[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 6
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Это в условии не прописано, но это возможно.
И второе: я уже доказал, что Петя - не "0", "1", "24" или "25". Так что любым Петя никак не получится.
You have not "proved" anything, colleague. You PROVIDED it. It was your APPROVAL - for clarity's sake. No f@@@ power can prove here - with that formulation of the problem - how Petya is different from Vasya in this class. And neither can you, colleague, I suppose. Petya has just noticed (post factum, as an OBSERVER) that the number of his friends is the same as one of his classmates, and that all the rest have DIFFERENT numbers. Can the solution of this problem depend on the observer?
What if Vasya noticed it EVERYTHING, a day before Petr? Then it is not Peter who has 12:13 friends (Ocean)?
Again: Petya did not notice that "he has the same number of friends as one of his classmates". He didn't care about that, it wasn't in the problem statement. But he did notice that the numbers of his friends were different.
Petya is singled out in a special way, it's his own view. Only one other person in the class could have exactly the same view. Everyone else will have a different view: the numbers of friends will not all be different.
This is solved using a similar approach.
Suppose there are 3 people in the class. Then the possible choices are 0,1,1 (Last Petya).
4 people: 0,1,2,1 and 1,2,3,2
5 persons: 0,1,2,3,2 and 1,2,3,4,2
6 persons: 0,1,2,3,4,2 and 1,2,3,4,5,3
7 persons: 0,1,2,3,4,5,3 and 1,2,3,4,5,6,4
etc.
i.e. we get a recurrent formula, when we exclude the most "friendly" one, we get cases when there is one less person in the class
Not finished yet...
Еще раз: Петя не заметил, что "у него количество друзей совпадает с одним из одноклассников". Ему на это наплевать, в условии задачи этого не было. Но он заметил, что у остальных числа друзей разные.
Петя спецом выделен, это его собственный взгляд. Только у одного другого человека в классе может быть точно такой же взгляд. У всех остальных он будет другой: количества друзей будут не все разными.
Uh-uh-uh-uh, no, that's not right. If Petya's number of friends does NOT match any of his classmates, then the problem is invalid, Petya is overextended in forex and is stupidly wrong in his analysis of class friendships. If it matches, then Petya can be anyone (because they are DIFFERENT by the terms of the problem).
The conditions are formulated in such a clever way (is this for 7th grade?!!!, BLEEP) that they should be understood as :
"Petya noticed that all of his 25 classmates (( NOT counting HIMSELF!!!). Which Petya is unique in that the number of his friends is the same as Vasya's - also unique))) different number of friends in this class. How many friends can Peter have?"
Right, it looks like it's going to be a bit hard to do without matinduction.
By the way, for 3 people {1,2}|1 is still possible.
Если у Пети число друзей НЕ СОВПАДАЕТ ни с одним из одноклассников - задача некорректна.
This condition is not in the problem, AlexEro! It may be a deduction from the logic in solving it, but it is not there in the first place! The incorrectness of the problem implies that its conditions are contradictory.
"Petya noticed that all his 25 classmates ((( NOT counting HIMSELF!!!! Which Petya is unique in that the number of his friends is the same as Vasya's - also unique))) different number of friends in this class. How many friends can Peter have?"
The highlighted in blue was not in the condition! Why is the original statement unclear?
"Petya noticed that all his 25 classmates have a different number of friends in this class. How many friends can Petya have?"
Right, it looks like it's going to be a bit hard to do without matinduction.
By the way, for 3 people {1,2}|1 is still possible.
Yeah, right.
but the main thing is that excluding the friendliest one we go to the previous step for which we already have a solution. Thereby it is proved that there are no other solutions, whatever number of people in the class, there are always two.
Now all that's left is to formalise it all.
Just don't start with Petya, leave Petya as an appetizer, and number his friends with X, and number the others with a series of numbers from 0 to 24 or from 1 to 25 - there are only TWO numbering options, there can't be any other, can there? Then you will see that the LAST number in any numbering option is either 24 or 25 ...... You want PETER! - Because for the last number (24 or 25), there simply aren't enough PEOPLE (if without Petty). But if someone (at least one) is friends with Petya, then Petya must have a number not 0, but at least 1, 2, 3,....24, 25, which are all already taken.
It's a piece of cake.
But you can't trick kids with tricky conditions. It's immoral. That's how you discourage maths.
So what is the solution, AlexEro?
P.S. This is clearly an Olympiad problem. No ordinary school would torture poor kids with it. But those who take part in olympiads (or study in PE schools), this problem will only excite them.