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Message: [QUOTE="Christer Nilsson, post: 4495150"] This was really a good and enjoyable exercise. I learnt a lot. One of the best solutions is made by Leavengood. I would like to point out a small error. 36.calc_divisors returns [1,2,3,4,6,6,9,12,18,36]. 6 is reckoned twice. This seems to have no implication on the result though. def calc_divisors res=[1] 2.upto(Math.sqrt(self).floor) do |i| if self % i == 0 res << i end end res.reverse.each do |i| res << self / i end res end The part I appreciate most is the beautiful and fast handling of the binary set: def sum_in_subset?(a) if self < 0 false elsif a.include?(self) true else if a.length == 1 false else f = a.first remaining = a[1..-1] (self - f).sum_in_subset?(remaining) or sum_in_subset?(remaining) end end end I also learnt that Ruby is really quick once you leave the debugger! Here is my solution. It is short but slow. def divisors(n) (1..n.div(2)).collect {|i| i if n.modulo(i)==0}.compact end def sum(arr) arr.inject(0) {|sum,element| sum+element} end def subset?(n, divisors) arr=[] divisors.each do |i| arr.concat arr.collect {|j| i+j} arr << i arr.uniq! return true if arr.member?(n) end false end def weird(n) return if n.modulo(2)==1 coll = divisors(n) diff = sum(coll)-n return if diff <= 0 return n unless subset?(diff,coll) end def all_weird(n) (1..n).collect {|i| weird(i)}.compact end assert_equal [1,2,3,4,6], divisors(12) assert_equal [1,2,5,7,10,14,35], divisors(70) assert_equal 16, sum(divisors(12)) assert_equal 74, sum(divisors(70)) assert_equal true, subset?(12,divisors(12)) assert_equal false, subset?(70,divisors(70)) assert_equal false, subset?(4,divisors(70)) assert_equal nil, weird(2) assert_equal nil, weird(12) assert_equal nil, weird(20) assert_equal 70, weird(70) assert_equal [70], all_weird(70) assert_equal [70,836], all_weird(836) Christer [/QUOTE]

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