Feat: Project Euler Problem 14

Author: domix80

DIRECTORY.md

@@ -153,6 +153,8 @@
     * [Sol2](https://github.com/TheAlgorithms/Ruby/blob/master/project_euler/problem_004/sol2.rb)
   * Problem 005
     * [Sol1](https://github.com/TheAlgorithms/Ruby/blob/master/project_euler/problem_005/sol1.rb)
+  * Problem 014
+    * [Sol1](https://github.com/TheAlgorithms/Ruby/blob/master/project_euler/problem_014/sol1.rb)    
   * Problem 020
     * [Sol1](https://github.com/TheAlgorithms/Ruby/blob/master/project_euler/problem_020/sol1.rb)
   * Problem 021

project_euler/problem_014/sol1.rb

@@ -0,0 +1,42 @@
+#Problem 14: https://projecteuler.net/problem=14
+
+#Problem Statement:
+#The following iterative sequence is defined for the set of positive integers:
+#
+#    n → n/2 (n is even)
+#    n → 3n + 1 (n is odd)
+#
+#Using the rule above and starting with 13, we generate the following sequence:
+#
+#    13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
+#
+#It can be seen that this sequence (starting at 13 and finishing at 1) contains
+#10 terms. Although it has not been proved yet (Collatz Problem), it is thought
+#that all starting numbers finish at 1.
+
+#Which starting number, under one million, produces the longest chain?
+
+def solution?()
+  index_best_result = 0
+  for num in 2..1000000
+    index_candidate = 0
+    n = num
+    while n > 1
+      if n%2 == 0
+        n = n / 2
+      else
+        n = (3*n) + 1
+      end
+      index_candidate +=1
+    end
+    if index_best_result < index_candidate
+      index_best_result = index_candidate
+      value = num
+    end
+  end
+  result = value
+end
+
+answer = solution?()
+p answer
+