[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way
OK, let's start to get a grip on something. Let's divide the class into two sets, {Petya} and {Other} (there are 25 of them). A person who has N friends, for convenience we will call "N".
Suppose Petya has 0 friends. Then {Other} can have from 0 to 24 without repetition (a person "25" cannot exist, since he must be friends with everyone, and we already have Petya, who is "0").
But there can't be a person "24" either, as we have two "0's" who are not friends with anyone, and therefore he is not friends with both of them either.
Consequently, for 25 {Others}, only options from 0 to 23 remain. Contradiction.
Similarly, it is proved that Petya cannot have 25 friends (if it were, then {Other} is from "1" to "25". But two people "25" and the existing "1" is a contradiction, since "1" would have to be friends with both "25").
More subtle reasoning shows that Petya cannot have and only 1 friend. And then I'm stalled.
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A professor asks a student:
P: what is the probability that you will meet a dinosaur when you walk out of the house?
С: 50%
P: why?
S: because I either meet it or I don't :)
Svetik, mistresses meet the same requirements as friends: if A is B's lover, then B is A's lover, so let's consider Petya an adult.
Continuing the binge. Obviously, there cannot be "0" and "25" people at the same time in the set {Others}. Consequently, {Others} can have only two possible configurations, either from "0" to "24" or from "1" to "25".
Продолжаем пьянку. Очевидно, что во множестве {Остальных} не может быть одновременно людей "0" и "25". Следовательно, {Остальные} могут иметь только две возможные конфигурации - либо от "0" до "24", либо от "1" до "25".
If binge drinking, 25-(I can't think of exactly 1 to 3). It seems to me that the condition "that all his 25 classmates have a different number of friends in that class" would be met in this case. But that's without taking into account the sexual relationships.
Or maybe not :o)
Petya has only one friend 26, because only the last 26 is friends with everyone, including Petya himself.
If anyone else is friends with Petya besides 26, then 26 himself would not have the same combination as the others.
.....Damn it's work to do ))))
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The problem is from the Mechmatov forum, here.
Петя заметил, что у всех его 25 одноклассников различное число друзей в этом классе. Сколько друзей может быть у Пети?
Comment:
1. Petya is also in this class, i.e. there are 26 people in total in the class.
2. If A is friends with B, then B is friends with A.
Find all solutions.
In the same branch the solution is given - 12 or 13.
Such a categorical answer is astonishing. I began to ponder at my leisure and came to some conclusions. But it's a long way to solving the problem. Whoever is interested, join me.
But please don't google and rable, or it will become uninteresting. Surely the problem is solved elementary.