Pure maths, physics, logic (braingames.ru): non-trade-related brain games - page 185
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It seems to me that the fourth wave will only accelerate the bird once, after which the bird will begin to move away from it.
There is no reason for them to meet a second time.
If you are interested, I will send the solution of the Leopold and the mouse problem to you.
Next:
Megamogg has invented a mechanical bird for his child. The bird flies vertically upwards and can instantly lose speed from a clap of the hand. The starting speed of the bird is 6 m/s. What is the maximum speed that can be given to the toy? Damping of sound in the air is neglected.
The weight of the problem is 3.
...
Bird speed 13122 m/s
Stopped by the branch on my own head... =)
Note: you won't find a similar rebuttal to the cat sequence 2,3,4,2,3,4. Don't even try (but you will anyway).
And what's wrong with "5,4,5,4,5,4" (or "4,5,4,5,4,5" if the sequence of moves is different)?
The answer is correct. I have removed it.
It seems to me that the fourth wave will accelerate the bird only once, after which the bird will begin to move away from it.
There is no reason for them to meet a second time.
Exactly, that's exactly right.
Well, for example, like this: to the cat sequence 5,4,5,4,5,4 the mouse responds with the following: 4,5,4,5,4,5.
That's just one option, the trickiest one. But there are plenty of other, simpler ones.
Here's the sequence: 2, 3, 4, 2, 3, 4.
It can be changed (e.g. 4, 3, 2, 2, 3, 4), but the essence of the strategy does not change.
Now try to explain the meaning of each move. That's the essence of the solution.
The simplest explanation is through parity.
The mouse changes parity on every pass. Leopold's first "anti-synchronous" pass guarantees the parity of finding the mouse, and the second one synchronizes by parity and eats lunch.
// Did you steal a problem from the market makers? ;)
Hello, everybody.
The tea/coffee problem was interesting, but the answer is bad - not very impressive.
If only one product is divided into slices, the result is strikingly different from the result of splitting both drinks into small doses and running two queues of slices towards each other.
The graph shows the temperature at the end of the procedure when slicing both (tea/coffee) and only one (tea 2/coffee 2) beverages, by increasing N from 1 to 1000
// For simplicity and clarity the starting temperatures are 100 and 0 degrees.
It is clearly visible that when splitting into two batches, in the limit the drinks fully exchange temperatures (at N -> ∞)
When splitting only one - not even close to this result (tried to estimate the limit - no brains to calculate analytically, the numerical result looks like (t1*2+t2)/3)
// In the trailer - calculation and graph in Excell
Couldn't help but post :)
The simplest explanation is through parity.
The mouse changes parity on every pass. Leopold's first "anti-synchronous" pass guarantees the parity of finding the mouse, and the second pass synchronizes with the parity and has lunch.
Well, finally I see the ideologically correct answer! I solved it this way:
/ Solution removed - Mathemat/.
I'll delete the solution in a few hours.
It says it's bearded. There's a similar problem about an artilleryman and a fighter. Only there are more burrows - as many as 60.
It would be better to take 95 and 5.
When dividing the tea into 2 parts (and not separating the coffee) I get tea-55 and coffee-45. So I'm wondering how much more you can increase the difference in the final temperature.
sanyooooook says it is bearded. There is a similar problem about an artilleryman and a fighter. Only there are more burrows there - as many as 60.
Well, the one I first saw was only five holes, and the link to which I threw there are 60, apparently mathematicians 5 is not enough)
ZZY: found "Science and Life" in 75 years, there is a math section with problems, the answers are unfortunately only from the previous issue )
ZZSY: I lost it as I found it), but there is one for 96 I will scrap it, it is on logic.