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diff --git a/challenge-285/paulo-custodio/python/ch-1.py b/challenge-285/paulo-custodio/python/ch-1.py
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+#!/usr/bin/env python3
+
+# Challenge 285
+#
+# Task 1: No Connection
+# Submitted by: Mohammad Sajid Anwar
+#
+# You are given a list of routes, @routes.
+#
+# Write a script to find the destination with no further outgoing connection.
+# Example 1
+#
+# Input: @routes = (["B","C"], ["D","B"], ["C","A"])
+# Output: "A"
+#
+# "D" -> "B" -> "C" -> "A".
+# "B" -> "C" -> "A".
+# "C" -> "A".
+# "A".
+#
+# Example 2
+#
+# Input: @routes = (["A","Z"])
+# Output: "Z"
+
+import sys
+
+def endpoints(routes):
+    endp = {}
+    for p in routes:
+        if not p[1] in endp:
+            endp[p[1]] = 1
+    for p in routes:
+        if p[0] in endp:
+            del endp[p[0]]
+    return sorted(list(endp.keys()))
+
+pairs = (" ".join(sys.argv[1:])).split(",")
+routes = [x.split() for x in pairs]
+print(", ".join(endpoints(routes)))
diff --git a/challenge-285/paulo-custodio/python/ch-2.py b/challenge-285/paulo-custodio/python/ch-2.py
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+#!/usr/bin/env python3
+
+# Challenge 285
+#
+# Task 2: Making Change
+# Submitted by: David Ferrone
+#
+# Compute the number of ways to make change for given amount in cents. By using
+# the coins e.g. Penny, Nickel, Dime, Quarter and Half-dollar, in how many
+# distinct ways can the total value equal to the given amount? Order of coin
+# selection does not matter.
+#
+# A penny (P) is equal to 1 cent.
+# A nickel (N) is equal to 5 cents.
+# A dime (D) is equal to 10 cents.
+# A quarter (Q) is equal to 25 cents.
+# A half-dollar (HD) is equal to 50 cents.
+#
+# Example 1
+#
+# Input: $amount = 9
+# Ouput: 2
+#
+# 1: 9P
+# 2: N + 4P
+#
+# Example 2
+#
+# Input: $amount = 15
+# Ouput: 6
+#
+# 1: D + 5P
+# 2: D + N
+# 3: 3N
+# 4: 2N + 5P
+# 5: N + 10P
+# 6: 15P
+#
+# Example 3
+#
+# Input: $amount = 100
+# Ouput: 292
+
+import sys
+
+COINS = [50,25,10,5,1]
+
+def compute_num_ways(amount, coins):
+    while len(coins) > 1 and coins[0] > amount:
+        coins = coins[1:]
+    if len(coins) == 1:
+        return 1
+    count = 0
+    for v in range(0, amount+1, coins[0]):
+        count += compute_num_ways(amount-v, coins[1:])
+    return count
+
+print(compute_num_ways(int(sys.argv[1]), COINS))