diff --git a/exercises-hello/hello.py b/exercises-hello/hello.py
index 142f68d..9636f59 100644
--- a/exercises-hello/hello.py
+++ b/exercises-hello/hello.py
@@ -8,4 +8,4 @@
 # Run ./test.sh to make sure your program matches our expected output.
 #
 # TODO: write your code below
-
+print "hello world"
diff --git a/exercises-hello/script.py b/exercises-hello/script.py
new file mode 100755
index 0000000..f88d80d
--- /dev/null
+++ b/exercises-hello/script.py
@@ -0,0 +1,2 @@
+#!/usr/bin/env python
+print "this is a python script!"
diff --git a/exercises-more/exercises.py b/exercises-more/exercises.py
index 3420fff..37540ef 100644
--- a/exercises-more/exercises.py
+++ b/exercises-more/exercises.py
@@ -2,55 +2,70 @@
 # 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)
 
 # PROB 2
 # Return the sum of all the numbers in lst. If lst is empty, return 0.
 def sum_list(lst):
-    return 0
+    if len(lst) > 0:
+        return sum(lst)
+    else:
+        return 0
 
 # 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
+    return len(set(lst))
 
 # 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 []
+    return reversed(lst)
 
 # 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 []
+    return reversed(sorted(lst))
 
 # 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):
+    for letter in s:
+        if letter in ['a','e','i','o','u']:
+            s = s.replace(letter,'')
     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
+    if len(lst) > 0:
+        return max(lst, key=len)
+    else:
+        return None
 
 # 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 {}
+    new_dict = {}
+    for word in lst:
+        if word in new_dict:
+            new_dict[word] += 1
+        else:
+            new_dict[word] = 1
+
 
 # PROB 10
 # Return the tuple (word, count) for the word that appears the most frequently
diff --git a/exercises-spellchecker/dictionary.py b/exercises-spellchecker/dictionary.py
index e71878a..8f6aa66 100644
--- a/exercises-spellchecker/dictionary.py
+++ b/exercises-spellchecker/dictionary.py
@@ -16,22 +16,28 @@ def load(dictionary_name):
     Each line in the file contains exactly one word.
     """
     # TODO: remove the pass line and write your own code
-    pass
+    f = open(dictionary_name, "r")
+    words = set()
+    for line in f:
+        line_stripped = line.strip()
+        words.add(line_stripped)
+    f.close()
+    return words
 
 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
+    dictionary.clear()