# Desperately seeking pandigitals

Also known as Project Euler Problem 38:

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?

This is a problem for which finding a practical approach took a lot longer than the implementation of said approach. It’s also a problem that forced me to think about the actual maths of the problem. I already have an `is_pandigital` function, which I wrote to help solve problem 32, so the main challenge here is to constrain the search enough that I can avoid endless looping.

So, a bit of analysis…

For n = 2,
9 * (1, 2) = 918
98 * (1, 2) = 98196
987 * (1, 2) = 9871974
9876 * (1, 2) = 987619752 <- That's 9 digits

Since the problem mentions the pandigital, 918273645, I plugged this into the calculator as well,
9182 * (1, 2) = 918218365 <- That's 9 digits again

For n = 3,
9 * (1, 2, 3) = 91827
98 * (1, 2, 3) = 98196294 <- 8 digits is not enough
987 * (1, 2, 3) = 98719742961 <- 11 digits is too many

So I now know that the number I am looking for is n * (1, 2) where n is between 9876 and 9182.

The implementation, therefore, is a very simple loop that checks `is_pandigital(int(str(n) + str(n * 2)))` for values of n from 9876 to 9182.

The result is almost instantaneous.