Help in refactoring this method - Ruby

Question

def divisible?(n)
  if n % 1 == 0 &&
    n % 2 == 0 &&
    n % 3 == 0 &&
    n % 4 == 0 &&
    n % 5 == 0 &&
    n % 6 == 0 &&
    n % 7 == 0 &&
    n % 8 == 0 &&
    n % 9 == 0 &&
    n % 10 == 0 &&
    n % 11 == 0 &&
    n % 12 == 0 &&
    n % 13 == 0 &&
    n % 14 == 0 &&
    n % 15 == 0 &&
    n % 16 == 0 &&
    n % 17 == 0 &&
    n % 18 == 0 &&
    n % 19 == 0 &&
    n % 20 == 0
    return true
  else 
    return false
  end
end

Answer 1

Ruby refactor: use Enumerable#all?:

def divisible?(n)
  (1..20).all? { |x| n % x == 0 }
end

Mathematical refactor:calculate the least common multiple of the integers in the range (Integer#lcm is available from Ruby 1.9):

def divisible?(n)
  n % (1..20).reduce(:lcm) == 0
end

This second snippet is, of course, more efficient once you pre-calculate (1..20).reduce(:lcm) only once and store it somewhere.

Answer 2

Note that you're doing this, which is a pretty egrigious use of if/else:

if boolean
  return true
else
  return false
end

If the branches of your if/else are return true and return false, you should just return the condition you're testing! In Ruby, you don't even need return, just let the condition "fall off" the end of the method.

Ignoring the ability to clean up your condition itself, a first pass at cleaning up your function could be:

def divisible?(n)
  n % 1 == 0 &&
  n % 2 == 0 &&
  n % 3 == 0 &&
  n % 4 == 0 &&
  n % 5 == 0 &&
  n % 6 == 0 &&
  n % 7 == 0 &&
  n % 8 == 0 &&
  n % 9 == 0 &&
  n % 10 == 0 &&
  n % 11 == 0 &&
  n % 12 == 0 &&
  n % 13 == 0 &&
  n % 14 == 0 &&
  n % 15 == 0 &&
  n % 16 == 0 &&
  n % 17 == 0 &&
  n % 18 == 0 &&
  n % 19 == 0 &&
  n % 20 == 0
end