This is part of my series on Elixir Learnings, where I share micro posts about everything I'm learning related to Elixir.

div and rem stand for division and remainder. We can get the result of a division between two numbers.

div 10, 2 # 5
div 10, 3 # 3
div 10, 4 # 2
div 10, 5 # 2

And the remaining:

rem 10, 2 # 0
rem 10, 3 # 1
rem 10, 4 # 2
rem 10, 5 # 0

While learning Elixir, I came across an exercise: I needed to build a function that returns the how many boxes of each type (big, medium, and small) I need to use to organize a number of matchsticks. And I also needed to return the remaining matchsticks.

The big box can be filled with 50 matchsticks. The medium: 20. And the small: 5.

To solve this problem, I start with the big box. For a number N, how many big boxes I can use? I can use the div function to get this information: div(N, 50). As I filled the big boxes, I need to get the remaining matchsticks to do the same thing with the medium and small boxes.

big = div(matchsticks, 50)
remaining_after_big = rem(matchsticks, 50)

medium = div(remaining_after_big, 20)
remaining_after_medium = rem(remaining_after_big, 20)

small = div(remaining_after_medium, 5)
remaining_matchsticks = rem(remaining_after_medium, 5)

With this, I have the big, the medium, the small, and the remaining_matchsticks. And I can return this whole information with a map object.

def boxes(matchsticks) do
  big = div(matchsticks, 50)
  remaining_after_big = rem(matchsticks, 50)

  medium = div(remaining_after_big, 20)
  remaining_after_medium = rem(remaining_after_big, 20)

  small = div(remaining_after_medium, 5)
  remaining_matchsticks = rem(remaining_after_medium, 5)

  %{
    big: big,
    medium: medium,
    small: small,
    remaining_matchsticks: remaining_matchsticks
  }
end

I can also build this function inside a module:

defmodule MatchstickFactory do
  def boxes(matchsticks) do
    big = div(matchsticks, 50)
    remaining_after_big = rem(matchsticks, 50)

    medium = div(remaining_after_big, 20)
    remaining_after_medium = rem(remaining_after_big, 20)

    small = div(remaining_after_medium, 5)
    remaining_matchsticks = rem(remaining_after_medium, 5)

    %{
      big: big,
      medium: medium,
      small: small,
      remaining_matchsticks: remaining_matchsticks
    }
  end
end

And test it:

MatchstickFactory.boxes(98) # %{big: 1, medium: 2, remaining_matchsticks: 3, small: 1}
MatchstickFactory.boxes(39) # %{big: 0, medium: 1, remaining_matchsticks: 4, small: 3}