Am I not good enough for you?












4












$begingroup$


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 decision-problem 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.










share|improve this question











$endgroup$












  • $begingroup$
    Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
    $endgroup$
    – Esolanging Fruit
    1 hour ago










  • $begingroup$
    @EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
    $endgroup$
    – Jo King
    1 hour ago










  • $begingroup$
    Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
    $endgroup$
    – Esolanging Fruit
    1 hour ago










  • $begingroup$
    @EsolangingFruit Good point. I've renamed the test cases to Imperfect/Perfect to make it clearer
    $endgroup$
    – Jo King
    1 hour ago
















4












$begingroup$


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 decision-problem 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.










share|improve this question











$endgroup$












  • $begingroup$
    Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
    $endgroup$
    – Esolanging Fruit
    1 hour ago










  • $begingroup$
    @EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
    $endgroup$
    – Jo King
    1 hour ago










  • $begingroup$
    Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
    $endgroup$
    – Esolanging Fruit
    1 hour ago










  • $begingroup$
    @EsolangingFruit Good point. I've renamed the test cases to Imperfect/Perfect to make it clearer
    $endgroup$
    – Jo King
    1 hour ago














4












4








4





$begingroup$


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 decision-problem 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.










share|improve this question











$endgroup$




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 decision-problem 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.







code-golf number decision-problem number-theory factoring






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 3 mins ago







Jo King

















asked 1 hour ago









Jo KingJo King

24.6k357126




24.6k357126












  • $begingroup$
    Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
    $endgroup$
    – Esolanging Fruit
    1 hour ago










  • $begingroup$
    @EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
    $endgroup$
    – Jo King
    1 hour ago










  • $begingroup$
    Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
    $endgroup$
    – Esolanging Fruit
    1 hour ago










  • $begingroup$
    @EsolangingFruit Good point. I've renamed the test cases to Imperfect/Perfect to make it clearer
    $endgroup$
    – Jo King
    1 hour ago


















  • $begingroup$
    Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
    $endgroup$
    – Esolanging Fruit
    1 hour ago










  • $begingroup$
    @EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
    $endgroup$
    – Jo King
    1 hour ago










  • $begingroup$
    Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
    $endgroup$
    – Esolanging Fruit
    1 hour ago










  • $begingroup$
    @EsolangingFruit Good point. I've renamed the test cases to Imperfect/Perfect to make it clearer
    $endgroup$
    – Jo King
    1 hour ago
















$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour ago




$begingroup$
Wait, so truthy is for values that aren't perfect, and falsey is for values that are?
$endgroup$
– Esolanging Fruit
1 hour ago












$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
1 hour ago




$begingroup$
@EsolangingFruit Yes, though the actual output values don't really matter, so you can output true for perfect numbers if you wish
$endgroup$
– Jo King
1 hour ago












$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
1 hour ago




$begingroup$
Fair enough, but wording the challenge as "output whether it is not perfect" makes the test cases slightly confusing if you interpret "truthy" as meaning "values corresponding to true".
$endgroup$
– Esolanging Fruit
1 hour ago












$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to Imperfect/Perfect to make it clearer
$endgroup$
– Jo King
1 hour ago




$begingroup$
@EsolangingFruit Good point. I've renamed the test cases to Imperfect/Perfect to make it clearer
$endgroup$
– Jo King
1 hour ago










7 Answers
7






active

oldest

votes


















3












$begingroup$


Japt -!, 4 bytes



¥â¬x


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



Try it online!






share|improve this answer











$endgroup$





















    2












    $begingroup$


    R, 33 bytes





    !2*(n=scan())-sum(which(!n%%1:n))


    Try it online!



    Returns TRUE for perfect numbers ans FALSE for imperfect ones.






    share|improve this answer









    $endgroup$













    • $begingroup$
      What do the 2 !s in a row get you?
      $endgroup$
      – CT Hall
      55 mins ago










    • $begingroup$
      @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
      34 mins ago





















    1












    $begingroup$


    CJam, 17 bytes



    ri_,(;{1$%!},:+=


    Try it online!






    share|improve this answer









    $endgroup$





















      1












      $begingroup$

      Javascript, 62



      n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)


      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!






      share|improve this answer











      $endgroup$













      • $begingroup$
        thanks! Added that in
        $endgroup$
        – zevee
        1 hour ago



















      1












      $begingroup$


      Python 3, 46 bytes





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


      Try it online!



      Brute force, sums the factors and checks for equality.






      share|improve this answer











      $endgroup$













      • $begingroup$
        Yeah, that was a typo. I'll fix it now.
        $endgroup$
        – Neil A.
        13 mins ago








      • 1




        $begingroup$
        Using the comprehension condition as a mask for your iteration variable would save a byte.
        $endgroup$
        – Jonathan Frech
        8 mins ago



















      1












      $begingroup$


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





      n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2


      Try it online!






      share|improve this answer











      $endgroup$





















        1












        $begingroup$


        Brachylog, 4 bytes



        fk+?


