Given the lengths of two different tracks, output how many laps it will take until two joggers meet again

number_of_labs.rb

# lcm => least common multiple

# one-liner
def nbr_of_laps(x,y)
  [x.lcm(y)/x, x.lcm(y)/y]
end

# simple
def nbr_of_laps(x, y)
  lcm = x.lcm(y)
  [lcm/x, lcm/y]
end

# alt
def nbr_of_laps(x,y)
  l = [x,y].reduce(:lcm)
  [l/x,l/y]
end