Math and Logic Puzzles: Redux

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chennisden
Jack of All Trades Joined: February 11, 2019
Location: milky way
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Doh

Tetrahedral: (n * (n + 1) * (n + 2)) / 6

Triangular: (n * (n + 1)) / 2

If these are equal n+2=3

chennisden
Jack of All Trades Joined: February 11, 2019
Location: milky way
Pronoun: He
Ugh people these days just think about alg and NT

and geo gets ignored chennisden
Jack of All Trades Joined: February 11, 2019
Location: milky way
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Hmm.

chennisden
Jack of All Trades Joined: February 11, 2019
Location: milky way
Pronoun: He
(n * (n + 1) * (n + 2)) / 6=(m * (m + 1)) / 2

n(n+1)(n+2)=3m(m+1)

Hmm.

There are only so many ways to get 3m(m+1) to be three consecutive numbers. Uh

chennisden
Jack of All Trades Joined: February 11, 2019
Location: milky way
Pronoun: He
In post 91, Kagami wrote: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.

i agree and now i feel worser about not finding it

chennisden
Jack of All Trades Joined: February 11, 2019
Location: milky way
Pronoun: He
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?

chennisden
Jack of All Trades Joined: February 11, 2019
Location: milky way
Pronoun: He
In post 90, BTD6_maker wrote:Part of the difficulty in this problem is that there is no easy way to make tetrahedral numbers into nth powers. I have a proof that, for any n >= 3, there are only finitely many triangular numbers that are nth powers, but I haven't yet got it to work for tetrahedral numbers. This is an interesting problem in itself, so you can try this problem (as well as my two previous combinatorics problems).

For geometry's geometry problem, let O be the centre of W and let it have radius R. Let one of the tangents from A touch W at C, and let theta be the angle AOC. Then OA' = OC cos theta and OC = OA cos theta so OA' OA = OC^2 = R^2. Likewise, OB' OB = R^2. Also, O, A, A' are collinear, and so are O, B, B'. Since OA' OA = OB' OB, we have AA'B'B is cyclic.

yup! A' is called the "inversion" of A about W. defined such that OA' * OA = r^2

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