TheAlgorithms · peteryao7 · Oct 8, 2020 · Oct 15, 2020 · Oct 16, 2020 · Oct 16, 2020
diff --git a/DIRECTORY.md b/DIRECTORY.md
@@ -691,6 +691,8 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_173/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 234
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_234/sol1.py)
   * Problem 551

diff --git a/project_euler/problem_206/__init__.py b/project_euler/problem_206/__init__.py
diff --git a/project_euler/problem_206/sol1.py b/project_euler/problem_206/sol1.py
@@ -0,0 +1,65 @@
+"""
+Project Euler 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. We can either start checking for the square root from
+101010103, which is the closest square root of 10203040506070809 that ends
+in 3 or 7, or 138902663, the closest square root of 1929394959697989. The
+problem says there's only 1 answer, so starting at either point is fine,
+but the result happens to be much closer to the latter.
+"""
+
+
+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
+
+
+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
+
+
+if __name__ == "__main__":
+    print(f"{solution() = }")