[Archive!] Pure mathematics, physics, chemistry, etc.: brain-training problems not related to trade in any way - page 598
You are missing trading opportunities:
- Free trading apps
- Over 8,000 signals for copying
- Economic news for exploring financial markets
Registration
Log in
You agree to website policy and terms of use
If you do not have an account, please register
Another one, a three-point scale. There is only one weighing. The scales allow you to see the exact difference in the weight of the coins.
There are exactly 50 fake coins among the 101 coins. The weights of all the real coins are the same; the weight of each counterfeit coin is 1 gram more or less than the real coin (counterfeit coins can have different weights). How can you determine in one weighing on a two-cup scale with an arrow and a scale (without weights) whether a given coin is false or not?
Let the statement be true:
Under the given conditions of the problem put equal number of coins on the scales. If the reading of the scales is even, then an even number of counterfeit coins are involved in the measurement, otherwise, an odd number.
Then the solution to the problem:
Put 50 coins each on the scales. If the reading of the scales is odd, then the number of counterfeit coins participating in the measurement is odd. That is, the coin on the scale is not false. Otherwise (the reading is even) all the counterfeits are on the scales, hence the coin not on the scales is real.
The proof of the statement is based on three obvious assertions.
1) If the same number of coins lie on the scales, then moving two arbitrary coins between the scales does not change the evenness of the scales.
2) Adding (removing) a real coin to each cup of the scales does not change the evenness of the scales.
3) If there are the same number of coins on the scales, all real coins on one and all fake coins on the other, the evenness of the scales corresponds to the evenness of the number of coins.
There was a puzzle they wanted badly. Here, solve it.
[The problem is rated 4 points, i.e. difficult.]
Black's move. Which piece stands on g4?
Another one. Bojan, but still can't solve it completely (partially solved, but it's an incomplete solution):
There are 10 prisoners sitting in a prison, each in solitary confinement. They cannot communicate with each other. One day the warden announced to them that he was giving everyone a chance to be released and offered the following conditions: "In the basement of the prison there is a room with a switch that has two states: ON/OFF (up/down). You will be brought in randomly one by one into this room and after a few minutes you will be taken out. While in the room, each of you can either change the position of the switch or do nothing with it. Prison staff will not touch this switch. At some point, one of you (any of you) must say that all the prisoners have been in the room. If he is right, everyone will be released; if he is wrong, you will stay in prison forever. I promise you that all the prisoners will be in the room and that every one of you will be brought back an unlimited number of times. The prisoners were then allowed to meet and discuss strategy, then separated into their cells. What do they have to do to be guaranteed release?
To clarify: the initial state of the switch is unknown. This makes the task very difficult. The SCs come into the room the way the jailers decide. They can't do anything other than turn the switch on/off. No notching, spitting or anything like that.Another one. Bojan, but still can't solve it completely (partially solved, but it's an incomplete solution):
There are 10 prisoners sitting in a prison, each in solitary confinement. They cannot communicate with each other. One day the warden announced to them that he was giving everyone a chance to be released and offered the following conditions: "In the basement of the prison there is a room with a switch that has two states: ON/OFF (up/down). You will be brought in randomly one by one into this room and after a few minutes you will be taken out. While in the room, each of you can either change the position of the switch or do nothing with it. Prison staff will not touch this switch. At some point, one of you (any of you) must say that all the prisoners have been in the room. If he is right, everyone will be released; if he is wrong, you will stay in prison forever. I promise you that all the prisoners will be in the room and that every one of you will be brought back an unlimited number of times. The prisoners were then allowed to meet and discuss strategy, then separated into their cells. What do they have to do to be guaranteed release?
To clarify: the initial state of the switch is unknown. This makes the task very difficult. The SCs come into the room the way the jailers decide. They can't do anything other than turn the switch on/off. No notching, spitting or anything like that.They have to agree that 5 people are responsible for ON and 5 people for OFF. Each person entering the cell, if the switch is not his, has to switch and count how many times he has hit the switch that is not his.
When someone counts to 20, everyone has been in the cell.
Nah, it's more complicated than that. There's only one person in charge. He's in charge.
And anyway, why is it under 20?
Nah, it's more complicated than that. There's only one person in charge. He's in charge.
9 can only switch on and 1 can only switch off. That is to reset the busy flag :)
When this one resets 9 times, it means all the SCs have been there.
There was a puzzle they wanted badly. Here, solve it.
[Problem is rated 4 points, i.e. difficult].
Black's move. Which piece stands on g4?
I'll begin...
1. How could Black's white-squared bishop get to a2? Obviously, only from square b1, where Black's passed pawn has turned into a bishop. With a bit of thinking it isn't hard to deduce the route of this pawn: e7 - d6 - c5 - b4 - a3 - a2 - b1F. Altogether we have 5 diagonal moves on its way, i.e. 5 captures, plus one of White's bishops eats a1, total 6 captures. We see that White is exactly six pieces short, so it immediately follows that only a black piece can be on g4.
2. How could white pawns g3 and h3 occupy their current positions? Black bishop on h2 suggests only one way - h2-h3, and then (after ...ch2) g2-g3. The variant White pawn hits h2-g3, then Black moves along the h-line and hits ...h2-g1, turning into bishop (and then White pawn hits someone g2-h3), is not suitable, because all 6 allowed captures of white pieces are already used up by Black.
3. from point 2 directly follows, that the passed pawn on b1 was the only passed pawn of Black, hence, pawns from lines f,g,h were either beaten by white pieces, or one of them (the one from the field g7) is on g4 now.
4. For g4 there is also the option of a knight and a white-squared bishop (not the one now on a2, but another one from the beginning of the game).
5. Black's move now. How did white just move? On reflection, we realise that the only permissible move would have been a long castling (if Le1-d1, then on the previous move Black's king is under check, and for Kr b1-c1 White is under check). But the rules of chess say that castling cannot be done via a broken square, so the bishop cannot be on g4. This leaves the options of knight and pawn.
6. Further on is still a jam. It is necessary to eliminate one of the options, I have not thought up how)))
when this one resets 9 times, it means that all the SCs have been there.
Another one. Bojan, but still can't solve it completely (partially solved, but it's an incomplete solution):
There are 10 prisoners sitting in a prison, each in solitary confinement. They cannot communicate with each other. One day the warden announced to them that he was giving everyone a chance to be released and offered the following conditions: "In the basement of the prison there is a room with a switch that has two states: ON/OFF (up/down). You will be brought in randomly one by one into this room and after a few minutes you will be taken out. While in the room, each of you can either change the position of the switch or do nothing with it. Prison staff will not touch this switch. At some point, one of you (any of you) must say that all the prisoners have been in the room. If he is right, everyone will be released; if he is wrong, you will stay in prison forever. I promise you that all the prisoners will be in the room and that every one of you will be brought back an unlimited number of times. The prisoners were then allowed to meet and discuss strategy, then separated into their cells. What do they have to do to be guaranteed release?
To clarify: The initial state of the switch is not known. This makes the task very difficult. The SCs come into the room the way the jailers decide. They can't do anything other than turn the switch on/off. No notching, spitting or anything like that.They have to choose one, let's call him 'The Chosen One'.
The Chosen One will count how many times the switch has been in the ON position when they visit the room and make sure to turn it OFF.
Each of the remaining 9 will only turn the switch ON once, they will never turn it OFF.
Accordingly, once the Chosen One counts nine ONs - everyone has been in the room.