# Statement

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?

# Solution

As there must be 2 multiplications concatenated at least the number chosen can't

be more than five digits long cause 2 multiplications would have more than 9 digits.

So I brute-forced for the first 9999 numbers:

from CommonFunctions import is_pandigital if __name__ == '__main__': result = "0" for x in range(2,10000): tmp = str(x * 1) + str(x * 2) n = 3 while len(tmp) < 9: tmp += str(x * n) n += 1 if is_pandigital(tmp): result = max(int(result), int(tmp)) print("The result is:", result)

The Python file is available for download here.