diff --git a/project_euler/problem_057/__init__.py b/project_euler/problem_057/__init__.py
new file mode 100644
index 000000000..e69de29bb
diff --git a/project_euler/problem_057/sol1.py b/project_euler/problem_057/sol1.py
new file mode 100644
index 000000000..04b6199f4
--- /dev/null
+++ b/project_euler/problem_057/sol1.py
@@ -0,0 +1,48 @@
+"""
+Project Euler Problem 57: https://projecteuler.net/problem=57
+It is possible to show that the square root of two can be expressed as an infinite
+continued fraction.
+
+sqrt(2) = 1 + 1 / (2 + 1 / (2 + 1 / (2 + ...)))
+
+By expanding this for the first four iterations, we get:
+1 + 1 / 2 = 3 / 2 = 1.5
+1 + 1 / (2 + 1 / 2} = 7 / 5 = 1.4
+1 + 1 / (2 + 1 / (2 + 1 / 2)) = 17 / 12 = 1.41666...
+1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) = 41/ 29 = 1.41379...
+
+The next three expansions are 99/70, 239/169, and 577/408, but the eighth expansion,
+1393/985, is the first example where the number of digits in the numerator exceeds
+the number of digits in the denominator.
+
+In the first one-thousand expansions, how many fractions contain a numerator with
+more digits than the denominator?
+"""
+
+
+def solution(n: int = 1000) -> int:
+    """
+    returns number of fractions containing a numerator with more digits than
+    the denominator in the first n expansions.
+    >>> solution(14)
+    2
+    >>> solution(100)
+    15
+    >>> solution(10000)
+    1508
+    """
+    prev_numerator, prev_denominator = 1, 1
+    result = []
+    for i in range(1, n + 1):
+        numerator = prev_numerator + 2 * prev_denominator
+        denominator = prev_numerator + prev_denominator
+        if len(str(numerator)) > len(str(denominator)):
+            result.append(i)
+        prev_numerator = numerator
+        prev_denominator = denominator
+
+    return len(result)
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")