[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 348
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Аборигенов Необитаемого Острова приглашаю на базу, я там отписал кое-что.
What is the island?
А что за остров?
DHARMA project...
Instead of a black one, a white swan arrives.
;)
Аборигенов Необитаемого Острова приглашаю на базу, я там отписал кое-что.
Base, base, I'm CharlieFoxtrot, get the coordinatesReception
This island has an email address, but it's secret for now. There aren't many inhabitants there, and they don't meet very often. Not like here. Most of the discussion is about theoretical and boring aspects of trading that are unlikely to be interesting for the majority of people here :).
Прием, Mischek.
У этого острова есть электронный адрес, но он пока секретен. Жителей там немного, да и встречаются они нечасто. Не то что здесь. Обсуждаются в-основном теоретические и скучные аспекты трейдинга, которые здесь большинству вряд ли интересны :)
Got it, I'm going back.)P.S. Ну так как, никто на лапу никогда не играл, что ли?
B (given to Kolya):
Let's number the cards 0,1,...,6
Grisha and Lyosha must tell each other the sums of their cards modulo 7
7-(A+B) mod 7 - the card that Kolya has.
(a) Have Grisha say, "I either have {names his cards} or {names three cards he doesn't have}". Then have Lyosha say, "I either {name my cards} or {name three of Grisha's cards if the second of the sets Grisha named does not match his set, and any other three cards he doesn't have, otherwise}". After that, each of them obviously knows the whole layout. Kolya, on the other hand, is unclear. Indeed three sets of cards are named: A, B and C. Sets B and C overlap by two cards, Grisha said: "I have either A or B", Lyosha said: "I have either A or C". This means that either Grisha has a set A and Lyosha has C, or Grisha has B and Lyosha has A. Of course, these layouts are different, and even a closed card cannot be determined.
b) Note that the previous method doesn't work: knowing the closed card, Kolya can determine everything. Let's number the cards with numbers from 0 to 6. Let Grisha and Lyosha take turns naming the remainder of the sum of the numbers of their cards divided by 7. Then they will know the deal: each of them must only add to their sum the sum of the other and find the remainder opposite to this total sum modulo 7 (that is, one that, when added to this sum, gives a number divisible by 7). This will be the number of the closed card. After that, it is easy to restore the deal. Let's check that Kolya hasn't learned anything. Consider the card with the number s. Let's show that it could get to Grisha, if he called the amount a. To do that, we need to supplement this card with two other cards with the sum of the numbers a-s. It is easy to see that there are three different pairs of numbers, giving the sum a-s. Of these, two are probably spoiled by the fact that they include a card with the number s or a closed card, but at least one pair remains. With it we will complete Grisha's set. The same reasoning shows that any card could also be in Alex's possession.
У Феди есть три палочки. Если из них нельзя сложить треугольник, Федя укорачивает самую длинную из палочек на сумму длин двух других. Если длина палочки не обратилась в нуль и треугольник снова нельзя сложить, то Федя повторяет операцию, и т. д. Может ли этот процесс продолжаться бесконечно?
Yes, if the lengths are three consecutive numbers of a sequence obeying the recurrence law x[n + 3] = x[n] + x[n + 1] + x[n + 2] and being simultaneously a geometric progression.
That is, they are three numbers of the form x ax a^2x, where a is the solution to the cubic equation a^3 - a^2 - a - 1 = 0