My solution to this is ugly, but like the tar baby the more I mess with it the worse it gets.

I originally tried to find all the 9 digit pandigitals to cycle through, but was able to cut down the processing by a TON after I figured out that there were only 2 possible digit combinations that could satisfy the problem. (x + xxxx = xxxx and xx + xxx = xxxx)

Enjoy!

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

def combo size, current = 0, stack = [], results = {} return results[stack.join.to_i] = stack.clone if size == 0 (1..9).each do |n| next if stack.include?(n) stack[current] = n combo(size - 1, current + 1, stack.clone, results) end return results end def pandigitals a, b, c results = [] $repo[a].each_pair do |a_num, a_arr| $repo[b].each_pair do |b_num, b_arr| product = a_num * b_num if $repo[c].include?(product) c_arr = $repo[c][product] results.push(product) if (a_arr + b_arr + c_arr).uniq.length == 9 end end end return results end $repo = { 1 => combo(1), 2 => combo(2), 3 => combo(3), 4 => combo(4) } results = pandigitals(1, 4, 4) + pandigitals(2, 3, 4) puts results.uniq.inject(:+)