Project EulerのSmallest multipleをElixirで解く ref: http://qiita.com/y-temp4/items/865b17837a4fb9eaf86f

file0.txt

LCM = \frac{a \times b}{GCD}

file1.txt

defmodule Problem5 do
  def gcd(a, b) do
    if rem(a, b) == 0 do
      b
    else
      gcd(b, rem(a, b))
    end
  end
end

file2.txt

IO.puts Problem5.gcd(824, 128) # 8

file3.txt

defmodule Problem5 do
  def gcd(a, b) do
    if rem(a, b) == 0 do
      b
    else
      gcd(b, rem(a, b))
    end
  end

  def solve(n, lcm, max) do
    if n == max do
      lcm
    else
      lcm_next = div(n * lcm, gcd(n, lcm))
      solve(n+1, lcm_next, max)
    end
  end
end

file4.txt

iex(1)> 6 / 2
3.0
iex(2)> div(6, 2)
3

file5.txt

IO.puts Problem5.solve(2, 1, 10) # 2520
IO.puts Problem5.solve(2, 1, 20) # 232792560

file6.txt

defmodule Problem5 do
  def gcd(x, 0), do: x
  def gcd(x, y), do: gcd(y, rem(x, y))

  def solve(max, lcm, max), do: lcm
  def solve(n, lcm, max) do
    lcm_next = div(n * lcm, gcd(n, lcm))
    solve(n+1, lcm_next, max)
  end
end

file7.txt

defmodule Problem5 do
  def gcd(x, 0), do: x
  def gcd(x, y), do: gcd(y, rem(x, y))

  def solve(max), do: solve(2, 1, max)
  def solve(max, lcm, max), do: div(max * lcm, gcd(max, lcm))
  def solve(n, lcm, max) do
    lcm_next = div(n * lcm, gcd(n, lcm))
    solve(n+1, lcm_next, max)
  end
end

file8.txt

IO.puts Problem5.solve(10) # 2520
IO.puts Problem5.solve(11) # 27720
IO.puts Problem5.solve(20) # 232792560