[superprismatic]

Suppose you permute the numbers 1 through 11 amongst themselves on

a clock face, leaving the 12 undisturbed. Then pick a starting

number. Begin a list with this number. Then, move clockwise

around the clockface the number of numbered spots given by the

last number on your list. Add this number to your list. Then

move clockwise the number of spots given by the last number on

the list. Continue in this way until you hit the 12 or a number

which is already on your list. Here's an example.

Suppose you number the clockface clockwise from the 12 as follows:

12,1,4,2,3,5,6,7,8,9,10,11. And you start at the 2. The next

number for the list will be 5, then 10, then 8, etc. Your

list will end up 2,5,10,8,3,7,4,6,12.

Or, suppose you number the clockface clockwise from the 12 as:

12,1,3,2,4,5,6,7,8,9,10,11. And you start at 1. Your list will

then be 1,3,5,10,8,4 and 8 would be next but it is already on the

list.

Can you permute the numbers between 1 and 11, and pick a starting

number, such that your list will contain all 12 numbers?

Interesting fact: Less than .0009% of random permutation/start

combinations will produce a list containing all 12 numbers.

[CaptainEd]

Another fascinating and odd puzzle, thanks!

Start with 1

11 9 7 5 3 1 10 8 6 4 2 12

[Guest]

Another fascinating and odd puzzle, thanks!

Start with 1

11 9 7 5 3 1 10 8 6 4 2 12

I didn't solve this, but after seeing your solution I realized another, you can bounce inwards instead of out (Start with 11)

11 9 7 5 3 1 12 10 8 6 4 2

[Guest]

Some of the solutions

1 10 3 5 2 4 12 8 6 7 9 11

2 10 11 3 6 4 12 1 9 7 5 8

3 4 8 5 9 2 12 11 6 7 1 10

4 7 11 8 1 5 12 2 10 6 9 3

5 7 8 3 10 11 12 4 1 6 9 2

I found that there are in all 3856 solutions.....

So wont the probability be ~= 8e-6

[Guest]

firstly, superprismatic, I just noticed the "not just prismatic" under your name and I just want to thank you for that. It really made me laugh. I also want to thank you for all the interesting and challenging puzzles you have come up with.

That being said, I would like to add something to this problem: Is there any permutation for which any starting number will result in a list of all 12 numbers? Prove it (and not by exhaustion).

[superprismatic]

Glad that you appreciated my little joke and like my puzzles. Thanks.

Now as to your addition:

I would like to add something to this problem: Is there any permutation for which any starting number will result in a list of all 12 numbers? Prove it (and not by exhaustion).

There is no such permutation. Any list which contains all 12 numbers must end in 12. Once we get to 12, the number 12 away is 12. Now suppose there is a number, N, which

begins a list containing all 12 numbers. It starts with N, ends with 12, and has 10 numbers in between. Now, starting with one of those 10, we can never get N on our

list because we will hit 12 first.

[Guest]

while i agree that's not possible, is it possible to hit every number but 12, with the last number going back to the first?