diff --git a/DIRECTORY.md b/DIRECTORY.md
index 7c6958921..0cecae28d 100644
--- a/DIRECTORY.md
+++ b/DIRECTORY.md
@@ -725,6 +725,8 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_174/sol1.py)
   * Problem 191
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_191/sol1.py)
+  * Problem 206
+    * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_206/sol1.py)
   * Problem 207
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_207/sol1.py)
   * Problem 234
diff --git a/project_euler/problem_206/__init__.py b/project_euler/problem_206/__init__.py
new file mode 100644
index 000000000..e69de29bb
diff --git a/project_euler/problem_206/sol1.py b/project_euler/problem_206/sol1.py
new file mode 100644
index 000000000..ffac2b32a
--- /dev/null
+++ b/project_euler/problem_206/sol1.py
@@ -0,0 +1,74 @@
+"""
+Project Euler Problem 206: https://projecteuler.net/problem=206
+
+Find the unique positive integer whose square has the form 1_2_3_4_5_6_7_8_9_0,
+where each “_” is a single digit.
+
+-----
+
+Instead of computing every single permutation of that number and going
+through a 10^9 search space, we can narrow it down considerably.
+
+If the square ends in a 0, then the square root must also end in a 0. Thus,
+the last missing digit must be 0 and the square root is a multiple of 10.
+We can narrow the search space down to the first 8 digits and multiply the
+result of that by 10 at the end.
+
+Now the last digit is a 9, which can only happen if the square root ends
+in a 3 or 7. From this point, we can try one of two different methods to find
+the answer:
+
+1. Start at the lowest possible base number whose square would be in the
+format, and count up. The base we would start at is 101010103, whose square is
+the closest number to 10203040506070809. Alternate counting up by 4 and 6 so
+the last digit of the base is always a 3 or 7.
+
+2. Start at the highest possible base number whose square would be in the
+format, and count down. That base would be 138902663, whose square is the
+closest number to 1929394959697989. Alternate counting down by 6 and 4 so the
+last digit of the base is always a 3 or 7.
+
+The solution does option 2 because the answer happens to be much closer to the
+starting point.
+"""
+
+
+def is_square_form(num: int) -> bool:
+    """
+    Determines if num is in the form 1_2_3_4_5_6_7_8_9
+
+    >>> is_square_form(1)
+    False
+    >>> is_square_form(112233445566778899)
+    True
+    >>> is_square_form(123456789012345678)
+    False
+    """
+    digit = 9
+
+    while num > 0:
+        if num % 10 != digit:
+            return False
+        num //= 100
+        digit -= 1
+
+    return True
+
+
+def solution() -> int:
+    """
+    Returns the first integer whose square is of the form 1_2_3_4_5_6_7_8_9_0
+    """
+    num = 138902663
+
+    while not is_square_form(num * num):
+        if num % 10 == 3:
+            num -= 6  # (3 - 6) % 10 = 7
+        else:
+            num -= 4  # (7 - 4) % 10 = 3
+
+    return num * 10
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")