Added solution for Project Euler problem 38. Fixes: #2695

Author: fpringle

project_euler/problem_38/__init__.py

@@ -0,0 +1,74 @@
+"""
+Take the number 192 and multiply it by each of 1, 2, and 3:
+
+192 × 1 = 192
+192 × 2 = 384
+192 × 3 = 576
+
+By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call
+192384576 the concatenated product of 192 and (1,2,3)
+
+The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5,
+giving the pandigital, 918273645, which is the concatenated product of 9 and
+(1,2,3,4,5).
+
+What is the largest 1 to 9 pandigital 9-digit number that can be formed as the
+concatenated product of an integer with (1,2, ... , n) where n > 1?
+"""
+
+
+def is_9_palindromic(n: int) -> bool:
+    """
+    Checks whether n is a 9-digit 1 to 9 pandigital number.
+    >>> is_9_palindromic(12345)
+    False
+    >>> is_9_palindromic(156284973)
+    True
+    >>> is_9_palindromic(1562849733)
+    False
+    """
+    s = str(n)
+    return len(s) == 9 and set(s) == set("123456789")
+
+
+def solution() -> int:
+    """
+    Return the largest 1 to 9 pandigital 9-digital number that can be formed as the
+    concatenated product of an integer with (1,2,...,n) where n > 1.
+
+    Solution:
+        Since n>1, the largest candidate for the solution will be a concactenation of
+        a 4-digit number and its double, a 5-digit number.
+        Let a be the 4-digit number.
+        a  has 4 digits  =>  1000 <=  a  < 10000
+        2a has 5 digits  => 10000 <= 2a  < 100000
+        =>  5000 <= a < 10000
+
+        The concatenation of a with 2a = a * 10^5 + 2a
+        so our candidate for a given a is 100002 * a.
+        We iterate through the search space 5000 <= a < 10000 in reverse order,
+        calculating the candidates for each a and checking if they are 1-9 pandigital.
+
+        In case there are no 4-digit numbers that satisfy this property, we check
+        the 3-digit numbers with a similar formula (the example a=192 gives a lower
+        bound on the length of a):
+        a has 3 digits, etc...
+        =>  100 <= a < 334, candidate = a * 10^6 + 2a * 10^3 + 3a
+                                      = 1002003 * a
+
+    """
+    for base_num in range(9999, 4999, -1):
+        candidate = 100002 * base_num
+        if is_9_palindromic(candidate):
+            return candidate
+
+    for base_num in range(333, 99, -1):
+        candidate = 1002003 * base_num
+        if is_9_palindromic(candidate):
+            return candidate
+
+    return 192384576
+
+
+if __name__ == "__main__":
+    print(solution())