Problem 37

Problem 38: Pandigital multiples

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?

Algorthim:

Another Pandigital question. The best way to tackle this problem is to restrict our search space. Since we want the largest pandigital number, we can utilize the given example to use 91827365. Let's think about the way we can create a concatenated 9 digit Number

3 Three digit numbers as seen in the example, without producing new digit
2 Four digit numbers. We can get one new digit, making 5 + 4 digit total

There are othere ways. However, if you take a look at the second list, we can control the first four digits manually. We can pick a four digit number, and find multiple of this four digit number tht yields and 9 Pandigitala number!

We start our search using 9183 since the example given to us is 918273645. Since we want the maximum digit, we start from 9183, while keeping it pandigital!

Code:

Result:


                        Answer: 932718654
                        
Runtime: 11 ms