[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 224
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And immediately - a new one, which may be of interest not only to the "advanced" (8th grade):
Cauchy I have managed to forget, I studied at the institute long ago, but my intuition tells me that you can not, if of course all the conditions of the problem are met.
The milk puzzle prompted another, quite original one about water. Hint: I recommend solving it by drawing on paper - it's easier. You can also do it in your head, but it is not easy to reproduce the solution afterwards.
There are three vials with volumes of 14, 9 and 5 litres. The first vessel is filled to overflowing with water. The other two are empty. Goal: pour water from one vessel to another to reach 7 litres in the first vessel. Special features: you cannot pour water out, you can only overflow the water by filling the vessel completely, not by overflowing it.
И сразу - новая, которая может заинтересовать не только "продвинутых" (8 класс):
The boy seems to be 18 years old, in the army and under the watchful eye of his grandfathers, hovering in his kitchen outfit:)))
Naturally, the boy is as immortal as a muzik (he can hardly do one operation faster than a second), the quantities in the glasses he aligns mathematically accurately, and the milk doesn't evaporate or spill.
Generally speaking, the problem is incorrect. It can be understood in two senses.
1. "Finite" problem: consider that his problem is solved if he has exactly equalized the quantities of milk in all glasses in a finite number of steps.
2. "Infinite" problem: Let's assume that the problem is solved in principle if for any predetermined inaccuracy epsilon he can specify such an algorithm which equalizes the amounts of milk with this accuracy.
The notion of limit is not yet known to eighth graders, so it is logical to assume that it has to be solved in the first sense.
For two glasses the problem is always solvable from the first step. But for three, how?
P.S. Mathematical formulation of the "final" problem - without boys and milk - is approximately as follows: There are 30 numbers a_1, a_2, ... a_30. At each step of any two can be replaced by their arithmetic mean. Is it possible to make all the numbers equal in a finite number of steps?
This is a strange task. For three glasses, equalise the largest and the smallest. repeat until satisfaction is achieved. Each operation increases the accuracy of the equation. Somewhere at the molecular level we can stop:)
Something this procedure reminds sorting.
No, no, nothing infinite, just a finite number of steps! Eighth-graders don't know the limit!
I think I know where to dig. I'm going to watch you guys struggle around here.
Try taking a closer look at the case of three glasses where two have 100 grams of milk and one has 130 grams. Can you make a finite number of overflows to equalise?
No, no, nothing infinite, just a finite number of steps! Eighth-graders don't know the limit!
I think I know where to dig. I'm going to watch you guys struggle around here.
Try taking a closer look at the case of three glasses where two have 100 grams of milk and one has 130 grams. Can you make it even out in a finite number of pourings?
Well if to a gram then yes, but in a thousand years. for very catastrophically the degree of equalisation in glasses of volume drops, well almost vertically.
ну если до грамма то да, но через тысячу лет. ибо уж очень катастрофически степень выравниваемости в стаканах объёма падает, ну почти вертикально.
Why do you have to go to a gram in a thousand? A gram can be done in ten minutes. But more precisely...
The correct answer is: If the number of atoms of each species is divisible by the number of glasses, then you can. Otherwise you can't.
;)
А для трех существуют такие неодинаковые числа, с которыми получается выравнивание за конечное количество шагов?
That's easy. For example: 2, 3, 4. In one step, turn it into 3, 3, 3.