Prime

Found the ultimate mate in 9 moves.

Not only I forgot about half dollar coin in my previous post, I also made few arithmetic errors in calculating probabilities.
Although, the answer who has a better probability is the same.

The code calculates the exact probability for the average payoff with the best strategy (unless I messed up.)
Both simulation and code must take into account the cards leaving the deck. Also, must discard invalid (in terms of strategy) variations. E.g., a draw like (13, 13, 8).
The criteria for the staying card on the third turn can be established analytically (see the spoiler.)

I suppose, blue eyed child also must play part in figuring out question 1. I only took it into account for figuring out question two, so my previous answer is wrong.

This way it's more interesting. I understand, one person has 4 times the money of his neighbor at the end of the game.

I am only aware of this general rule.

That is not to say there aren't any other possibilities for finding perfect numbers.

I am uncertain, what uncertain means in this context.
Cole's first shot strategy is clear. He must shoot himself in the foot thus exiting the duel with a non-fatal injury. (Hopefully, he does not miss.)

Actually, we should split the difference and go with 27.

Going with the following assumptions as answers to my questions:
1. Leading zeros are allowed.
2. We work with the sum of numbers, not digits.
3. The school No. 7 is in Moscow in the vicinity of MSU (МГУ) and Vernadski Prospect.



I have a different proof of the same, which gives even more insight into divisibility rules, but it would take longer to explain.
We have another Divisibility puzzle going here: http://brainden.com/forum/index.php/topic/15583-more-divisibility/?p=328196]More divisibility

I see, I mixed up the adavance of the hour hand with the elapsed time for each cycle. And I calculated period incorrectly for the second part in my previous post.
Still, after the corrections, I get a different answer. My hour and minute hands will meet a lot sooner.

Why rejecting the occurences where the clock hands meet at 12 in a mid-cycle?
I see nothing in the OP prohibiting that. In fact, the clock hands will meet again at 12 on a mid-cycle before they do so after a whole number of cycles.

Nice problem.

Time's up.

I’ll try again.

Same as Kingly Tour.

But they don't need to be integers, those 11 periodic values.

I found couple solutions, but not with as many digits, I'm afraid.

Found the ultimate mate in 9 moves.

It seems to stand the reason...