The smallest number is going to be the one with the least digits. We need to make use of all the digits as efficiently as possible. Therefore, we should:
Therefore, the solution is
I’ll admit I used a computer (despite the no-computers tag) to generate the output, since a Bash command is much faster and more accurate than typing that up by hand.
The three other answers have already listed the correct years. What I'd like to add is a clean way of finding them:
Writing $d(x)$ for the digit sum of $x$ it is easily verified that
Indeed, expanding $d(x)^2$ yields
$x_1^2+x_2^2+x_3^2+\ldots+2x_1x_2+2x_1x_3+\ldots$ where $x_1$ is the first digit, $x_2$ the second and so forth
Doing the long multiplication $...
Assuming that Bo and Jo cannot set up secret stashes from each other (as otherwise the problem is fairly trivial), but assuming that they can set up a shared stash, one strategy they could use is to:
Bo and Jo must:
Bo's travel time is:
Jo's travel time is:
Bars 1 and 2 are the easiest to solve:
From here things get interesting, but essentially:
Which means we must calculate:
Bars 3 and 4:
Summary of Train Station Arrivals
Missed the 5 minutes early being important.
T. Linnell has it done properly.
Bigger question, how they get the bars on the train quickly without someone pushing the other out at the last second?
Here is the finished tiling:
To get started:
Looking at the left pear:
With those chokepoints:
For the next step I took a guess:
From there, I just looked at the pieces I had left and found something that worked.
Since black has many checks available, we can
This limits our options to four possible moves. Nf7+ seems particularly promising, so
Black has only one move that doesn't immediately end in a smothered checkmate at Nf6, so we check the checks (heh) after that move to find
after which we can somewhat incredibly finish with either