        Try it online!



        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.





        share|improve this answer









        $endgroup$













          Your Answer





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          7 Answers
          7






          active

          oldest

          votes








          7 Answers
          7






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$


          Japt -!, 4 bytes



          ¥â¬x


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



          Try it online!






          share|improve this answer











          $endgroup$


















            3












            $begingroup$


            Japt -!, 4 bytes



            ¥â¬x


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



            Try it online!






            share|improve this answer











            $endgroup$
















              3












              3








              3





              $begingroup$


              Japt -!, 4 bytes



              ¥â¬x


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



              Try it online!






              share|improve this answer











              $endgroup$




              Japt -!, 4 bytes



              ¥â¬x


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



              Try it online!







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited 1 hour ago

























              answered 1 hour ago









              Luis felipe De jesus MunozLuis felipe De jesus Munoz

              5,60821670




              5,60821670























                  2












                  $begingroup$


                  R, 33 bytes





                  !2*(n=scan())-sum(which(!n%%1:n))


                  Try it online!



                  Returns TRUE for perfect numbers ans FALSE for imperfect ones.






                  share|improve this answer









                  $endgroup$













                  • $begingroup$
                    What do the 2 !s in a row get you?
                    $endgroup$
                    – CT Hall
                    55 mins ago










                  • $begingroup$
                    @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
                    34 mins ago


















                  2












                  $begingroup$


                  R, 33 bytes





                  !2*(n=scan())-sum(which(!n%%1:n))


                  Try it online!



                  Returns TRUE for perfect numbers ans FALSE for imperfect ones.






                  share|improve this answer









                  $endgroup$













                  • $begingroup$
                    What do the 2 !s in a row get you?
                    $endgroup$
                    – CT Hall
                    55 mins ago










                  • $begingroup$
                    @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
                    34 mins ago
















                  2












                  2








                  2





                  $begingroup$


                  R, 33 bytes





                  !2*(n=scan())-sum(which(!n%%1:n))


                  Try it online!



                  Returns TRUE for perfect numbers ans FALSE for imperfect ones.






                  share|improve this answer









                  $endgroup$




                  R, 33 bytes





                  !2*(n=scan())-sum(which(!n%%1:n))


                  Try it online!



                  Returns TRUE for perfect numbers ans FALSE for imperfect ones.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 59 mins ago









                  GiuseppeGiuseppe

                  16.8k31052




                  16.8k31052












                  • $begingroup$
                    What do the 2 !s in a row get you?
                    $endgroup$
                    – CT Hall
                    55 mins ago










                  • $begingroup$
                    @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
                    34 mins ago




















                  • $begingroup$
                    What do the 2 !s in a row get you?
                    $endgroup$
                    – CT Hall
                    55 mins ago










                  • $begingroup$
                    @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
                    34 mins ago


















                  $begingroup$
                  What do the 2 !s in a row get you?
                  $endgroup$
                  – CT Hall
                  55 mins ago




                  $begingroup$
                  What do the 2 !s in a row get you?
                  $endgroup$
                  – CT Hall
                  55 mins ago












                  $begingroup$
                  @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
                  34 mins ago






                  $begingroup$
                  @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
                  34 mins ago













                  1












                  $begingroup$


                  CJam, 17 bytes



                  ri_,(;{1$%!},:+=


                  Try it online!






                  share|improve this answer









                  $endgroup$


















                    1












                    $begingroup$


                    CJam, 17 bytes



                    ri_,(;{1$%!},:+=


                    Try it online!






                    share|improve this answer









                    $endgroup$
















                      1












                      1








                      1





                      $begingroup$


                      CJam, 17 bytes



                      ri_,(;{1$%!},:+=


                      Try it online!






                      share|improve this answer









                      $endgroup$




                      CJam, 17 bytes



                      ri_,(;{1$%!},:+=


                      Try it online!







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 1 hour ago









                      Esolanging FruitEsolanging Fruit

                      8,50932674




                      8,50932674























                          1












                          $begingroup$

                          Javascript, 62



                          n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)


                          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!






                          share|improve this answer











                          $endgroup$













                          • $begingroup$
                            thanks! Added that in
                            $endgroup$
                            – zevee
                            1 hour ago
















                          1












                          $begingroup$

                          Javascript, 62



                          n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)


                          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!






                          share|improve this answer











                          $endgroup$













                          • $begingroup$
                            thanks! Added that in
                            $endgroup$
                            – zevee
                            1 hour ago














                          1












                          1








                          1





                          $begingroup$

                          Javascript, 62



                          n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)


                          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!






                          share|improve this answer











                          $endgroup$



                          Javascript, 62



                          n=>n==[...Array(n).keys()].filter(a=>n%a<1).reduce((a,b)=>a+b)


                          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!







