magic_number.rb

# Find the whole number _n_ that consists of 10 digits
# (e.g. 3451209876) for which the first _k_ digits 
# can be divided by _k_ with `2 <= _k_ <= 9`
#
# Examples:
# The first 2 digits of _n_ can be divided by 2 (`34` => true)
# The first 3 digits of _n_ can be divided by 3 (`345` => true)
# The first 4 digits of _n_ can be divided by 4 (`3451` => false)
# etc.
#
# How many can you find?

def magic_number?(candidate)
  return false if candidate.size < 10
  return false if candidate.join.to_i % 10 != 0
  return false if candidate[0,9].join.to_i % 9 != 0
  return false if candidate[0,8].join.to_i % 8 != 0
  return false if candidate[0,7].join.to_i % 7 != 0
  return false if candidate[0,6].join.to_i % 6 != 0
  return false if candidate[0,5].join.to_i % 5 != 0
  return false if candidate[0,4].join.to_i % 4 != 0
  return false if candidate[0,3].join.to_i % 3 != 0
  return false if candidate[0,2].join.to_i % 2 != 0
  true
end

(0..9).to_a.permutation.each do |candidate|
  puts candidate.join.to_i if magic_number? candidate
end

This was written for a math assignment. I wonder if it can be made more efficient (both in terms of byte size and speed).