroup\$ That's not a TIO issue; U doesn't get auto-inserted before !. \$\endgroup\$ – Shaggy Mar 12 '19 at 9:31

Python 3, 46 bytes

lambda x:sum(i for i in range(1,x)if x%i<1)==x

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Brute force, sums the factors and checks for equality.

  • 2
    \$\begingroup\$ Using the comprehension condition as a mask for your iteration variable would save a byte. \$\endgroup\$ – Jonathan Frech Mar 12 '19 at 4:04
  • \$\begingroup\$ Since you can return truthy for an imperfect number, lambda x:sum(i for i in range(1,x)if x%i<1)^x should work as well. \$\endgroup\$ – nedla2004 Mar 12 '19 at 13:13

Python, 45 bytes

lambda n:sum(d*(n%d<1)for d in range(1,n))==n

True for perfect; False for others (switch this with == -> !=)

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 44 42  41 bytes (-2 thanks to ovs) if we may output using "truthy vs falsey":

f=lambda n,i=1:i/n or-~f(n,i+1)-(n%i<1)*i

(falsey (0)) for perfect; truthy (a non-zero integer) otherwise

  • \$\begingroup\$ If the second output format is valid, this can be done in 42 bytes. \$\endgroup\$ – ovs Mar 12 '19 at 11:43
  • \$\begingroup\$ @ovs ah, nicely done. \$\endgroup\$ – Jonathan Allan Mar 12 '19 at 12:06
  • \$\begingroup\$ @ovs ..and another saved from that - thanks! \$\endgroup\$ – Jonathan Allan Mar 12 '19 at 12:16

Octave, 25 bytes


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@(n)                        % Define anonymous function with input n
             1:n            % Row vector [1,2,...,n]
           t=               % Store in variable t
     mod(n,     )           % n modulo [1,2,...,n], element-wise. Gives 0 for divisors
    ~                       % Logical negate. Gives 1 for divisors
                  t'        % t transposed. Gives column vector [1;2;...;n]
                 *          % Matrix multiply
                      2*n   % Input times 2
                    ==      % Equal? This is the output value

JavaScript, 38 bytes


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(Last testcase timeout on TIO.)

  • \$\begingroup\$ @Arnauld Just forgot to remove the f= after converting from a recursive function. \$\endgroup\$ – tsh Mar 12 '19 at 10:07
  • \$\begingroup\$ Just out of curiosity, why not going with a recursive version? (It would be 34 bytes.) \$\endgroup\$ – Arnauld Mar 12 '19 at 10:09
  • \$\begingroup\$ @Arnauld because recursive version would simply failed for larger testcase due to stack overflow. Maybe I need some environments default to strict mode to make it work. \$\endgroup\$ – tsh Mar 12 '19 at 10:17
  • 2
    \$\begingroup\$ Fair enough, but your program doesn't have to complete the larger test cases (which I think is the default rule, anyway). \$\endgroup\$ – Arnauld Mar 12 '19 at 10:22

C# (Visual C# Interactive Compiler), 46 bytes


Returns 0 if perfect, otherwise returns a positive number. I don't know if outputting different types of integers are allowed in place of two distinct truthy and falsy values, and couldn't find any discussion on meta about it. If this is invalid, I will remove it.

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C# (Visual C# Interactive Compiler), 49 47 bytes


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Labyrinth, 80 bytes

+puts; "
}zero: "
"negI"  _~

The Latin characters perfect puts zero else neg I are actually just comments*.
i.e. if the input is perfect a 0 is printed, otherwise -1 is.

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* so this or this work too...

?::`}:("(!@               ?::`}:("(!@
       :                  BEWARE :
{:{:;%"}                  {:{:;%"}
+    ; "                  +LAIR; "
}    : "                  } OF : "
}    {(:                  }MINO{(:
"    "  _~                "TAUR"  _~
""""""{{{"!@              """"""{{{"!@


Takes as an input a positive integer n and places an accumulator variable of -n onto the auxiliary stack, then performs a divisibility test for each integer from n-1 down to, and including, 1, adding any which do divide n to the accumulator. Once this is complete if the accumulator variable is non-zero a -1 is output, otherwise a 0 is.

