A month ago I set a programming challenge but have had no response. This could be not enough visitors (about 15-20 a day), or the nature of me posting daily on a WordPress blog where entries vanish.
That challenge was to identify the 100,000th number in a sequence of nine digits numbers that start at 1234567879 and finish at 987654321 where each digit only appears once. It’s relatively easy to program.
It’s also not that difficult to do manually if you think of factorials. As each digit appears once there 9! different numbers possible. There are 9 possibilities for the first digit then 8 for the second and so on. 362,880 in total. (I think…).
So of all the numbers in the range the first 8! (40,320) must start with 1, the 2nd 8! with 2, the 3rd 8! with 3. So the 100,000th is going to start with a 3 as it occurs somewhere between the 80,640th and the 120,960th. You can then apply the same logic to work out where the 2nd digit is going to lie and so on. I did it on a piece of paper and it took me about 20 minutes to figure.
You don’t need maths (or math as US readers will be familiar with) to program computers but on occasions it can come in handy. Especially with games and random numbers, combinations and so on.