diff --git a/src/Python/Problems001-050/Problem038.py b/src/Python/Problems001-050/Problem038.py index bb38c86..125bb5b 100644 --- a/src/Python/Problems001-050/Problem038.py +++ b/src/Python/Problems001-050/Problem038.py @@ -1,4 +1,4 @@ -#!/usr/bin/env python3 +#!/usr/bin/env python """ Created on 03 Jun 2021 @@ -17,25 +17,24 @@ def compute(): """ Take the number 192 and multiply it by each of 1, 2, and 3: - 192 × 1 = 192 - 192 × 2 = 384 - 192 × 3 = 576 + 192 x 1 = 192 + 192 x 2 = 384 + 192 x 3 = 576 - By concatenating each product we get the 1 to 9 pandigital, - 192384576. We will call 192384576 the concatenated product of + 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 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 + be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1? - """ - results = [] + ans = [] # Number must 4 digits (exactly) to be pandigital # if n > 1 for i in range(1, 10_000): @@ -43,14 +42,13 @@ def compute(): number = "" while len(number) < 9: number += str(integer * i) - if sorted(number) == list('123456789'): - results.append(number) - + if sorted(number) == list("123456789"): + ans.append(number) + integer += 1 - - return max(results) + + return max(ans) if __name__ == "__main__": - - print(f"Result for Problem 38: {compute()}") + print(f"Result for Problem 38 is {compute()}")