[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 282
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Кстати, вот приведенное в задачнике решение задачи о 5 числах (и не только о 5):
That's funny. I'll think about it at my leisure.
For an arbitrary number of numbers n the divisibility of the sums is obvious.
Mutual simplicity:
For i#k (both numbers <= n), the numbers i*n!+1 and k*n!+1 are not divisible by 2, 3, ..., n (numbers i, k are in the same group up to n). Hence, if they are not mutually prime, then all their non-trivial common prime divisors must be greater than n.
On the other hand, their difference (i-k)*n! must be divisible by the NOD of the two numbers.
But (i-k)*n! is not divisible by any prime greater than n. So the NOD is trivial, i.e. it equals 1.
Spent two hours on your task. Couldn't do it. Got drunk. Didn't help. Well, not everyone's a PhD :)
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Mystery. This thing has everything to do with the beautiful half of humanity, all the ladies have seen it, but not all the ladies have seen it like this.
The question is, what is it and what do they need it for?
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Anything, even a cutting board. There's too little information.
Richie, where's the brain break here, please tell me?
Anything, even a cutting board. There's too little information.
Richie, where's the brain break here, please tell me?
Mathemat, if you turn this thing upside down, any man can guess what it is, but looking at this side, only a woman can guess :)
And then, why break your brain on a holiday?
Here I am reminded of an aphorism by W. Hugo: You cannot tell a woman anything that is difficult for her to understand. She begins to think about it, and her thoughts often take a bad turn :)
Another aphorism, it seems Wilde: A woman is created to be loved, not understood :)
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By the way, almost all 8th graders know what it is :) When you find out, you'll laugh.
Кстати, почти все 8-ми классники знают, что это такое :) Когда вы узнаете, вам будет смешно.
A rectangular parallelepiped. Heehee.
// Sergei, not everyone is funny when they look in the mirror.
I am reminded of a problem that hardly anyone has said anything sensible about. I myself don't know how to solve it yet.
Probably, the angle is not Pi (otherwise any odd function that doesn't necessarily satisfy (a) ) would do.
MetaDriver, shall we torture her? Not necessarily right away. The problem seems to be a serious one. 10th grade.
Mathemat писал(а) >>
The angle is probably not Pi (any odd function).
Why not? There is nothing about it in the problem. I.e. maybe you are complicating it for nothing (possibly to the point of being unsolvable).
Let's start without this addition. Then we can consider that you have already answered the second question.
The first remains - prove that for any odd function there exists exactly one fixed point.
// (as well as for any other satisfying the condition, if any).
You can't prove it. Here is an odd function defined on R: y = 3*x^3 - 2*x. It has three fixed points: 0, +1, -1.
Дык не докажешь. Вот тебе нечетная функция, определенная на R: y = 3*x^3 - 2*x. У нее три неподвижные точки: 0, +1, -1.
Yes, I've already found a bunch myself while smoking. Could be infinitely many: y=x; y=sin(x)+x etc.
// In short, I don't quite understand the condition then. Well if with your assumption... now... I'll go think about it some more.