Math and Logic Puzzles: Redux

[Kagami]

puzzle!

155619367777177344298579121640577930358466048997080741489502115154736052423438639656322566701951512888971130859442675019505459646950828457775128

divided by

814295694336200301695257325143637445297122644140195688542610298982742658367669506161728178242191616818890832959609127053599963426497407079941861

Answer is nine letters and the name of a San Francisco based company.

[Kagami]

With regard to 70, the argument for taking both boxes is silly. If we assume the super-being is indeed a super-being with super prediction powers, then we accept that she can predict what our philosophical response is to the dilemma when it's presented and so our ultimate decision. It doesn't matter that she can't move it unless the decider is provided some kind of unpredictable input that could influence their decision.

We take both boxes only if we assume she's a fraud. Since this is a philosophical problem that demands we accept the good faith presentation, you take box B and only box B.

[Kagami]

Regardless of whether you have prepared a strategy, the super-being can divine the strategy you will arrive at once presented with the box, and she will arrange the contents knowing that strategy. It presupposes that she is not a super predicting super-being to say that she can't do so.

[Kagami]

I'm treating the evidence of her being a super being in the article as designed for entertainment purposes, and that this should be approached philosophically, accepting her superness as a given.

If this were a practical exercise, I assume her to be a fraud and any prior predictive prowess she's shown is likely driven by survivorship bias (we wouldn't be in this situation if she had been documented being wrong in the past). The correct choice would be to take both if I had to and assume I'm getting 1k pounds, or, if allowed, to sell the chance to choose to a more gullible individual.

[Kagami]

The index of the tetrahedral and triangular number don't have to be the same.

[Kagami]

The triangular number = tetrahedral number proof is not something anyone should be banging their head over.

I don't think (please correct me if I'm wrong) anyone's presented an elegant proof of it, and the inelegant proof is arduous with steps that are far from straightforward.

[Kagami]

here's a nice problem:

A secret spy organization needs to spread some secret knowledge to all of its members. In the beginning, only 1 member is informed. Every informed spy will call an uninformed spy such that every informed spy is calling a different uninformed spy. After being called, an uninformed spy becomes informed. The call takes 1 minute, but since the spies are running low on time, they call the next spy directly afterward. However, to avoid being caught, after the third call an informed spy makes, the spy stops calling. How many minutes will it take for every spy to be informed, provided that the organization has 600 spies?

S0 = 1
S1 = 2
S2 = 4
S3 = 8
and
Si = 2*Si-1 - Si-4 for i > 3

So you get a goofy fibonacci-esque recursive sequence. For something as low as 600, you can just do it manually or programatically to get the sequence:

1, 2, 4, 8, 15, 28, 52, 96, 177, 326, 600, 1104, 2031, 3736, 6872, 12640, 23249, 42762, 78652, 144664, 266079, 489396, 900140, 1655616

And you see that you've informed exactly 600 spies in 10 minutes. To solve it for an arbitrary N, there should be a closed form solution of the form sum(c_i a_i ^ N) with at most four terms, which wouldn't be terribly difficult to sort out.