From 0f78cd6a55083c62ceec241b6ec22d3e3d056d7b Mon Sep 17 00:00:00 2001
From: Thejus-Paul <thejuspaul@protonmail.ch>
Date: Sun, 12 Nov 2017 23:56:18 +0530
Subject: [PATCH] Project Euler Solution Added

---
 Project Euler/Problem 14/sol1.py | 20 ++++++++++++++++++++
 Project Euler/README.md          |  7 +++++++
 2 files changed, 27 insertions(+)
 create mode 100644 Project Euler/Problem 14/sol1.py

diff --git a/Project Euler/Problem 14/sol1.py b/Project Euler/Problem 14/sol1.py
new file mode 100644
index 000000000..c5050b561
--- /dev/null
+++ b/Project Euler/Problem 14/sol1.py	
@@ -0,0 +1,20 @@
+largest_number = 0
+pre_counter = 0
+
+for input1 in range(750000,1000000):
+    counter = 1
+    number = input1
+
+    while number > 1:
+        if number % 2 == 0:
+            number /=2
+            counter += 1
+        else:
+            number = (3*number)+1
+            counter += 1
+
+    if counter > pre_counter:
+        largest_number = input1
+        pre_counter = counter
+
+print('Largest Number:',largest_number,'->',pre_counter,'digits')
diff --git a/Project Euler/README.md b/Project Euler/README.md
index 7d8bd2d8d..5d7238e40 100644
--- a/Project Euler/README.md	
+++ b/Project Euler/README.md	
@@ -42,3 +42,10 @@ PROBLEMS:
     a^2 + b^2 = c^2
     There exists exactly one Pythagorean triplet for which a + b + c = 1000.
     Find the product abc.
+
+14. 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
+    Which starting number, under one million, produces the longest chain?