How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Elixir programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Elixir programming language

Table of Contents

Problem Statement

These define three classifications of positive integers based on their   proper divisors. Let   P(n)   be the sum of the proper divisors of   n   where the proper divisors are all positive divisors of   n   other than   n   itself.

6   has proper divisors of   1,   2,   and   3. 1 + 2 + 3 = 6,   so   6   is classed as a perfect number.

Calculate how many of the integers   1   to   20,000   (inclusive) are in each of the three classes. Show the results here.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Abundant, deficient and perfect number classifications step by step in the Elixir programming language

Source code in the elixir programming language

defmodule Proper do
  def divisors(1), do: []
  def divisors(n), do: [1 | divisors(2,n,:math.sqrt(n))] |> Enum.sort
  
  defp divisors(k,_n,q) when k>q, do: []
  defp divisors(k,n,q) when rem(n,k)>0, do: divisors(k+1,n,q)
  defp divisors(k,n,q) when k * k == n, do: [k | divisors(k+1,n,q)]
  defp divisors(k,n,q)                , do: [k,div(n,k) | divisors(k+1,n,q)]
end

{abundant, deficient, perfect} = Enum.reduce(1..20000, {0,0,0}, fn n,{a, d, p} ->
  sum = Proper.divisors(n) |> Enum.sum
  cond do
    n < sum -> {a+1, d, p}
    n > sum -> {a, d+1, p}
    true    -> {a, d, p+1}
  end
end)
IO.puts "Deficient: #{deficient}   Perfect: #{perfect}   Abundant: #{abundant}"


  

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