[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 11
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The original problem is unique precisely in its brevity and elementary wording, without any "exceptions".
And Petya is not an outsider after all: he is in this class and is friends with some of them.
Mischek >> Не зависимо от N всегда будет двое с одинаковым количеством друзей
Why? Why put it as a condition of the problem, if it follows from its analysis?
You're all counting here.... :)
there is no concrete solution to this problem... there's only probability...
The original problem is unique precisely in its brevity and elementary wording, without any "exceptions".
And Petya is not an outsider after all: he is in this class and is friends with some people.
I have a feeling the solution will be brilliantly simple too.
Первоначальная задача уникальна именно краткостью и элементарностью формулировки, без всяких "за исключением".
И Петя все же не со стороны приперся: он в этом классе учится и дружит с некоторыми.
Почему? Зачем выносить это как условие задачи, если это вытекает из ее анализа?
No, no, no, no, no.
I'll add quotation marks.
Indeed, it is. I somehow forgot that the connection is two-sided and the graph is not so branchy :/
Действительно, так и есть. Я как то забыл, что связь то двусторонняя по условию и граф не такой ветвистый получается :/
By the way, as a BEGIN to check the correctness of the statement and the existence of the solution, one can proceed from the parity-transitivity property mentioned by Matemat: at any numbering these will be the terms of the arithmetic progression and their sum must be even. I can see that this will not always be the case, and the inclusions of Petya's friends (the possible recurrence of one number in the progression) are also relevant. Sorry, no time today, won't be able to calculate.
https://ru.wikipedia.org/wiki/%D0%90%D1%80%D0%B8%D1%84%D0%BC%D0%B5%D1%82%D0%B8%D1%87%D0%B5%D1%81%D0%BA%D0%B0%D1%8F_%D0%BF%D1%80%D0%BE%D0%B3%D1%80%D0%B5%D1%81%D1%81%D0%B8%D1%8F
All right, I give up, my answer is the number of pupils divided by two, unless you count the maniac pupil :)
Всё нафиг, сдаюсь, мой ответ - количество учеников, делёное на два, если не считать ученика-маньяка :)
If there is a person in the class who is not friends with anyone, the answer is 12.
If there is no such person, i.e. everyone is friends with someone, then the answer is 13.
I can prove it very simply, without induction, binomial, graphs etc. And for the general case of N students. (12 and 13 naturally for N=25)