[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 79
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No offence, Mischek, I've already apologised :)
Не обижайтесь, Mischek, я уже извинился :)
I'm not offended by anything, I'm just talking about the thread, about us ...
By the way, in your last answer, you didn't forget anything, like the proof
By the way, in your last answer, aren't you forgetting something, like proof
I'm thinking about it right now. It seems to be a combinatorial problem.
Если сумма любых 5-и чисел из 21-го является положительной, значит все эти 21 числа являются положительными, а следовательно их сумма не может быть отрицательной.
This means that there are no more than 4 negative numbers among them, and the smallest positive number exceeds the modulus of their sum. Correspondingly, if there are 3 negative numbers, then their sum is less (modulo) than the sum of the two smallest positive ones. And so on. Clearly, by adding the remaining positive sums to these, we get a positive number.
P.S. Oops, too late :)
Well, good for you, Matemat. He'll write a one-line problem and you can't fucking solve it :)
As far as I remember combinatorics, the number of placements for 21 elements in 5 elements:
21!/(21-5)!=21*20*19*18*17=2441880
Consequently, there can be 24441880 combinations of numbers and by convention all these combinations
yield positive results.
Keep thinking.
Although, the condition doesn't say that these numbers can't be equal.
OK, I have a different solution. For some reason I didn't get to the Dirichlet principle, although it's the right one here.
Take all the numbers in some given order and write this sequence 5 times in a row, then sum up all 105 elements. On the one hand, it is the sum of the original 21, and on the other hand, it is the sum of 21 fives.
The next one is a little more complicated, also from 9th grade:
There is a square. We intersect it with 9 lines, each of which divides it by area in the ratio 3:2. Prove that at least three of them intersect at the same point.
Минимальная плошадь полученной фигуры 2/5. следовательно с помощью паралельных прямых таких фигур можно разместить 2. 9 прямых - имеется в виду несовпадающих ведь? Следовательно уже третья прямая будет непаралельна первым двум - вот и три пересекающихся.
we have to be at one point