diff --git a/DIRECTORY.md b/DIRECTORY.md
index 866e0d658bf660c9c65bc826a368a7b2090d4483..f0f494e7a3b41aaf046c97ffa71bcbd255341d7d 100644
--- a/DIRECTORY.md
+++ b/DIRECTORY.md
@@ -671,6 +671,8 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_062/sol1.py)
   * Problem 063
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_063/sol1.py)
+  * Problem 065
+    * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_065/sol1.py)
   * Problem 067
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_067/sol1.py)
   * Problem 069
diff --git a/project_euler/problem_065/__init__.py b/project_euler/problem_065/__init__.py
new file mode 100644
index 0000000000000000000000000000000000000000..e69de29bb2d1d6434b8b29ae775ad8c2e48c5391
diff --git a/project_euler/problem_065/sol1.py b/project_euler/problem_065/sol1.py
new file mode 100644
index 0000000000000000000000000000000000000000..229769a77d07f0845a0c55f398a348d13956f2c2
--- /dev/null
+++ b/project_euler/problem_065/sol1.py
@@ -0,0 +1,99 @@
+"""
+Project Euler Problem 65: https://projecteuler.net/problem=65
+
+The square root of 2 can be written as an infinite continued fraction.
+
+sqrt(2) = 1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / (2 + ...))))
+
+The infinite continued fraction can be written, sqrt(2) = [1;(2)], (2)
+indicates that 2 repeats ad infinitum. In a similar way, sqrt(23) =
+[4;(1,3,1,8)].
+
+It turns out that the sequence of partial values of continued
+fractions for square roots provide the best rational approximations.
+Let us consider the convergents for sqrt(2).
+
+1 + 1 / 2 = 3/2
+1 + 1 / (2 + 1 / 2) = 7/5
+1 + 1 / (2 + 1 / (2 + 1 / 2)) = 17/12
+1 + 1 / (2 + 1 / (2 + 1 / (2 + 1 / 2))) = 41/29
+
+Hence the sequence of the first ten convergents for sqrt(2) are:
+1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, ...
+
+What is most surprising is that the important mathematical constant,
+e = [2;1,2,1,1,4,1,1,6,1,...,1,2k,1,...].
+
+The first ten terms in the sequence of convergents for e are:
+2, 3, 8/3, 11/4, 19/7, 87/32, 106/39, 193/71, 1264/465, 1457/536, ...
+
+The sum of digits in the numerator of the 10th convergent is
+1 + 4 + 5 + 7 = 17.
+
+Find the sum of the digits in the numerator of the 100th convergent
+of the continued fraction for e.
+
+-----
+
+The solution mostly comes down to finding an equation that will generate
+the numerator of the continued fraction. For the i-th numerator, the
+pattern is:
+
+n_i = m_i * n_(i-1) + n_(i-2)
+
+for m_i = the i-th index of the continued fraction representation of e,
+n_0 = 1, and n_1 = 2 as the first 2 numbers of the representation.
+
+For example:
+n_9 = 6 * 193 + 106 = 1264
+1 + 2 + 6 + 4 = 13
+
+n_10 = 1 * 193 + 1264 = 1457
+1 + 4 + 5 + 7 = 17
+"""
+
+
+def sum_digits(num: int) -> int:
+    """
+    Returns the sum of every digit in num.
+
+    >>> sum_digits(1)
+    1
+    >>> sum_digits(12345)
+    15
+    >>> sum_digits(999001)
+    28
+    """
+    digit_sum = 0
+    while num > 0:
+        digit_sum += num % 10
+        num //= 10
+    return digit_sum
+
+
+def solution(max: int = 100) -> int:
+    """
+    Returns the sum of the digits in the numerator of the max-th convergent of
+    the continued fraction for e.
+
+    >>> solution(9)
+    13
+    >>> solution(10)
+    17
+    >>> solution(50)
+    91
+    """
+    pre_numerator = 1
+    cur_numerator = 2
+
+    for i in range(2, max + 1):
+        temp = pre_numerator
+        e_cont = 2 * i // 3 if i % 3 == 0 else 1
+        pre_numerator = cur_numerator
+        cur_numerator = e_cont * pre_numerator + temp
+
+    return sum_digits(cur_numerator)
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")