The ?::`}:( is only executed once, at the beginning of execution:

?::`}:(                                                      Main,Aux
?       - take an integer from STDIN and place it onto Main  [[n],[]]
 :      - duplicate top of Main                            [[n,n],[]]
  :     - duplicate top of Main                          [[n,n,n],[]]
   `    - negate top of Main                            [[n,n,-n],[]]
    }   - place top of Main onto Aux                       [[n,n],[-n]]
     :  - duplicate top of Main                          [[n,n,n],[-n]]
      ( - decrement top of Main                        [[n,n,n-1],[-n]]

The next instruction, ", is a no-op, but we have three neighbouring instructions so we branch according to the value at the top of Main, zero takes us forward, while non-zero takes us right.

If the input was 1 we go forward because the top of Main is zero:

(!@                                                          Main,Aux
(   - decrement top of Main                             [[1,1,-1],[-1]]
 !  - print top of Main, a -1
  @ - exit the labyrinth

But if the input was greater than 1 we turn right because the top of Main is non-zero:

:}                                                           Main,Aux
:  - duplicate top of Main                         [[n,n,n-1,n-1],[-n]]
 } - place top of Main onto Aux                        [[n,n,n-1],[-n,n-1]]

At this point we have a three-neighbour branch, but we know n-1 is non-zero, so we turn right...

"%                                                           Main,Aux
"  - no-op                                             [[n,n,n-1],[-n,n-1]]
 % - place modulo result onto Main                   [[n,n%(n-1)],[-n,n-1]]
   - ...i.e we've got our first divisibility indicator n%(n-1), an
   -    accumulator, a=-n, and our potential divisor p=n-1:
   -                                                 [[n,n%(n-1)],[a,p]]

We are now at another three-neighbour branch at %.

If the result of % was non-zero we go left to decrement our potential divisor, p=p-1, and leave the accumulator, a, as it is:

;:{(:""}"                                                    Main,Aux
;          - drop top of Main                                [[n],[a,p]]
 :         - duplicate top of Main                         [[n,n],[a,p]]
  {        - place top of Aux onto Main                  [[n,n,p],[a]]
           - three-neighbour branch but n-1 is non-zero so we turn left
   (       - decrement top of Main                     [[n,n,p-1],[a]]
    :      - duplicate top of Main                 [[n,n,p-1,p-1],[a]]
     ""    - no-ops                                [[n,n,p-1,p-1],[a]]
       }   - place top of Main onto Aux                [[n,n,p-1],[a,p-1]]
        "  - no-op                                     [[n,n,p-1],[a,p-1]]
         % - place modulo result onto Main           [[n,n%(p-1)],[a,p-1]]
           - ...and we branch again according to the divisibility
           -    of n by our new potential divisor, p-1

...but if the result of % was zero (for the first pass only when n=2) we go straight on to BOTH add the divisor to our accumulator, a=a+p, AND decrement our potential divisor, p=p-1:

;:{:{+}}""""""""{(:""}                                       Main,Aux
;                      - drop top of Main                    [[n],[a,p]]
 :                     - duplicate top of Main             [[n,n],[a,p]]
  {                    - place top of Aux onto Main      [[n,n,p],[a]]
   :                   - duplicate top of Main         [[n,n,p,p],[a]]
    {                  - place top of Aux onto Main  [[n,n,p,p,a],[]]
     +                 - perform addition            [[n,n,p,a+p],[]]
      }                - place top of Main onto Aux      [[n,n,p],[a+p]]
       }               - place top of Main onto Aux        [[n,n],[a+p,p]]
        """""""        - no-ops                            [[n,n],[a+p,p]]
                       - a branch, but n is non-zero so we turn left
               "       - no-op                             [[n,n],[a+p,p]]
                {      - place top of Aux onto Main      [[n,n,p],[a+p]]
                       - we branch, but p is non-zero so we turn right
                 (     - decrement top of Main         [[n,n,p-1],[a+p]]
                  :    - duplicate top of Main     [[n,n,p-1,p-1],[a+p]]
                   ""  - no-ops                    [[n,n,p-1,p-1],[a+p]]
                     } - place top of Main onto Aux    [[n,n,p-1],[a+p,p-1]]

At this point if p-1 is still non-zero we turn left:

"%                                                           Main,Aux
"  - no-op                                             [[n,n,p-1],[a+p,p-1]]
 % - modulo                                          [[n,n%(p-1)],[a+p,p-1]]
   - ...and we branch again according to the divisibility
   -    of n by our new potential divisor, p-1

...but if p-1 hit zero we go straight up to the : on the second line of the labyrinth (you've seen all the instructions before, so I'm leaving their descriptions out and just giving their effect):

