Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 111
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Try a stack of five bricks. How much more can you shift the topmost one relative to the bottom one?
For five you get an offset of a whole brick.
And for three it is also a brick, so the conclusion is that a post is considered a stack of at least three bricks ))
The problem doesn't seem to have an explicit solution, because the condition is not explicit.
The condition is explicit. You have five bricks. They are lying on top of each other. You have the ability to move them relative to the lowest one (it is attached to the ground). What is the maximal distance you can move the fifth one relative to the first one?
When you solve the problem for five bricks, try solving it for six, seven... ten... a hundred.
I'm such a dork, it turns out everyone has bricks in their house.
Shh. I believe you, you know the solution.
P.S. I forgot to add: the length of one brick is 1.
I will make a contribution.
Maybe who has already solved it, keep quiet.
On the end of a 1 m long rubber hose sits a muzik. The same end is attached to a tree. The other end is pulled at a speed of 1m/s, at the same moment the mutsik starts crawling towards the opposite end of the cord at a speed of 1 cm/s. Will the mucic crawl to the end of the cord? If no, prove, if yes, in what time?
The condition is explicit. You have five bricks. They are lying on top of each other. You have the ability to move them relative to the lowest one (it is attached to the ground). What is the maximal distance you can shift the fifth one relative to the first one?
When you've solved the problem for five bricks, try solving it for six, seven... ten... a hundred.
In short, it turns out, as long as the centre of gravity of the pole is within the base of that pole.
In short my brain goes up in smoke, it can not solve such problems.)
In short, it turns out as long as the centre of gravity of the pole is within the base of that pole.
There's a trick to solving the problem in a general way. You have to solve the problem from the end :)
P.S. 2 Mischek: you have a good memory (about Muzik). I myself was deeply shaken by this problem - perhaps that's why I remember it.
There's a trick to solving the problem in a general way. You have to solve the problem from the end :)
P.S. 2 Mischek: you have a good memory (about Muzik). I myself was deeply shaken by that task - perhaps that's why I remember it.
So I got a number of 1/2+1/3+1/4+1/5+1/6+1/7....., etc. (if the bricks are put from heaven to earth), which, as it animatedly hints a number of friends do not converge.
That is, if the bricks will not crumble (by the will of his naimakaroniesth omnipotence) then the shift on the infinitely high tower, turns out quite infinite.
// Pisa is smoking.