project_euler.problem_038.sol1¶
Project Euler Problem 38: https://projecteuler.net/problem=38
Take the number 192 and multiply it by each of 1, 2, and 3:
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 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?
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
Functions¶
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Checks whether n is a 9-digit 1 to 9 pandigital number. |
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Return the largest 1 to 9 pandigital 9-digital number that can be formed as the |
Module Contents¶
- project_euler.problem_038.sol1.is_9_pandigital(n: int) bool¶
Checks whether n is a 9-digit 1 to 9 pandigital number. >>> is_9_pandigital(12345) False >>> is_9_pandigital(156284973) True >>> is_9_pandigital(1562849733) False
- project_euler.problem_038.sol1.solution() int | None¶
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.