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On May 09 2012 21:03 frietjeman wrote:Make this sum right by using a single straight line in any way. 5+5+5=550 I don't know the answer  Saw this on a children's drawing plate and I was a bit amazed that I actually couldn't figure it out. Maybe the makers made a mistake or I'm just missing something really obvious here.
answer:
+ Show Spoiler +
545 +5=550
or
5+5+5(not equals sign)550
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Pandemona
Charlie Sheens House51481 Posts
Easy one :D Which i like...a Riddle too
What is once in a minute, twice in a moment, and never in a thousand years?
It's a very old simple riddle but i like ones like these
Hint:
+ Show Spoiler +Look Closely at the sentence
Answer:
+ Show Spoiler +The Answer Is........"M" once in a Minute twice in a MoMent and never in a thousand years :3
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+ Show Spoiler +Oh you can turn a + sign into a 4 I suppose. Hadn't thought of that ^^.
I don't think you can remove the equals sign though cause then you're not really solving the sum anymore, just removing it.
Didn't expect such a fast answer wow.
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On May 09 2012 21:20 Pandemona wrote:Easy one :D Which i like...a Riddle too What is once in a minute, twice in a moment, and never in a thousand years? It's a very old simple riddle but i like ones like these
+ Show Spoiler +
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man that pingpong ball one was really interesting. I had my mind set on + Show Spoiler + as well for a while, but I think it was because I had just read the riddle about "the family with 2 kids, and a boy answers the door, what's the probability the other child is also a boy." and seeing as that one is actually + Show Spoiler + I had a hard time changing my mind about the ping pong balls. I'm glad I kept reading all the comments! and to say I actually used to be good at stats... lol
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Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually.
Here is the problem again:
+ Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
+ Show Spoiler +At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
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Here some riddles for you guys to kill some time...
The Clock:
Sam and John was trapped in a room with only a digital clock in front of them. A speaker from nowhere told them that they can get out only if they solved the "Clock Riddle". Here it is: The clock will beep every time it reads 3 consecutive same number(e.g. 01:11). How many beeps will the clock do 2 days from now? At the beginning the clock reads 12:00am.
Numbers:
How can you place the numbers 1 through 9 in a 3x3 grid such that every row, column, and the two diagonals all add up to 15?
Assassination Attempt (This one is a bit harder):
The King of a small country invites 1000 senators to his annual party. As gifts, each senator brings the King a bottle of wine, for a grand total of 1000 bottles of wine. Each bottle is signed by the senator who gave it.
At the end of the party, the Queen tells the King that one of the senators is trying to assassinate him, and has put deadly poison in the bottle of wine he gave as a gift. Unfortunately, the Queen doesn't know which senator is the traitor (and thus doesn't know which bottle of wine has the poison in it).
The King has 10 servants. He views them as expendable, and does not care if they live or die. He decides to use them to figure out which bottle is poisoned, which will then indicate which senator is trying to assassinate him.
His plan is to make each servant drink from zero or more of the bottles of wine. The King knows that the poison is such that if a servant drinks it, he will feel fine until noon on the next day, at which point he will instantly drop dead.
The King must know for sure who the traitor is by noon on the day after the party, or else the traitor will try to find another way to assassinate him. This essentially means that he has one shot to make his servants drink the wine in order to figure out which is the poison wine.
Note that the King can make any of the servants drink from any of the wine bottles. He does not need to make all of the servants drink wine if he doesn't want to. Any servant who drinks from the poisoned bottle will die the next day at noon.
How can the King figure out for sure who the traitor is by noon on the following day?
Critical Hint (Don't check if you want to solve everything on your own):
+ Show Spoiler +Think of a grid, with 1000 rows representing the bottles of wine, and 10 columns representing the servants.
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On May 10 2012 05:11 Misder wrote:Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually. Here is the problem again: + Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng.
Well, I guess this is what I'm using my 500th post on.
+ Show Spoiler +
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random
P(B)=0.5 since there are equal number of balls of each color and they are drawn at random.
P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
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On May 10 2012 05:50 Aelfric wrote:
Numbers:
How can you place the numbers 1 through 9 in a 3x3 grid such that every row, column, and the two diagonals all add up to 15?
+ Show Spoiler +
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On May 10 2012 05:50 Aelfric wrote:Assassination Attempt (This one is a bit harder):The King of a small country invites 1000 senators to his annual party. As gifts, each senator brings the King a bottle of wine, for a grand total of 1000 bottles of wine. Each bottle is signed by the senator who gave it. At the end of the party, the Queen tells the King that one of the senators is trying to assassinate him, and has put deadly poison in the bottle of wine he gave as a gift. Unfortunately, the Queen doesn't know which senator is the traitor (and thus doesn't know which bottle of wine has the poison in it). The King has 10 servants. He views them as expendable, and does not care if they live or die. He decides to use them to figure out which bottle is poisoned, which will then indicate which senator is trying to assassinate him. His plan is to make each servant drink from zero or more of the bottles of wine. The King knows that the poison is such that if a servant drinks it, he will feel fine until noon on the next day, at which point he will instantly drop dead. The King must know for sure who the traitor is by noon on the day after the party, or else the traitor will try to find another way to assassinate him. This essentially means that he has one shot to make his servants drink the wine in order to figure out which is the poison wine. Note that the King can make any of the servants drink from any of the wine bottles. He does not need to make all of the servants drink wine if he doesn't want to. Any servant who drinks from the poisoned bottle will die the next day at noon. How can the King figure out for sure who the traitor is by noon on the following day? Critical Hint (Don't check if you want to solve everything on your own):+ Show Spoiler +Think of a grid, with 1000 rows representing the bottles of wine, and 10 columns representing the servants.
