# Tag Info

1

WORK IN PROGRESS: It must be around .... .... give or take a small increment. Pattern:

1

Not an extremely elegant solution, but The next term is

8

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. I ...

9

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 \$...

6

For the last ten times that this has happened, they are: The next ten years are going to be: My thinking process:

1

I don't know how to do this without a computer, but with a computer, I found the next ten such years to be The previous ten years with the property were I find it interesting that

5

Going down: Going up:

2

My best shot: And for 10:

3

Here is one possible arrangement: Proof of optimality: In view of the answer by loopy walt, this also proves optimality for the case of decagon.

7

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: Proof: Addendum:

1

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: Bar 5: Bar 6: Bar 7: Summary of Train Station Arrivals

1

Missed the 5 minutes early being important. T. Linnell has it done properly. Solution? Bigger question, how they get the bars on the train quickly without someone pushing the other out at the last second?

5

Here is the finished tiling: To get started: Next step: 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.

6

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 or

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