:":}"":({):""}"%;:{:{+}}"""""""{{{                           Main,Aux
:                                  -                   [[n,n,0,0],[a,0]]
 "                                 -                   [[n,n,0,0],[a,0]]
                                   - top of Main is zero so we go straight
                                   -  ...but we hit the wall and so turn around
  :                                -                 [[n,n,0,0,0],[a,0]]
   }                               -                   [[n,n,0,0],[a,0,0]]
                                   - top of Main is zero so we go straight
    ""                             -                   [[n,n,0,0],[a,0,0]]
      :                            -                 [[n,n,0,0,0],[a,0,0]]
       (                           -                [[n,n,0,0,-1],[a,0,0]]
        {                          -              [[n,n,0,0,-1,0],[a,0]]
                                   - top of Main is zero so we go straight
                                   -  ...but we hit the wall and so turn around
         (                         -             [[n,n,0,0,-1,-1],[a,0]]
          :                        -          [[n,n,0,0,-1,-1,-1],[a,0]]
           ""                      -          [[n,n,0,0,-1,-1,-1],[a,0]]
             }                     -             [[n,n,0,0,-1,-1],[a,0,-1]]
                                   - top of Main is non-zero so we turn left
              "                    -             [[n,n,0,0,-1,-1],[a,0,-1]]
               %                   - (-1)%(-1)=0     [[n,n,0,0,0],[a,0,-1]]
                ;                  -                   [[n,n,0,0],[a,0,-1]]
                 :                 -                 [[n,n,0,0,0],[a,0,-1]]
                  {                -              [[n,n,0,0,0,-1],[a,0]]
                   :               -           [[n,n,0,0,0,-1,-1],[a,0]]
                    {              -         [[n,n,0,0,0,-1,-1,0],[a]]
                     +             -           [[n,n,0,0,0,-1,-1],[a]]
                      }            -              [[n,n,0,0,0,-1],[a,-1]]
                       }           -                 [[n,n,0,0,0],[a,-1,-1]]
                        """""""    -                 [[n,n,0,0,0],[a,-1,-1]]
                                   - top of Main is zero so we go straight
                               {   -              [[n,n,0,0,0,-1],[a,-1]]
                                {  -           [[n,n,0,0,0,-1,-1],[a]]
                                 { -         [[n,n,0,0,0,-1,-1,a],[]]

Now this { has three neighbouring instructions, so...

...if a is zero, which it will be for perfect n, then we go straight:

"!@                                                          Main,Aux
"   -                                        [[n,n,0,0,0,-1,-1,a],[]]
    - top of Main is a, which is zero, so we go straight
 !  - print top of Main, which is a, which is a 0
  @ - exit the labyrinth

...if a is non-zero, which it will be for non-perfect n, then we turn left:

_~"!@                                                        Main,Aux
_     - place a zero onto Main             [[n,n,0,0,0,-1,-1,a,0],[]]
 ~    - bitwise NOT top of Main (=-1-x)   [[n,n,0,0,0,-1,-1,a,-1],[]]
  "   -                                   [[n,n,0,0,0,-1,-1,a,-1],[]]
      - top of Main is NEGATIVE so we turn left
   !  - print top of Main, which is -1
    @ - exit the labyrinth

Ruby, 33 bytes


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TI-BASIC (TI-84), 30 23 bytes


Horribly inefficient, but it works.
Reducing the bytecount sped up the program by a lot.
Input is in Ans.
Output is in Ans and is automatically printed out when the program completes.

(TI-BASIC doesn't have comments, so just assume that ; makes a comment)

:2Ans=sum(seq(Ans/Xnot(remainder(Ans,X)),X,1,Ans    ;Full program

 2Ans                                               ;double the input
          seq(                                      ;generate a list
                                         X,          ;using the variable X,
                                           1,        ;starting at 1,
                                             Ans     ;and ending at the input
                                                     ;with an implied increment of 1
              Ans/X                                 ;from the input divided by X
                   not(                ),           ;multiplied by the negated result of
                       remainder(Ans,X)              ;the input modulo X
                                                     ;(result: 0 or 1)
      sum(                                          ;sum up the elements in the list
     =                                              ;equal?



Note: The byte count of a program is evaluated using the value in [MEM]>[2]>[7] (36 bytes) then subtracting the length of the program's name, CDGF2, (5 bytes) and an extra 8 bytes used for storing the program:

36 - 5 - 8 = 23 bytes


Java (JDK), 54 bytes

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

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Though for a strict number by number matching, the following will return the same values, but is only 40 bytes.


