Add dummy files to test the script

Author: dhruvmanila

project_euler/problem_001/sol8.py

@@ -0,0 +1,37 @@
+"""
+Project Euler Problem 1: https://projecteuler.net/problem=1
+
+Multiples of 3 and 5
+
+If we list all the natural numbers below 10 that are multiples of 3 or 5,
+we get 3, 5, 6 and 9. The sum of these multiples is 23.
+
+Find the sum of all the multiples of 3 or 5 below 1000.
+"""
+
+
+def solution(n: int = 1000) -> int:
+    """
+    Returns the sum of all the multiples of 3 or 5 below n.
+
+    >>> solution(3)
+    0
+    >>> solution(4)
+    3
+    >>> solution(10)
+    23
+    >>> solution(600)
+    83700
+    """
+
+    result = 0
+    for i in range(n):
+        if i % 3 == 0:
+            result += i
+        elif i % 5 == 0:
+            result += i
+    return result
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")

project_euler/problem_014/sol3.py

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+"""
+Problem 14: https://projecteuler.net/problem=14
+
+Problem Statement:
+The following iterative sequence is defined for the set of positive integers:
+
+    n → n/2 (n is even)
+    n → 3n + 1 (n is odd)
+
+Using the rule above and starting with 13, we generate the following sequence:
+
+    13 → 40 → 20 → 10 → 5 → 16 → 8 → 4 → 2 → 1
+
+It can be seen that this sequence (starting at 13 and finishing at 1) contains
+10 terms. Although it has not been proved yet (Collatz Problem), it is thought
+that all starting numbers finish at 1.
+
+Which starting number, under one million, produces the longest chain?
+"""
+
+
+def solution(n: int = 1000000) -> int:
+    """Returns the number under n that generates the longest sequence using the
+    formula:
+    n → n/2 (n is even)
+    n → 3n + 1 (n is odd)
+
+    # The code below has been commented due to slow execution affecting Travis.
+    # >>> solution(1000000)
+    # 837799
+    >>> solution(200)
+    171
+    >>> solution(5000)
+    3711
+    >>> solution(15000)
+    13255
+    """
+    largest_number = 0
+    pre_counter = 0
+
+    for input1 in range(n):
+        counter = 1
+        number = input1
+
+        while number > 1:
+            if number % 2 == 0:
+                number /= 2
+                counter += 1
+            else:
+                number = (3 * number) + 1
+                counter += 1
+
+        if counter > pre_counter:
+            largest_number = input1
+            pre_counter = counter
+    return largest_number
+
+
+if __name__ == "__main__":
+    print(solution(int(input().strip())))