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited 1 hour ago

























                          answered 1 hour ago









                          zeveezevee

                          22016




                          22016












                          • $begingroup$
                            thanks! Added that in
                            $endgroup$
                            – zevee
                            1 hour ago


















                          • $begingroup$
                            thanks! Added that in
                            $endgroup$
                            – zevee
                            1 hour ago
















                          $begingroup$
                          thanks! Added that in
                          $endgroup$
                          – zevee
                          1 hour ago




                          $begingroup$
                          thanks! Added that in
                          $endgroup$
                          – zevee
                          1 hour ago











                          1












                          $begingroup$


                          Python 3, 46 bytes





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


                          Try it online!



                          Brute force, sums the factors and checks for equality.






                          share|improve this answer











                          $endgroup$













                          • $begingroup$
                            Yeah, that was a typo. I'll fix it now.
                            $endgroup$
                            – Neil A.
                            13 mins ago








                          • 1




                            $begingroup$
                            Using the comprehension condition as a mask for your iteration variable would save a byte.
                            $endgroup$
                            – Jonathan Frech
                            8 mins ago
















                          1












                          $begingroup$


                          Python 3, 46 bytes





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


                          Try it online!



                          Brute force, sums the factors and checks for equality.






                          share|improve this answer











                          $endgroup$













                          • $begingroup$
                            Yeah, that was a typo. I'll fix it now.
                            $endgroup$
                            – Neil A.
                            13 mins ago








                          • 1




                            $begingroup$
                            Using the comprehension condition as a mask for your iteration variable would save a byte.
                            $endgroup$
                            – Jonathan Frech
                            8 mins ago














                          1












                          1








                          1





                          $begingroup$


                          Python 3, 46 bytes





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


                          Try it online!



                          Brute force, sums the factors and checks for equality.






                          share|improve this answer











                          $endgroup$




                          Python 3, 46 bytes





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


                          Try it online!



                          Brute force, sums the factors and checks for equality.







                          share|improve this answer














                          share|improve this answer



                          share|improve this answer








                          edited 13 mins ago

























                          answered 14 mins ago









                          Neil A.Neil A.

                          1,278120




                          1,278120












                          • $begingroup$
                            Yeah, that was a typo. I'll fix it now.
                            $endgroup$
                            – Neil A.
                            13 mins ago








                          • 1




                            $begingroup$
                            Using the comprehension condition as a mask for your iteration variable would save a byte.
                            $endgroup$
                            – Jonathan Frech
                            8 mins ago


















                          • $begingroup$
                            Yeah, that was a typo. I'll fix it now.
                            $endgroup$
                            – Neil A.
                            13 mins ago








                          • 1




                            $begingroup$
                            Using the comprehension condition as a mask for your iteration variable would save a byte.
                            $endgroup$
                            – Jonathan Frech
                            8 mins ago
















                          $begingroup$
                          Yeah, that was a typo. I'll fix it now.
                          $endgroup$
                          – Neil A.
                          13 mins ago






                          $begingroup$
                          Yeah, that was a typo. I'll fix it now.
                          $endgroup$
                          – Neil A.
                          13 mins ago






                          1




                          1




                          $begingroup$
                          Using the comprehension condition as a mask for your iteration variable would save a byte.
                          $endgroup$
                          – Jonathan Frech
                          8 mins ago




                          $begingroup$
                          Using the comprehension condition as a mask for your iteration variable would save a byte.
                          $endgroup$
                          – Jonathan Frech
                          8 mins ago











                          1












                          $begingroup$


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





                          n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2


                          Try it online!






                          share|improve this answer











                          $endgroup$


















                            1












                            $begingroup$


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





                            n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2


                            Try it online!






                            share|improve this answer











                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$


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





                              n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2


                              Try it online!






                              share|improve this answer











                              $endgroup$




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





                              n=>Enumerable.Range(1,n).Sum(x=>n%x<1?x:0)==n*2


                              Try it online!







                              share|improve this answer














                              share|improve this answer



                              share|improve this answer








                              edited 13 mins ago

























                              answered 18 mins ago









                              Embodiment of IgnoranceEmbodiment of Ignorance

                              1,588124




                              1,588124























                                  1












                                  $begingroup$


                                  Brachylog, 4 bytes



                                  fk+?


                                  Try it online!



                                  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.





                                  share|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$


                                    Brachylog, 4 bytes



                                    fk+?


                                    Try it online!



                                    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.





                                    share|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$


                                      Brachylog, 4 bytes



                                      fk+?


                                      Try it online!



                                      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.





                                      share|improve this answer









                                      $endgroup$




                                      Brachylog, 4 bytes



                                      fk+?


                                      Try it online!



                                      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.






                                      share|improve this answer












                                      share|improve this answer



                                      share|improve this answer










                                      answered 11 mins ago









                                      Unrelated StringUnrelated String

                                      91118




                                      91118






























                                          draft saved

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