Solution to problem 43

src/Python/Problem043.py

@@ -0,0 +1,55 @@
+#!/usr/bin/env python3
+"""
+Created on 03 Aug 2021
+
+@author: David Doblas Jiménez
+@email: daviddoji@pm.me
+
+Solution for problem 43 of Project Euler
+https://projecteuler.net/problem=43
+"""
+
+from itertools import permutations
+from utils import timeit
+
+
+@timeit("Problem 43")
+def compute():
+    """
+    The number, 1406357289, is a 0 to 9 pandigital number because 
+    it is made up of each of the digits 0 to 9 in some order, but 
+    it also has a rather interesting sub-string divisibility property.
+
+    Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this
+    way, we note the following:
+
+        d2d3d4=406 is divisible by 2
+        d3d4d5=063 is divisible by 3
+        d4d5d6=635 is divisible by 5
+        d5d6d7=357 is divisible by 7
+        d6d7d8=572 is divisible by 11
+        d7d8d9=728 is divisible by 13
+        d8d9d10=289 is divisible by 17
+
+    Find the sum of all 0 to 9 pandigital numbers with this property.
+    """
+    ans = []
+    pandigital = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']
+
+    for n in permutations(pandigital):
+        n_ = "".join(n)
+        if n_[0] != "0" and sorted("".join(n_)) == pandigital:
+            if int(n_[7:]) % 17 == 0:
+                if int(n_[6:9]) % 13 == 0:
+                    if int(n_[5:8]) % 11 == 0:
+                        if int(n_[4:7]) % 7 == 0:
+                            if int(n_[3:6]) % 5 == 0:
+                                if int(n_[2:5]) % 3 == 0:
+                                    if int(n_[1:4]) % 2 == 0:
+                                        ans.append(int(n_))
+
+    return sum(ans)
+
+if __name__ == "__main__":
+
+    print(f"Result for Problem 43: {compute()}")