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michaelkpfeifer's solution

to Prime Factors in the Elixir Track

Published at Jun 12 2020 · 0 comments
Test suite

Compute the prime factors of a given natural number.

A prime number is only evenly divisible by itself and 1.

Note that 1 is not a prime number.


What are the prime factors of 60?

  • Our first divisor is 2. 2 goes into 60, leaving 30.
  • 2 goes into 30, leaving 15.
    • 2 doesn't go cleanly into 15. So let's move on to our next divisor, 3.
  • 3 goes cleanly into 15, leaving 5.
    • 3 does not go cleanly into 5. The next possible factor is 4.
    • 4 does not go cleanly into 5. The next possible factor is 5.
  • 5 does go cleanly into 5.
  • We're left only with 1, so now, we're done.

Our successful divisors in that computation represent the list of prime factors of 60: 2, 2, 3, and 5.

You can check this yourself:

  • 2 * 2 * 3 * 5
  • = 4 * 15
  • = 60
  • Success!

Running tests

Execute the tests with:

$ mix test

Pending tests

In the test suites, all but the first test have been skipped.

Once you get a test passing, you can unskip the next one by commenting out the relevant @tag :pending with a # symbol.

For example:

# @tag :pending
test "shouting" do
  assert Bob.hey("WATCH OUT!") == "Whoa, chill out!"

Or, you can enable all the tests by commenting out the ExUnit.configure line in the test suite.

# ExUnit.configure exclude: :pending, trace: true

If you're stuck on something, it may help to look at some of the available resources out there where answers might be found.


The Prime Factors Kata by Uncle Bob http://butunclebob.com/ArticleS.UncleBob.ThePrimeFactorsKata

Submitting Incomplete Solutions

It's possible to submit an incomplete solution so you can see how others have completed the exercise.


defmodule PrimeFactorsTest do
  use ExUnit.Case

  # @tag :pending
  test "1" do
    assert PrimeFactors.factors_for(1) == []

  @tag :pending
  test "2" do
    assert PrimeFactors.factors_for(2) == [2]

  @tag :pending
  test "3" do
    assert PrimeFactors.factors_for(3) == [3]

  @tag :pending
  test "4" do
    assert PrimeFactors.factors_for(4) == [2, 2]

  @tag :pending
  test "6" do
    assert PrimeFactors.factors_for(6) == [2, 3]

  @tag :pending
  test "8" do
    assert PrimeFactors.factors_for(8) == [2, 2, 2]

  @tag :pending
  test "9" do
    assert PrimeFactors.factors_for(9) == [3, 3]

  @tag :pending
  test "27" do
    assert PrimeFactors.factors_for(27) == [3, 3, 3]

  @tag :pending
  test "625" do
    assert PrimeFactors.factors_for(625) == [5, 5, 5, 5]

  @tag :pending
  test "901255" do
    assert PrimeFactors.factors_for(901_255) == [5, 17, 23, 461]

  @tag :pending
  test "93819012551" do
    assert PrimeFactors.factors_for(93_819_012_551) == [11, 9539, 894_119]

  @tag :pending
  # @tag timeout: 2000
  # The timeout tag above will set the below test to fail unless it completes
  # in under two sconds. Uncomment it if you want to test the efficiency of your
  # solution.
  test "10000000055" do
    assert PrimeFactors.factors_for(10_000_000_055) == [5, 2_000_000_011]


ExUnit.configure(exclude: :pending, trace: true)
defmodule PrimeFactors do
  @doc """
  Compute the prime factors for 'number'.

  The prime factors are prime numbers that when multiplied give the desired

  The prime factors of 'number' will be ordered lowest to highest.
  @spec factors_for(pos_integer) :: [pos_integer]
  def factors_for(1) do

  def factors_for(number) do
    factors_for(number, 2, [])

  def factors_for(number, lower_bound, primes) when lower_bound * lower_bound > number do
    Enum.reverse([number | primes])

  def factors_for(number, lower_bound, primes) when rem(number, lower_bound) == 0 do
    factors_for(div(number, lower_bound), lower_bound, [lower_bound | primes])

  def factors_for(number, lower_bound, primes) do
    factors_for(number, lower_bound + 1, primes)

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