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Background:

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.

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:

Imperfect:
1,12,13,18,20,1000,33550335

Perfect:
6,28,496,8128,33550336,8589869056

Rules

  • 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.
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    \$\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, 2019 at 7:40
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    \$\begingroup\$ Are we allowed to assume there are no odd perfect numbers? \$\endgroup\$
    – Grimmy
    Mar 12, 2019 at 8:56
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    \$\begingroup\$ @Grimy Only if you can prove so. Good luck! (though I'm wondering how that would save bytes) \$\endgroup\$
    – Jo King
    Mar 12, 2019 at 9:15
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    \$\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, 2019 at 9:18
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    \$\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\$ Mar 12, 2019 at 10:15

66 Answers 66

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Raku, 24 bytes

{$_==sum grep $_%%*,^$_}

Try it online!

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Desmos, 45 bytes


f(n)=\{\sum_{N=1}^nN\{\mod(n,N)=0,0\}=2n,0\}

\$f(n)\$ returns \$1\$ if \$n\$ is a perfect number, \$0\$ is \$n\$ is an imperfect number.

Try It On Desmos!

Try It On Desmos! - Prettified

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Alice, 17 bytes

/O\B!d v
@I/-?+&<

Try it online!

Outputs 0 if the number is perfect, non 0 if imperfect

Flattened

/I\B!d&+?-/O@
/I\             Reads the input
   B            Get all the divisors in ascending order
    !           Remove the input number and save it on the tape
     d&+        Sum all the remaining divisors of the stack
        ?       Get the input from the tape
         -      Subtract from the sum
          /O@   Output and exit
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APL (Dyalog/GNU), 18 bytes, ⎕IO=0

{⍵=+/(0=⍵|⍨⍳⍵)/⍳⍵}

Try it online! (The in-browser version of APL has too small a workspace to handle the larger test cases.)

Explanation:

  • {...} creates an anonymous function (dfn), whose argument becomes the parameter . We'll examine the expression inside the braces from right to left.
    1. ⍳N generates a vector of the first N numbers starting with the index origin. Since that is set to 0, it goes up to N-1. For this function to work we only need to go up to N/2, but only accepts integers, and making the division result an integer would take more characters.
    2. / with two vector operands produces a new vector with each element of the right vector repeated a number of times equal to the corresponding element of the left vector. In this case we're going to build a vector of N 0s and 1s with 1s only at the positions corresponding to the proper divisors of N; the result will therefore be a vector containing only those divisors.
    3. The expression inside parentheses builds the selector; going right to left again, we start with another ⍳⍵. (We could have built a DRYer version that didn't need to repeat that, but it would have taken more characters.)
    4. | is the divide-and-take-remainder operator (mod). By itself it takes its operands on the opposite order from e.g. C's %, but we are transposing it with to flip that, which removes the need for another pair of parentheses. ⍵|⍨⍳⍵ divides each number from 0 to ⍵-1 into and returns a vector of the remainders. (Fortunately for this use, 0|N just returns N rather than throwing a division-by-zero error.)
    5. 0= is what it looks like: a simple equality test with 0. Applied to a vector, it returns a vector with 1's wherever the original vector had a 0, and 0's everywhere else - exactly what we want for our selector.
    6. Having built our vector of the proper divisors, we then compute the sum of its elements with +/ (reduction via addition).
    7. Finally, ⍵= returns our desired result: 1 if the sum is equal to itself, 0 otheriwse.
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Scala, 72 71 bytes

saved 1 byte thanks to the comment.

Try it online!

def f(n:Int):Boolean={val F=for(i<-1 until n if n%i<1)yield i;F.sum==n}
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Thunno 2, 4 bytes

FṫS=

Attempt This Online!

Explanation

FṫS=  # Implicit input     ->  28
F     # Factors of input   ->  [1,2,4,7,14,28]
 ṫ    # Without the tail   ->  [1,2,4,7,14]
  S   # Sum the list       ->  28
   =  # Equals the input?  ->  1
      # Implicit output
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