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  • \$\begingroup\$ The rules say Your program doesn't have to complete the larger test cases, if there's memory or time constraints, but it should be theoretically able to if it were given more memory/time. \$\endgroup\$ – Jo King Mar 13 '19 at 14:28
  • \$\begingroup\$ @JoKing Does that mean that I can't use a Java int at all, but rather a BigInteger? Because Java has BigIntegers, but it won't ever have an int that's more than 31 bits as signed, which can't hold any other value than those represented here... \$\endgroup\$ – Olivier Grégoire Mar 13 '19 at 14:30
  • \$\begingroup\$ no, but if the program should still work if the int type was unbounded \$\endgroup\$ – Jo King Mar 13 '19 at 21:16
  • 1
    \$\begingroup\$ @JoKing Ok, I switched the two solutions again to have the computation first. \$\endgroup\$ – Olivier Grégoire Mar 13 '19 at 22:35

x86 Assembly, 45 43 Bytes.

6A 00 31 C9 31 D2 41 39  C1 7D 0B 50 F7 F9 58 85
D2 75 F1 51 EB EE 31 D2  59 01 CA 85 C9 75 F9 39
D0 75 05 31 C0 40 EB 02  31 C0 C3

Explaination (Intel Syntax):

PUSH $0          ; Terminator for later
XOR ECX, ECX        ; Clear ECX
    XOR EDX, EDX    ; Clear EDX
    CMP ECX, EAX    ; divisor >= input number?
    JGE .factordone ; if so, exit loop.
    PUSH EAX        ; backup EAX
    IDIV ECX        ; divide EDX:EAX by ECX, store result in EAX and remainder in EDX
    POP EAX         ; restore EAX
    TEST EDX, EDX   ; remainder == 0?
    JNZ .factor     ; if not, jump back to loop start
    PUSH ECX        ; push factor
    JMP .factor     ; jump back to loop start
XOR EDX, EDX        ; clear EDX
    POP ECX         ; pop divisor
    ADD EDX, ECX    ; sum into EDX
    TEST ECX, ECX   ; divisor == 0?
    JNZ .sum        ; if not, loop.
CMP EAX, EDX        ; input number == sum?
JNE .noteq          ; if not, skip to .noteq
    XOR EAX, EAX    ; clear EAX
    INC EAX         ; increment EAX (sets to 1)
JMP .return         ; skip to .return
    XOR EAX, EAX    ; clear EAX

Input should be provided in EAX.
Function sets EAX to 1 for perfect and to 0 for imperfect.

EDIT: Reduced Byte-Count by two by replacing MOV EAX, $1 with XOR EAX, EAX and INC EAX

  • 1
    \$\begingroup\$ I use a macro assembly so I don't know for sure but the comment"; divisor > input number" for me would be "; divisor >= input number" \$\endgroup\$ – user58988 Mar 14 '19 at 16:54
  • \$\begingroup\$ Assembly has easy operations one could reduce instructions length puts all in a line, use indentation and comment every 10 20 asm instructions.... \$\endgroup\$ – user58988 Mar 14 '19 at 16:58
  • \$\begingroup\$ @RosLuP I've fixed the comment in the code (thanks), but I don't know what you mean with your second comment. \$\endgroup\$ – Fayti1703 Mar 14 '19 at 19:23

CJam, 17 bytes


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Javascript, 62


Explanation (although it's pretty simple)

n=> //return function that takes n
  n== //and returns if n is equal to
    [...Array(n).keys()] //an array [0..(n-1)]...
      .filter(a=>n%a<1) //where all of the elements that are not divisors of n are taken out...
      .reduce((a,b)=>a+b) //summed up

Thanks to Jo King for the improvement!


05AB1E, 4 bytes


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  O    # the sum
Ñ      # of the divisors of the input
 ¨     # with the last one removed
   Q   # equals the input

Powershell, 46 bytes 43 bytes


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Edit: -3 bytes thanks to @AdmBorkBork

  • \$\begingroup\$ 43 bytes by rolling the accumulator into the loop and checking against 2*$i to eliminate the subtract-one parens. \$\endgroup\$ – AdmBorkBork Mar 12 '19 at 12:42

C (gcc), 41 bytes


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1: 0
12: 0
13: 0
18: 0
20: 0
1000: 0
33550335: 0
6: 1
28: 1
496: 1
8128: 1
33550336: 1
-65536: 0 <---- Unable to represent final test case with four bytes, fails

Let me know if that failure for the final case is an issue.

