coderjack
diff --git a/‎project_euler/problem_301/__init__.py‎ b/‎project_euler/problem_301/__init__.py‎
diff --git a/‎project_euler/problem_301/sol1.py‎
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+"""
+Project Euler Problem 301: https://projecteuler.net/problem=301
+
+Problem Statement:
+Nim is a game played with heaps of stones, where two players take
+it in turn to remove any number of stones from any heap until no stones remain.
+
+We'll consider the three-heap normal-play version of
+Nim, which works as follows:
+- At the start of the game there are three heaps of stones.
+- On each player's turn, the player may remove any positive
+  number of stones from any single heap.
+- The first player unable to move (because no stones remain) loses.
+
+If (n1, n2, n3) indicates a Nim position consisting of heaps of size
+n1, n2, and n3, then there is a simple function, which you may look up
+or attempt to deduce for yourself, X(n1, n2, n3) that returns:
+- zero if, with perfect strategy, the player about to
+  move will eventually lose; or
+- non-zero if, with perfect strategy, the player about
+  to move will eventually win.
+
+For example X(1,2,3) = 0 because, no matter what the current player does,
+the opponent can respond with a move that leaves two heaps of equal size,
+at which point every move by the current player can be mirrored by the
+opponent until no stones remain; so the current player loses. To illustrate:
+- current player moves to (1,2,1)
+- opponent moves to (1,0,1)
+- current player moves to (0,0,1)
+- opponent moves to (0,0,0), and so wins.
+
+For how many positive integers n <= 2^30 does X(n,2n,3n) = 0?
+"""
+
+
+def solution(exponent: int = 30) -> int:
+    """
+    For any given exponent x >= 0, 1 <= n <= 2^x.
+    This function returns how many Nim games are lost given that
+    each Nim game has three heaps of the form (n, 2*n, 3*n).
+    >>> solution(0)
+    1
+    >>> solution(2)
+    3
+    >>> solution(10)
+    144
+    """
+    # To find how many total games were lost for a given exponent x,
+    # we need to find the Fibonacci number F(x+2).
+    fibonacci_index = exponent + 2
+    phi = (1 + 5 ** 0.5) / 2
+    fibonacci = (phi ** fibonacci_index - (phi - 1) ** fibonacci_index) / 5 ** 0.5
+
+    return int(fibonacci)
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")