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
  • 3
    \$\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
  • 4
    \$\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

48 Answers 48


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.
  • 2
    \$\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

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

Jelly, 3 bytes


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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

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

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

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

Ruby, 33 bytes


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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

Husk, 5 bytes


Yields 1 for a perfect argument and 0 otherwise.

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=¹ΣhḊ - Function: integer, n   e.g. 24
    Ḋ - divisors                    [1,2,3,4,6,8,12,24]
   h  - initial items               [1,2,3,4,6,8,12]
  Σ   - sum                         36
 ¹    - first argument              24
=     - equal?                      0

Actually, 5 bytes


Full program taking input from STDIN which prints 1 for perfect numbers and 0 otherwise.

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;÷Σ½= - full program, implicitly place input, n, onto the stack
;     - duplicate top of stack
 ÷    - divisors (including n)
  Σ   - sum
   ½  - halve
    = - equal?

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

MathGolf, 5 4 bytes


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─       get divisors (includes the number itself)
 ╡      discard from right of list (removes the number itself)
  Σ     sum(list)
   =    pop(a, b), push(a==b)

Since MathGolf returns divisors rather than proper divisors, the solution is 1 byte longer than it would have been in that case.


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

MATL, 5 bytes


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      % Implicitly grab the input as an integer
      %    STACK: { 6 }
E     % Multiply by two
      %    STACK: { 12 }
G     % Grab the input again
      %    STACK: { 12,  6 }
Z\    % Compute all divisors (including itself)
      %    STACK: { 12,  [1, 2, 3, 6] }
s     % Sum up these divisors
      %    STACK: { 12, 12 }
=     % Check that the two elements on the stack are equal
      %    STACK: { 1 }
      % Implicitly display the result

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)

Gaia, 4 bytes


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1 for perfect, 0 for imperfect.

d	| implicit input, n, push divisors
 Σ	| take the sum
  ⁻	| subtract n
   =	| equal to n?


Regex (.NET), 47 bytes


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This is based on my 44 byte abundant numbers regex, with is in turn based on Martin Ender's 45 byte abundant numbers regex. It was trivial to adapt to matching perfect numbers; just two small changes were needed, with a third change made for aesthetic purposes.

# For the purpose of these comments, the input number will be referred to as N.

^(?=                  # Attempt to add up all the divisors of N.
  (x                  # Cycle through all positive values of tail that are less than N,
                      # testing each one to see if it is a divisor of N. Start at N-1.
    (?=               # Do the below operations in a lookahead, so that upon popping back
                      # out of it our position will remaing the same as it is here.
      (x*$)           # \2 = tail, a potential divisor; go to end to that the following
                      # lookbehind can operate on N as a whole.
      (?<=            # Switch to right-to-left evaluation so that we can operate both
                      # on N and the potential divisor \2. This requires variable-length
                      # lookbehind, a .NET feature. Please read these comments in the
                      # order indicated, from [Step 1] to [Step 4].
        (?(^\2+)      # [Step 1] If \2 is a divisor of N, then...
          (           # [Step 2] Add it to \3, the running total sum of divisors:
                      #          \3 = \3 + \2
            \2        # [Step 4] Iff we run out of space here, i.e. iff the sum would
                      #          exceed N at this point, the match will fail, and the
                      #          summing of divisors will be halted. It can't backtrack,
                      #          because everything in the loop is done atomically.
            (?>\3?)   # [Step 3] Since \3 is a nested backref, it will fail to match on
                      #          the first iteration. The "?" accounts for this, making
                      #          it add zero to itself on the first iteration. This must
                      #          be done before adding \2, to ensure there is enough room
                      #          for the "?" not to cause the match to become zero-length
                      #          even if \3 has a value.
  )+                  # Using "+" instead of "*" here is an aesthetic choice, to make it
                      # clear we don't want to match zero. It doesn't actually make a
                      # difference (besides making the regex ever so slightly more
                      # efficient) because \3 would be unset anyway for N=0.
  $                   # We can only reach this point if all proper divisors of N, all the
                      # way down to \2 = 1, been successfully summed into \3.
\3$                   # Require the sum of divisors to be exactly equal to N.