  • 1
    \$\begingroup\$ 41 bytes \$\endgroup\$ – tsh Mar 12 '19 at 9:00
  • 2
    \$\begingroup\$ "Output can be two distinct and consistent values through any allowed output format." You're not returning any two distinct values. \$\endgroup\$ – Olivier Grégoire Mar 12 '19 at 9:29
  • 2
    \$\begingroup\$ @OlivierGrégoire Fortunately that can be readily fixed by replacing the space with an exclamation mark! \$\endgroup\$ – Neil Mar 12 '19 at 9:33
  • 1
    \$\begingroup\$ @Neil Better yet, it can be fixed with n=!s; instead of return!s; to save 5 bytes. \$\endgroup\$ – user77406 Mar 12 '19 at 9:50
  • \$\begingroup\$ @OlivierGrégoire ahh, I forgot that point. Also, updated the with the improved code. I tried something similar, but the idiot I am I did s=s which more than likely got optimized out. \$\endgroup\$ – Marcos Mar 13 '19 at 0:26

Smalltalk, 34 bytes


Forth (gforth), 45 bytes

: f 0 over 1 ?do over i mod 0= i * - loop = ;

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Loops over every number from 1 to n-1, summing all values that divide n perfectly. Returns true if sum equals n

Code Explanation

: f                \ start word definition
  0 over 1         \ create a value to hold the sum and setup the bounds of the loop
  ?do              \ start a counted loop from 1 to n. (?do skips if start = end)
    over           \ copy n to the top of the stack
    i mod 0=       \ check if i divides n perfectly
    i * -          \ if so, use the fact that -1 = true in forth to add i to the sum
  loop             \ end the counted loop
  =                \ check if the sum and n are equal
;                  \ end the word definition

Pyth, 9 13 bytes


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Thank you to the commentors for the golf help

Finds all the factors of the input, sums them, and compares that to the original input.

  • \$\begingroup\$ A few golfs for you - q0 can be replaced with !, and SQ produces the range [1-Q], so the range [1-Q) can be generated using StQ. As the Qs are now at the end of the program they can both be omitted. Fettled version, 9 bytes - qsf!%QTSt \$\endgroup\$ – Sok Mar 12 '19 at 10:35

VDM-SL, 101 bytes

f(i)==s([x|x in set {1,...,i-1}&i mod x=0])=i;s:seq of nat+>nat
s(x)==if x=[]then 0 else hd x+s(tl x) 

Summing in VDM is not built in, so I need to define a function to do this across the sequences, this ends up taking up the majority of bytes

A full program to run might look like this:

f(i)==s([x|x in set {1,...,i-1}&i mod x=0])=i;s:seq of nat+>nat
s(x)==if x=[]then 0 else hd x+s(tl x)

Batch, 81 bytes

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

Takes n as a command-line parameter and outputs 1 if it is a perfect number. Brute force method, starts the sum at -n so that it can include n itself in the loop.


Charcoal, 13 bytes


Try it online! Link is to verbose version of code. Outputs - for perfect numbers. Uses brute force. Explanation:

Nθ              Numeric input
     Φθ         Filter on implicit range
        ι       Current value (is non-zero)
       ∧        Logical And
           θ    Input value
          ﹪     Modulo
            ι   Current value
         ¬      Is zero
    Σ           Sum of matching values
  ⁼             Equals
   θ            Input value

Wolfram Language (Mathematica), 14 bytes


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  • \$\begingroup\$ Yes, Mathematica! Another built-in. \$\endgroup\$ – tsh Mar 12 '19 at 10:21

Pyth, 8 bytes


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qs{*MPyPQQ   Implicit: Q=eval(input())
             Trailing QQ inferred
       PQ    Prime factors of Q
      y      Powerset
     P       Remove last element - this will always be the full prime factorisation
   *M        Take product of each
  {          Deduplicate
 s           Sum
q        Q   Is the above equal to Q? Implicit print

Retina 0.8.2, 44 bytes


Try it online! Uses brute force, so link only includes the faster test cases. Explanation:


Convert to unary.


Match all factors of the input. This uses overlapping mode, which in Retina 0.8.2 requires all of the matches to start at different positions, so the matches are actually returned in descending order, starting with the original input.


Subtract the proper factors from the input.


Test whether the result is zero.


Java 8, 66 bytes

Someone has to use the stream API at some point, even if there's a shorter way to do it


Try it online!


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