[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 3
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Факт. Если А друг Б, не означает что Б друг А. Но вот если А спит с Б, автоматически означает, что Б спит с А, даже если при этом Б спит с остальными буквами алфавита, ну или я очень отстал от жизни)
Eh, mathematicians.
Have you decided for B yourself?)
Vitya, only one friend of Petya is also impossible. But first a simple observation: as "friends" ("lovers") is a reciprocal category, the sum of indices "N" (number of friends) over the whole class is necessarily even.
Suppose it is so, i.e. Petya is "1". Then there is only one variant of configuration {Other} - from "1" to "25" (because the sum of the arithmetic progression from 1 to 25 is 26*25/2 = 13*25, i.e. is odd. It is odd + 1 is even).
So we have two people "1" (Petya and someone else) and one person "25". These two people "1" are only friends with "25" and no one else. Contradiction: the guy "24" does not exist because none of the guys "1" is friends with him.
Swetten >> Вот уж совершенно не факт! :)
So, B does not know about it (or pretends to) - even with the obvious fact of physiology on A's part. Well, yes, it is possible and may even be nice :)
1-Петя
....
1(Петя)
It seems to me that if from the second/third they don't start being friends with Petya, that simply violates the conditions of the problem. So that's 25-2=gdThat's 20, 21, well maybe 23, ...
Figaro, can you show me the graphical solution?
2 Farnsworth: but the answer is 12 or 13.
ОК, начнем, чтобы за что-то зацепиться. Разделим класс на два множества - {Петя} и {Остальные} (их 25 человек). Человека, имеющего N друзей, для удобства назовем "N".
Допустим, у Пети 0 друзей. Тогда у {Остальных} может быть от 0 до 24 без повторений (человека "25" не может быть, так как он должен дружить со всеми, а у нас уже есть Петя, который есть "0").
Но и человека "24" тоже не может быть, т.к. у нас есть двое "0", которые ни с кем не дружат, и, следовательно, он с ними обоими не дружит тоже.
Следовательно, на 25 {Остальных} остаются только варианты от 0 до 23. Противоречие.
Аналогично доказывается, что у Пети не может быть 25 друзей (если бы было так, то {Остальные} - это от "1" до "25". Но два чела "25" и существующий "1" - это противоречие, т.к. "1" должен был бы дружить с обоими "25").
Более тонкое рассуждение показывает, что у Пети не может быть и только 1 друг. А дальше я застопорился.
I don't quite understand the 0 thing. So Peter has no friends and one out of 25 has no friends in this class. What's so unusual about that? The others are all friends.
Only one friend Petya has is also impossible. But first, a simple observation: since "friends" ("lovers") is a reciprocal category, the sum of indices "N" (number of friends) over the whole class is necessarily even.
Suppose it is so, i.e. Petya is "1". Then there is only one variant of configuration {Other} - from "1" to "25" (because the sum of the arithmetic progression from 1 to 25 is 26*25/2 = 13*25, i.e. is odd. This is odd + 1 is even).
So we have two people "1" (Petya and someone else) and one person "25". These two people "1" are friends only with "25" and with no one else. Contradiction: the person "24" does not exist, because none of the people "1" are friends with him.
Nobody has ever said that the number of friends of Pete can't be equal to the number of friends of someone else. The point was that Petya's classmates have a different number of friends. Petya's number of classmates' friends includes.
Не совсем понял про 0. Ну нет У Пети друзей и ещё у одного из 25 нет друзей в этом классе. что тут необычного? Остальные все попередружились.
Let's stick to the terms of the problem, Constantine. Two people "0" and an existing "24" are logically incompatible.
have forex accounts and they all trade and he doesn't....))))
1-Петя
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
2:3
3:24
4:235
5:2346
6:23457
7:234568
8:2345679
9:234567810
10:2345678911
11:234567891012
12:23456789101113
13:2345678910111214
14:234567891011121315
15:23456789101112131416
16:2345678910111213141517
17:234567891011121314151618
18:23456789101112131415161719
19:2345678910111213141516171820
20:234567891011121314151617181921
21:23456789101112131415161718192022
22:2345678910111213141516171819202123
23:234567891011121314151617181920212224
24:23456789101112131415161718192021222325
25:2345678910111213141516171819202122232426
26:2345678910111213141516171819202122232425 1(Петя)
I don't understand, people, why you don't like this solution (above): i.e. "from zero to 25"?
The problem boils down to simply NUMBERING all the students with numbers from 0 to 25 (there are 26 numbers in total, from 0 to 25, numbers not repeated). The number assigned means the number of friends. The numbers vary. Maximum can be 25, so minimum can be 0 (a loner who is not friends with anyone). Including the name "Petya" only obscures the problem, because the person "Petya" is in no way distinguished from the others in the problem, except for the fact that the number of friends is different, which allows each student to be numbered according to the number of his friends.
Давай соблюдать условия задачи, Константин. Два чела "0" и существующий "24" несовместимы логически.
Why, because it says that the number of friends is different? Well, who says that 0 for one of those 25 is not a "different" number of friends? It's not quite clear, 0 doesn't count as a number anymore?