diff --git a/exercises-hello/hello.py b/exercises-hello/hello.py
index 142f68d..6697b3b 100644
--- a/exercises-hello/hello.py
+++ b/exercises-hello/hello.py
@@ -9,3 +9,4 @@
 #
 # TODO: write your code below
 
+print "hello world"
\ No newline at end of file
diff --git a/exercises-hello/script.py b/exercises-hello/script.py
new file mode 100755
index 0000000..1b9f121
--- /dev/null
+++ b/exercises-hello/script.py
@@ -0,0 +1,2 @@
+#!/usr/bin/env python
+print "this is a python script!"
\ No newline at end of file
diff --git a/exercises-more/exercises.py b/exercises-more/exercises.py
index 3420fff..7d81d40 100644
--- a/exercises-more/exercises.py
+++ b/exercises-more/exercises.py
@@ -2,62 +2,83 @@
 # Return the number of words in the string s. Words are separated by spaces.
 # e.g. num_words("abc def") == 2
 def num_words(s):
-    return 0
+    return len(s.split());
 
 # PROB 2
 # Return the sum of all the numbers in lst. If lst is empty, return 0.
 def sum_list(lst):
-    return 0
+    return sum(lst)
 
 # PROB 3
 # Return True if x is in lst, otherwise return False.
 def appears_in_list(x, lst):
-    return False
+    return x in lst
 
 # PROB 4
 # Return the number of unique strings in lst.
 # e.g. num_unique(["a", "b", "a", "c", "a"]) == 3
 def num_unique(lst):
-    return 0
+    unique = set(lst)
+    return len(unique)
 
 # PROB 5
 # Return a new list, where the contents of the new list are lst in reverse order.
 # e.g. reverse_list([3, 2, 1]) == [1, 2, 3]
 def reverse_list(lst):
-    return []
+    # http://stackoverflow.com/questions/3940128/how-can-i-reverse-a-list-in-python
+    return lst[::-1]
 
 # PROB 6
 # Return a new list containing the elements of lst in sorted decreasing order.
 # e.g. sort_reverse([5, 7, 6, 8]) == [8, 7, 6, 5]
 def sort_reverse(lst):
-    return []
+    # http://stackoverflow.com/questions/11750469/python-list-sort-doesnt-seem-to-work
+    return reverse_list( sorted(lst, key=int) )
 
 # PROB 7
 # Return a new string containing the same contents of s, but with all the
 # vowels (upper and lower case) removed. Vowels do not include 'y'
 # e.g. remove_vowels("abcdeABCDE") == "bcdBCD"
 def remove_vowels(s):
+    # http://stackoverflow.com/questions/3939361/remove-specific-characters-from-a-string-in-python
+    vowels = "aeouiAEOUI"
+    for char in vowels:
+        s = s.replace(char, "")
     return s
 
 # PROB 8
 # Return the longest word in the lst. If the lst is empty, return None.
 # e.g. longest_word(["a", "aaaaaa", "aaa", "aaaa"]) == "aaaaaa"
 def longest_word(lst):
-    return None
+    word = None
+    for item in lst:
+        if word == None or len(item) > len(word):
+            word = item 
+    return word
 
 # PROB 9
 # Return a dictionary, mapping each word to the number of times the word
 # appears in lst.
 # e.g. word_frequency(["a", "a", "aaa", "b", "b", "b"]) == {"a": 2, "aaa": 1, "b": 3}
 def word_frequency(lst):
-    return {}
+    # http://stackoverflow.com/questions/2161752/how-to-count-the-frequency-of-the-elements-in-a-list
+    word_freq = {}
+    unique = set(lst)
+    for element in unique:
+        word_freq[element] = lst.count(element)
+    return word_freq
 
 # PROB 10
 # Return the tuple (word, count) for the word that appears the most frequently
 # in the list, and the number of times the word appears. If the list is empty, return None.
 # e.g. most_frequent_word(["a", "a", "aaa", "b", "b", "b"]) == ("b", 3)
 def most_frequent_word(lst):
-    return None
+    word = None
+    unique = set(lst)
+    for element in unique:
+        if word == None or lst.count(element) > lst.count(word[0]):   
+            word = (element, lst.count(element))
+    return word
 
 # PROB 11
 # Compares the two lists and finds all the positions that are mismatched in the list.
@@ -65,10 +86,18 @@ def most_frequent_word(lst):
 # mismatched positions in the list.
 # e.g. find_mismatch(["a", "b", "c", "d", "e"], ["f", "b", "c", "g", "e"]) == [0, 3]
 def find_mismatch(lst1, lst2):
-    return []
+    mismatches = []
+    for i in range(len(lst1)):
+        if lst1[i] != lst2[i]:
+            mismatches.append(i)
+    return mismatches
 
 # PROB 12
 # Returns the list of words that are in word_list but not in vocab_list.
 def spell_checker(vocab_list, word_list):
-    return []
+    words = []
+    for item in word_list:
+        if (item in vocab_list) == False:
+            words.append(item)
+    return words
 
diff --git a/exercises-spellchecker/dictionary.py b/exercises-spellchecker/dictionary.py
index e71878a..fea22c3 100644
--- a/exercises-spellchecker/dictionary.py
+++ b/exercises-spellchecker/dictionary.py
@@ -16,22 +16,35 @@ def load(dictionary_name):
     Each line in the file contains exactly one word.
     """
     # TODO: remove the pass line and write your own code
-    pass
+    
+    # Get an array of the file content
+    with open(dictionary_name) as f:
+        content = f.readlines()
+
+    # Make an empty set
+    dictionary = set([])
+
+    # Cleaning each word and adding them to the set
+    for word in content:
+        dictionary.add( word.strip().strip('\n') )
+
+    return dictionary
+
 
 def check(dictionary, word):
     """
     Returns True if `word` is in the English `dictionary`.
     """
-    pass
+    return (word in dictionary)
 
 def size(dictionary):
     """
     Returns the number of words in the English `dictionary`.
     """
-    pass
+    return len(dictionary)
 
 def unload(dictionary):
     """
     Removes everything from the English `dictionary`.
     """
-    pass
+    return set([])