[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 172
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На острове Серобуромалин обитают 13 серых, 15 бурых и 17 малиновых хамелеонов. Если встречаются два хамелеона разного цвета, то они одновременно меняют свой цвет на третий. Может ли случиться так, что через некоторое время все хамелеоны будут одного цвета?
No. I don't know how to prove it. But if one is drowned, it will work.
Интуиция мне подсказывает, что нет. Но, нужно думать.
I don't have to think about it, 17 crimson ones are removed, 13 grey and 15 brown ones are left, a total of 13 meetings to become crimson, 2 brown ones remain and those 17 crimson + 26 transformed = 45 (then they will meet again until 1 crimson and 46 grey remain)
Obviously, if there were the same number of chameleons, this would be possible. Then let's rephrase the problem:
remove 13 chameleons of each colour: 2 brown and 4 crimson remain.
My answer is that it is not possible.
About the answer - Mischek has already answered. What I mean is that a rigorous conclusion from zero is unlikely to work - too cumbersome an expression to find the optimum.
At first (when solving for the very first time) I didn't know the answer, so I had to prove it in two steps and it was hard.
Knowing the answer, getting the analytical formulas is of course not difficult and easier to prove.
I think that if you draw three straight lines through the points where the circles touch each other, then 1) they will intersect in one point, 2) they will divide the hatched area into three non-intersecting parts, 3) the problem will be reduced to the need to prove that the part of the hatched area which remains between the circle and the Mischek tangent is minimal compared to other tangents.
We need rigorous proof. We cannot remove chameleons without substantiating that we can do so.
Ah, I see.
These chameleon numbers are
3*x 3*x-1 and 3*x+1, i.e. the numbers have different residuals from division by 3.
The described action (colour change) does not change the properties of the group and the residuals remain different, i.e. the two groups can never have the same number due to the difference of the residuals.
Нужно строгое доказательство. Мы не можем убрать хамелеонов, не обосновав, что можем это сделать.
It is possible that 17 crimson chameleons will never meet 13 grey or 15 brown (theoretically) and then when the island is left with 2 brown and 45 crimson, the crimson will slowly meet the brown
OK, the chameleons have been cut. But the next one I don't understand myself, as if it were stupid:
There is a fraction 10/97 on the board. It is allowed to add the same number to the numerator and denominator, or multiply the numerator and denominator by the same number. Is it possible to get a fraction equal to a) 1/2; b) 1 as a result of several such operations?
I think I got it, sorry.
2 sanyooooook: well, it's possible, but does it cover all cases?
ОК, хамелеонов порешили. А вот следующую сам не понимаю, как будто бы глупость:
На доске записана дробь 10/97. Разрешается прибавлять к числителю и знаменателю одно и то же число или умножать числитель и знаменатель на одно и то же число. Можно ли в результате нескольких таких действий получить дробь, равную а) 1/2; б) 1?
numbers are only integers?
What a challenge - a Polish scientist proved that God exists. Quote - "Geller has developed a complex formula that allows everything, even randomness, to be explained by mathematical calculations".
I think there are overlaps here. The newspapermen, as always, have got a lot mixed up and are trying to "retell" things they cannot understand in principle. In particular, he received the Templeton Foundation Prize back in March 2008. And it is £820k, not euros. Some of the newspapermen only stumbled upon it now, published a "hot" fact, and the rest rushed to reprint it. In the said article the previous sentence is much more interesting: "The professor's theories not only contain proof of the existence of God, but they also make one question the material existence of the world around us".
А мужик конечно серъезный. http://www.wikiznanie.ru/ru-wz/index.php/%D0%9C%D0%B8%D1%85%D0%B0%D0%BB%D1%8C_%D0%93%D0%B5%D0%BB%D0%BB%D0%B5%D1%80
It would be interesting to look at this "formula".