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diff --git a/challenge-069/paulo-custodio/python/ch-1.py b/challenge-069/paulo-custodio/python/ch-1.py
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+#!/usr/bin/env python3
+
+# Challenge 069
+#
+# TASK #1 > Strobogrammatic Number
+# Submitted by: Mohammad S Anwar
+# A strobogrammatic number is a number that looks the same when looked at upside
+# down.
+#
+# You are given two positive numbers $A and $B such that 1 <= $A <= $B <= 10^15.
+#
+# Write a script to print all strobogrammatic numbers between the given two
+# numbers.
+#
+# Example
+# Input: $A = 50, $B = 100
+#     Output: 69, 88, 96
+
+import sys
+
+A, B = map(int, sys.argv[1:3])
+out = []
+
+def is_strobogrammatic(n):
+    if not all(c in '0689' for c in str(n)):
+        return False
+    inv = str(n).translate(str.maketrans('69', '96'))
+    return int(inv[::-1]) == n
+
+for n in range(A, B + 1):
+    if is_strobogrammatic(n):
+        out.append(n)
+
+print(", ".join(map(str, out)))
diff --git a/challenge-069/paulo-custodio/python/ch-2.py b/challenge-069/paulo-custodio/python/ch-2.py
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+#!/usr/bin/env python3
+
+# Challenge 069
+#
+# TASK #2 > 0/1 String
+# Submitted by: Mohammad S Anwar
+# A 0/1 string is a string in which every character is either 0 or 1.
+#
+# Write a script to perform switch and reverse to generate S30 as described
+# below:
+#
+# switch:
+#
+# Every 0 becomes 1 and every 1 becomes 0. For example, "101" becomes "010".
+#
+# reverse:
+#
+# The string is reversed. For example, "001" becomes "100".
+# UPDATE (2020-07-13 17:00:00):
+# It was brought to my notice that generating S1000 string would be nearly
+# impossible. So I have decided to lower it down to S30. Please follow the rule
+# as below:
+#
+# S0 = ""
+# S1 = "0"
+# S2 = "001"
+# S3 = "0010011"
+#
+# SN = SN-1 + "0" + switch(reverse(SN-1))
+
+# Note: modified to S20, as S30 was taking forever
+
+N = 20
+
+def bits_switch(s):
+    return s.translate(str.maketrans('01', '10'))
+
+def bits_reverse(s):
+    return ''.join(reversed(s))
+
+prev = ""
+s = ""
+for _ in range(1, N + 1):
+    s = prev + "0" + bits_switch(bits_reverse(prev))
+    prev = s
+
+print(s)