117. The task of ten digits. Arrange all ten digits $1,2,3,4,5,6,7,8,9,0$ in such an order that the resulting number is divisible by all numbers from 2 to 18. For example, if the digits are placed in the sequence 1274953680, the resulting number will be divisible by $2,3,4,5$ and so on up to 16, but will not be divisible by 17.
117. There are four solutions: 2438195760, 3785942160, 4753869120, 4876391520. The last digit must be zero. With any arrangement of the digits with an even digit before the zero, the number is divisible by \(2, 3, 4, 5, 6, 9, 10, 12, 15\), and 18. It remains to consider only \(7, 11, 13, 16\), and 17. (Divisibility by 8 and 14 follows from divisibility by 16 and 7.) For divisibility by 11, the sum of the digits in even positions must be 28, and the sum of the digits in odd positions must be 17, or vice versa. For the number to be divisible by \(7 \times 11 \times 13 = 1001\), the number formed by the first three digits and the number formed by the last three digits (excluding the zero) must sum to the number formed by the middle three digits. (Note that the third case given is actually: 474-1386-912, where 1 is carried forward and added to 4.) However, the best we can do is multiply the least common multiple (LCM) of our divisors (12,252,240) by the smallest number (82) such that the product (1,004,683,680) contains 10 digits, and then add the LCM until all digits are distinct.

Multiplying the LCM by 199 gives the first solution, by 309 the second, by 388 the third, and by 398 the fourth solution. The calculations can be significantly shortened by skipping over groups of numbers where the digits obviously repeat. All answers can be quickly obtained using a computer.