diff --git a/Project Euler/Problem 21/sol1.py b/Project Euler/Problem 21/sol1.py
new file mode 100644
index 000000000..6d137a7d4
--- /dev/null
+++ b/Project Euler/Problem 21/sol1.py	
@@ -0,0 +1,42 @@
+#-.- coding: latin-1 -.-
+from __future__ import print_function
+from math import sqrt
+'''
+Amicable Numbers
+Problem 21
+
+Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
+If d(a) = b and d(b) = a, where a ≠ b, then a and b are an amicable pair and each of a and b are called amicable numbers.
+
+For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
+
+Evaluate the sum of all the amicable numbers under 10000.
+'''
+try:
+	xrange		#Python 2
+except NameError:
+	xrange = range	#Python 3
+
+def sum_of_divisors(n):
+	total = 0
+	for i in xrange(1, int(sqrt(n)+1)):
+		if n%i == 0 and i != sqrt(n):
+			total += i + n//i
+		elif i == sqrt(n):
+			total += i
+
+	return total-n
+
+sums = []
+total = 0
+
+for i in xrange(1, 10000):
+	n = sum_of_divisors(i)
+	
+	if n < len(sums):
+		if sums[n-1] == i:
+			total += n + i
+
+	sums.append(n)
+
+print(total)
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diff --git a/Project Euler/Problem 76/sol1.py b/Project Euler/Problem 76/sol1.py
new file mode 100644
index 000000000..2832f6d7a
--- /dev/null
+++ b/Project Euler/Problem 76/sol1.py	
@@ -0,0 +1,35 @@
+from __future__ import print_function
+'''
+Counting Summations
+Problem 76
+
+It is possible to write five as a sum in exactly six different ways:
+
+4 + 1
+3 + 2
+3 + 1 + 1
+2 + 2 + 1
+2 + 1 + 1 + 1
+1 + 1 + 1 + 1 + 1
+
+How many different ways can one hundred be written as a sum of at least two positive integers?
+'''
+try:
+	xrange		#Python 2
+except NameError:
+	xrange = range	#Python 3
+
+def partition(m):
+	memo = [[0 for _ in xrange(m)] for _ in xrange(m+1)]
+	for i in xrange(m+1):
+		memo[i][0] = 1
+
+	for n in xrange(m+1):
+		for k in xrange(1, m):
+			memo[n][k] += memo[n][k-1]
+			if n > k:
+				memo[n][k] += memo[n-k-1][k]
+
+	return (memo[m][m-1] - 1)
+
+print(partition(100))
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