Solution to problem 1 part 2 in Python

src/P1.py

@@ -37,13 +37,52 @@
 # to 2020; what do you get if you multiply them together?
 
 with open("files/P1.txt", "r") as f:
-    numbers = [int(l) for l in f.read().strip().split("\n")]
+    numbers = [int(number) for number in f.read().strip().split("\n")]
 
-for n in numbers:
-    for i in range(len(numbers)):
-        if n + numbers[i] == 2020:
-            print(f"{n} times {numbers[i]} is {n*numbers[i]}")
+
+def part_1() -> None:
+    for n in numbers:
+        for i in range(len(numbers)):
+            if n + numbers[i] == 2020:
+                print(f"{n} times {numbers[i]} is {n*numbers[i]}")
+                break
+        else:
+            continue
+        break
+
+
+# --- Part Two ---
+
+# The Elves in accounting are thankful for your help; one of them even offers
+# you a starfish coin they had left over from a past vacation. They offer you
+# a second one if you can find three numbers in your expense report that meet
+# the same criteria.
+
+# Using the above example again, the three entries that sum to 2020 are 979,
+# 366, and 675. Multiplying them together produces the answer, 241861950.
+
+# In your expense report, what is the product of the three entries that sum to
+# 2020?
+
+
+def part_2() -> None:
+    for n in numbers:
+        for i in range(len(numbers)):
+            for j in range(len(numbers)):
+                if n + numbers[i] + numbers[j] == 2020:
+                    print(
+                        f"{n} times {numbers[i]} times {numbers[j]} is "
+                        f"{n*numbers[i]*numbers[j]}"
+                    )
+                    break
+            else:
+                continue
             break
-    else:
-        continue
-    break
+        else:
+            continue
+        break
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()