Slightly prettier

README.md

@@ -7,3 +7,8 @@ Some of the code will be somewhat quick and dirty when I first upload it, since
 For the first challenge, part one, one had to review a sequence of digits and find the `sum` of all the digits, which match the next digit in the list. This list is circular, thus the last digit will be able to match with the first digit, such that $`1221`$ will have a `sum` of $`3`$.
 
 For the second part, one now had to, instead of considering the next digit in the list, consider the digit *halfway around* the circular list, so $`1212`$ would produce a `sum` of $`6`$.
+
+## The 2nd of December
+For the second challenge, part one, the objective was to look through a bunch of lists, find the maximum and minimum value and then subtract these for each list. In the end, the results of the different subtractions had to be summed up and this would be the result.
+
+For the second part, the calculation of the checksum had changed to now being calculated from finding the only pair of evenly divisible numbers from each list and then sum up the results of the divisions.

checksum.rb

@@ -1,3 +1,9 @@
+class Integer
+  def is_lcm?(other)
+    self.lcm(other) == self and other != self
+  end
+end
+
 a = "515	912	619	2043	96	93	2242	1385	2110	860	2255	621	1480	118	1230	99
 161	6142	142	1742	237	6969	211	4314	5410	4413	3216	6330	261	3929	5552	109
 1956	4470	3577	619	105	3996	128	1666	720	4052	108	132	2652	306	1892	1869
@@ -17,11 +23,11 @@ a = "515	912	619	2043	96	93	2242	1385	2110	860	2255	621	1480	118	1230	99
 
 splitted = a.split("\n").map! {|i| i.split("\t").map(&:to_i)}
 puts splitted.reduce(0) {|sum, row| sum += row.max-row.min}
-
+    
 result = 0
 splitted.each do |row|
   row.permutation(2).to_a.find do |ele|
-    if (ele[0].lcm(ele[1]) == ele[0]) then
+    if ele[0].is_lcm?(ele[1]) then
       result += ele[0] / ele[1]
     end
   end
@@ -30,3 +36,4 @@ end
 puts result
 
 
+