[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 5
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Figaro, ты можешь показать графическое решение?
2 Farnsworth: но ведь ответ-то - 12 или 13.
May I ask why the answer is -12 or 13Я понимаю, что тут надо как-то рассматривать симметричные матрицы отношений людей. Но мы договорились не выходить за элементарные категории.
Easy.
Suppose there is Petya and a class, and let the class consist of 2 people. This means that
1 <-> 2
And the second one should start being friends with Petya
2 <-> 1
2 <-> П
Otherwise, the condition is violated. Add one more.
3 <-> 1
3 <-> 2
3 <-> П
But if we add only one person at the beginning, we break the condition:o)
Можно узнать почему ответ -12 или 13Why is this answer attributed to me? I have a different answer :o/
Почему этот ответ приписывают мне? У меня другой ответ :о/
I'm not attributing it to you.
You can see it was written by Alexei.
Mischek, I don't know why 12 or 13 myself. But I have reason to believe the person who wrote this answer.
OK, narrowing down the possible alternatives.
Suppose Petya is "24". Then due to the even number of friendship relations in the class, it turns out that {Others} has the following configuration: from "0" to "24" without repetition. So, we have two people "24" in our class - Petya and someone else. They are friends with everyone except the "0" person.
Let's look at '1', who must also be in the class. He must be friends with both "24". Contradiction.
So far we have excluded 4 options for Petya:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 13 14 15 16 17 18 19 20 21 22 23 24 25
The task is incorrect, I think. Nothing complicated: number 26 objects with numbers from 0 to 25 or 1 to 26. From 0 to 25 is impossible under the conditions of "friendship definition" - then 25 doesn't fit, and from 1 to 26 is impossible because 26 friends can't be among 25 people.
There really shouldn't be a zero number in the numbering by number of friends, because then there is a contradiction: there just aren't enough numbers for DIFFERENT numbers.
Petya made a mistake.
I want to stress again that {Others} can in principle only have two configs - {"0", "1",... "23", "24"} or {"1", "2",... "24", "25"}. This is very important.
Proof:
"0" and "25" cannot be present in {Others} at the same time ("25" must be friends with all, including "0"). Consequently, one of these numbers must be absent from {Others}. You can remove one of them, eliminating this contradiction of simultaneity, in two possible ways (by removing "0" or by removing "25"), and then we get exactly 25 remaining ones, as there were 26 possible numbers before - from 0 to 25.
AlexEro, the task is absolutely correct. You should number {Others} and Petya separately, and only then analyze.
Я хочу подчеркнуть еще раз, что у {Остальных} принципиально может быть только две конфиги - {"0","1",..."23","24"} либо {"1","2",..."24","25"}. Это очень важно.
Доказательство:
"0" и "25" не могут присутствовать в {Остальных} одновременно ("25" должен дружить со всеми, включая "0"). Следовательно, одно из этих чисел должно отсутствовать у {Остальных}. Убрать какое-то одно из них можно двумя возможными способами, мы получаем ровно 25 оставшихся, т.к. до этого было 26 чисел - от 0 до 25.
AlexEro, задача абсолютно корректна. Нумеровать нужно {Остальных} и Петю отдельно, а только потом анализировать.
Oh, that's it!
So the problem should be understood as "Petya has the same number of friends as one of his classmates"? Well, then what's easier: the problem is correct, all of them are numbered from 1 to 25 (or from 0 to 24), and Petya is assigned a number ANYTHING from 1 to 25 (or from 0 to 24).
То есть условие задачи следует понимать как "у Пети количество друзей совпадает с одним из одноклассников"? Ну тогда чего проще: задача корректна, все они нумеруются от 1 до 25, а Пете присваиваем номер ЛЮБОЙ от 1 до 25.
It's not spelled out in the condition, but it's possible.
And two: I have already proved that Petya is not "0", "1", "24" or "25". So there's no way any Petya can be any Petya. Look at my calculations, if you don't mind, and tell me where I'm wrong.