Regex (.NET), 48 bytes


This is based on my 43 byte abundant numbers regex. It was interesting and somewhat tricky to adapt into a fully golfed perfect numbers regex.

                   # No anchor needed, because \1 cannot be initially captured if
                   # matching is started anywhere besides the beginning.
# Sum as many divisors of N as we can (using the current position as the sum),
# without skipping any of them:
    (?=            # Atomic lookahead:
        .*(\1x+$)  # \2 = the smallest number larger than the previous value of \1
                   # (which is identical to the previous value of \2) for which the
                   # following is true. We can avoid using "x+?" because the ".*"
                   # before this implicitly forces values to be tested in increasing
                   # order from the smallest.
        (?<=^\2+)  # Assert that \2 is a divisor of N
    \2             # Add \2 to the running total sum of divisors
    ^x             # This must always be the first step. Add 1 to the sum of divisors,
                   # thus setting \1 = 1 so the next iteration can keep going
    \B             # Exclude 1 as a divisor if N == 1. We don't have to do this for
                   # N > 1, because this loop will never be able to add N to the
                   # already summed divisors (as 1 will always be one of them).
)+$                # Require that the sum equal N exactly. If this fails, the loop
                   # won't be able to find a match by backtracking, because everything
                   # inside the loop is done atomically.
# Assert that there are no more proper divisors of N that haven't already been summed:
(?<!               # Assert that the following, evaluated right-to-left, is not true:
    ^\3+           # [Step 2] is a proper divisor of N
    (x+\2)         # [Step 1] Any value of \3 > \2

Regex (PCRE2), 64 bytes

This is a direct port of the 48 byte .NET regex, emulating variable-length lookbehind using a recursive subroutine call.


Try it on regex101

            ((?<=          # (?3) calls this
                (?=        # Circumvent the constant-width limitation of lookbehinds
                           # in PCRE by using a lookahead inside the lookbehind
                    ^\2+$  # This is the payload of the emulated variable-length
                           # lookbehind, same as the one in the .NET regex
                    (?3)   # Recursive call - this is the only alternative that can
                           # match until we reach the beginning of the string
                .          # Go back one character at a time, trying the above
                           # lookahead for a match each time

It isn't possible to directly port the 47 byte regex to PCRE, because it changes the value of a capture group inside the lookbehind, and upon the return of any subroutine in PCRE, all capture groups are reset to the values they had upon entering the subroutine.

Regex (ECMAScript), 53 bytes – Even perfect numbers

If it is ever proved that there are no odd perfect numbers, the following would be a robust answer, but for now it is noncompeting, as it only matches even perfect numbers. It was written by Grimmy on 2019-03-15.


Try it online!

This uses the fact that every even perfect number is of the form \$2^{n-1} (2^n-1)\$ where \$2^n-1\$ is prime. For \$2^n-1\$ to be prime, it is necessary that \$n\$ itself be prime, so the regex does not need to do a \$log_2\$ test on \$2^n\$ to verify \$n\$ is prime. (I have used \$n\$ here instead of \$p\$ to make that clear.)

^                     # N = tail = input
(?=(x(x*?))(\1\1)+$)  # Assert that the largest odd divisor of N is >= 3 (due to
                      # having a "+" here instead of "*"; this is not necessary, but
                      # speeds up the regex's non-match when N is a a power of 2);
                      # \1 = N / {largest odd divisor of N}
                      #    == {largest power of 2 divisor of N}; \2 = \1 - 1
((x*)(?=\5$))+        # tail = N / {largest power of 2 divisor of N}
                      #      == {largest odd divisor of N}
(?!(xx+)\6+$)         # Assert tail is prime
\1\2$                 # Assert tail == \1*2 - 1; if this fails to match, it will
                      # backtrack into the "((x*)(?=\5$))+" loop (effectively
                      # multiplying by 2 repeatedly), but this will always fail
                      # to match because every subsequent match attempt will be
                      # an even number, and "\1\2$" can only be odd.

CJam, 17 bytes


Try it online!


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


Try it online!


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

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

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