[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 293

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Of course, we don't mean just ones and zeros, but first the ones and then the zeros.
P.S. The problem is reduced to the proof that for any number N which is not divisible by 2 and 5, one can find a number with only ones which is divisible by N.
 
Back to problem 22 from here: https://www.mql5.com/ru/forum/123519/page291.
The problem was very simple. I spent a day and a half with the solution, I still couldn't find it. The solution came from a construction similar to that for ordinary Fibs, which is their general formula.
It is enough to prove that for any integer n

(5+sqrt(26))^n + (5-sqrt(26))^n

- is an integer. This is obvious, since odd degrees of the root of 26 will be mutually reduced from different brackets, and even degrees will give integers.
Then, since |5-sqrt(26)| = 1/(sqrt(26)+5) < 1/10, we obtain that the second term is always less than 10^(-n) modulo. Proved.
 
Mathemat >>:
Возвращаемся к задаче 22 вот отсюда: https://www.mql5.com/ru/forum/123519/page291.
Задачка оказалась очень простой. Крутился около решения полтора дня, все никак не мог подобраться. Решение пришло из конструкции, аналогичной конструкции для обычных Фиб, являющейся их общей формулой.
Достаточно доказать, что при любом целом n число

(5+sqrt(26))^n + (5-sqrt(26))^n

- целое. Это очевидно, т.к. нечетные степени корня из 26 будут взаимно сокращаться из разных скобок, а четные будут давать целые.
Тогда, т.к. |5-sqrt(26)| = 1/(sqrt(26)+5) < 1/10, получаем, что второй член всегда меньше 10^(-n) по модулю. Доказано.

OK. I'll also add "combinatorial-playing", so as not to get bogged down in number theory alone. :)

--

On the board there is a field for the game "in numbers": (((((((((_?_)?_)?_)?_)?_)?_)?_)?_).Two players take turns playing. The first player writes a number on the place of the first (leftmost) space (_). Each subsequent move consists of writing the digit in the place of the next space and replacing the question mark (?) on the left with an addition or multiplication sign. None of the digits must occur twice. At the end of the game calculate the value of the expression. If the number is even, the first player wins, if it is odd, the second player wins. Who wins if the game is played correctly?

--

// Corrected, changed the asterisks to questions, that's better. The problem is ancient, in those days they didn't know that multiplication on our computers would be indicated by an asterisk.

 
Yes, I've noticed that people are more into combinatorial problems. OK, let's think about what to do. Maybe we'll play the game again when we have ideas :) But not right now.
P.S. I decided, wrote here, but then erased. Wrote you in private, MetaDriver. Let the others to suffer. Who saw the solution - do not tell!
 
So, MetaDriver, are we going to post the solution to this problem or not? In the meantime, I'll look for something else that's fun - combinatorial or geometric.
 
My son had a problem in first class today:
Vasya bent a triangle out of wire, the sides of which are 2, 3, 3.
If he bends a square from the wire, what will be its side equal?
 
Not bad for a first grader. Although, in principle, a clever first grader who knows what division is would be able to do it. But usually they don't.
 

joo, but at least you posted a problem I can solve :)

 
And we haven't had division yet, although the 2100 programme is like the most
We have asterisk problems like this, for cleverness, with no consequences.
 
Mathemat >>:
Неплохо для первого класса. Хотя в принципе сообразительный первоклашка, знающий, что такое деление, решит. Но обычно не знают ведь.

I was totally freaking out. They haven't done any division yet. And they only count to 10! :) My son did.

New fucking program. :)

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