feat: add Project Euler problem 073 solution 1 (#6273)

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Author: 3 people

Diff for: DIRECTORY.md

@@ -839,6 +839,8 @@
   * Problem 072
     * [Sol1](project_euler/problem_072/sol1.py)
     * [Sol2](project_euler/problem_072/sol2.py)
+  * Problem 073
+    * [Sol1](project_euler/problem_073/sol1.py)
   * Problem 074
     * [Sol1](project_euler/problem_074/sol1.py)
     * [Sol2](project_euler/problem_074/sol2.py)

Diff for: project_euler/problem_073/__init__.py

@@ -0,0 +1,46 @@
+"""
+Project Euler Problem 73: https://projecteuler.net/problem=73
+
+Consider the fraction, n/d, where n and d are positive integers.
+If n<d and HCF(n,d)=1, it is called a reduced proper fraction.
+
+If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size,
+we get:
+
+1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3,
+5/7, 3/4, 4/5, 5/6, 6/7, 7/8
+
+It can be seen that there are 3 fractions between 1/3 and 1/2.
+
+How many fractions lie between 1/3 and 1/2 in the sorted set
+of reduced proper fractions for d ≤ 12,000?
+"""
+
+from math import gcd
+
+
+def solution(max_d: int = 12_000) -> int:
+    """
+    Returns number of fractions lie between 1/3 and 1/2 in the sorted set
+    of reduced proper fractions for d ≤ max_d
+
+    >>> solution(4)
+    0
+
+    >>> solution(5)
+    1
+
+    >>> solution(8)
+    3
+    """
+
+    fractions_number = 0
+    for d in range(max_d + 1):
+        for n in range(d // 3 + 1, (d + 1) // 2):
+            if gcd(n, d) == 1:
+                fractions_number += 1
+    return fractions_number
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")