Finished 23, pretty darn slow and ugly solution

22/main.py

@@ -2,6 +2,9 @@ import os
 import math
 import numpy as np
 
+# Calculate the sum of all scores of the names in the textfile, where 
+# the score is the product of the sorted position and 
+# sum of alphabetic values of its letters
 
 def charScore(c):
 	assert len(c) == 1

23/main.py

@@ -0,0 +1,60 @@
+import os
+import math
+import numpy as np
+
+# Find the sum of all positive integers which cannot be written as the sum
+# of two abundant numbers. 
+
+# A number is abundant if its properdivisors sum to 
+# more than the original number, e.g. 12 is abundant, since 12 can be divided
+# by 1, 2, 3, 4, 6 = 16
+
+# It can be proven that all numbers above 28123 can be written as the sum
+# of 2 abundant numbers
+
+def getDivisors(n):
+	divs = [True] * math.ceil(n/2)
+
+	for x in range(1, len(divs)):
+		if divs[x] and n % (x + 1) != 0:
+			y = x
+			while y < len(divs):
+				divs[y] = False
+				y = y + x + 1
+
+	res = []
+	for x in range(len(divs)):
+		if divs[x]:
+			res.append(x + 1)
+	return res
+
+def isAbundant(n):
+	return sum(getDivisors(n)) > n
+
+
+def main():
+	print("Hello, this is Patrick")
+
+	m = 28123
+
+	a = list(filter(isAbundant, range(1, m+1)))
+
+	# We are assuming that the abundants are sorted
+	sums = set([])
+	for i in range(len(a)):
+		for j in range(i, len(a)):
+			s = a[i] + a[j]
+			if s > m:
+				break
+			else:
+				#print(s)
+				sums.add(s)
+
+	nonsums = set(range(1, m+1))
+	nonsums.difference_update(sums)
+	# print(nonsums)
+	print(sum(nonsums))
+
+
+if __name__ == "__main__":
+	main()