+ Show Spoiler +
I really don't want to do a truth table for 10 servants and 1000 bottles, so i'll do one for 3 and 8. The principle is the same and will always work if the number of bottles is equal or smaller than 2^(number of servants).
B is the bottle with the corresponding number, S1-S3 are the servants. 1 means "drink from bottle", 0 "don't drink from bottle".
B D1 D2 D3
1 1 1 1
2 1 1 0
3 1 0 1
4 1 0 0
5 0 1 1
6 0 1 0
7 0 0 1
8 0 0 0
If 3 servants die the next day, bottle 1 was poisoned. If D1 and D2 die, bottle 2 was poisoned. And so on...
For 1000 bottles and 10 servants, the first servant would have to drink the first 500 bottles, the second would have to drink bottles 0-250 and 500-750, and so on.
edit: writing this down took forever and it still looks kind of bad. Oh well, hope it's understandable anyway.
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On May 10 2012 06:18 frogrubdown wrote:Show nested quote +On May 10 2012 05:11 Misder wrote:Honestly, I'm really surprised that the ping-pong ball problem is still in discussion, and I'm glad actually. Here is the problem again: + Show Spoiler +On April 21 2012 03:07 TanGeng wrote:
A friend packed six ping pong balls for you, 3 orange, 3 white. He's placed them in three pockets of your sports bag. One pocket has two orange balls. One pocket has two white balls. One pocket has one white and one orange. You reach into one random pocket and pull out an orange ping pong ball. What's the probability of the other ball in the pocket being orange?
At first, I thought the probability was 2/3, but after doing some research inspired by the post above, I now believe the answer is 1/2. Take a look at this wikipedia article: http://en.wikipedia.org/wiki/Boy_or_Girl_paradoxThe question posed by TanGeng is definitely more related to the paradox in the wiki article than the Monty Hall problem. The explanation given there is very similar to the explanations given in this thread for why it is 1/2 but in much more detail. A lot of you guys are interpreting the question as the first interpretation in the Second Question section given in the wiki article, which is not the way the question is framed by TanGeng. Well, I guess this is what I'm using my 500th post on. + Show Spoiler +
The difference between the two cases is that in the ping pong case, there is only one mixed pocket whereas in the boy/girl case there are two different mixed possibilities (older girl and younger boy or older boy and younger girl). The latter case gives 50% chance if the child who is a boy was selected at random because even though there are two ways for there to be a girl sibling, each way is half as likely to result in a boy being randomly selected as the two boy case. In the ping pong case, the mixed hole still gives half the chance of returning the orange ball as the double orange hole, but since there's only one mixed hole this means that the chance that the other ball is white is half the chance that the other is orange.
There really isn't any reason to have to argue by vague gestures to analogous situations. You can just learn Bayes' Theorem and learn how to use it. Let 'A' stand for proposition that you drew from the double-orange hole and 'B' for the proposition that the drawn ball was orange. By Bayes' theorem:
P(A|B)=P(B|A) x (P(A)/P(B))
P(A)=1/3 since there are three pockets and the draw is random
P(B)=0.5 since there are equal number of balls of each color and they are drawn at random.
P(B|A)=1 since any draw from the double orange hole yields an orange.
Therefore,
P(A|B)=1 x (1/3)/0.5=2/3
+ Show Spoiler +zzzzzzzz. I'm upset. I spent so much effort convincing myself that the answer was 1/2 that I completely forgot that P(orange-orange) =/= P(orange-white) given orange is chosen, which, of course, is the basis of the problem anyway. herp derp. And I try Bayes'... but completely botched it when I did it earlier cause I got 1/2...
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On May 10 2012 05:50 Aelfric wrote:
Numbers:
How can you place the numbers 1 through 9 in a 3x3 grid such that every row, column, and the two diagonals all add up to 15?
In addition to this answer (as there are many if you count rotational symmetry), here's a quick and easy method to answer that for any odd-numbered square.
+ Show Spoiler +start at the top row, in the center with the number 1. Assume rows and columns wrap around (e.g. going "up" from the first row ==> last row, and so on). Go up and right, and place the next number there. If a number exists in that position already, then instead of going up and right, go down once instead. Continue until square is filled.
In this case, this would result in the following square
816
357
492
If you'll notice, that's the same square as frog's except rotated 90* ccw and flipped horizontally!
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On May 10 2012 07:05 JeeJee wrote:Show nested quote +On May 10 2012 05:50 Aelfric wrote:
Numbers:
How can you place the numbers 1 through 9 in a 3x3 grid such that every row, column, and the two diagonals all add up to 15?
In addition to this answer (as there are many if you count rotational symmetry), here's a quick and easy method to answer that for any odd-numbered square. + Show Spoiler +start at the top row, in the center with the number 1. Assume rows and columns wrap around (e.g. going "up" from the first row ==> last row, and so on). Go up and right, and place the next number there. If a number exists in that position already, then instead of going up and right, go down once instead. Continue until square is filled.
In this case, this would result in the following square
816
357
492
If you'll notice, that's the same square as frog's except rotated 90* ccw and flipped horizontally!
+ Show Spoiler +There aren't that many solutions, there are only eight 3x3 magic squares. There's only one, really, and then its seven symmetric counterparts (rotate it 90 degrees three times to get the next three of them, then take its mirror image and rotate that four times to get the other ones. For n>3, however, there are lots and lots of ways to arrange the numbers from 1 to n^2 in a grid with row/column/diagonal sums equal. For example, there are 880 (7040 if you count symmetric copies as different) 4x4 such squares, and so on.
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On May 10 2012 05:50 Aelfric wrote:Here some riddles for you guys to kill some time... Assassination Attempt (This one is a bit harder):The King of a small country invites 1000 senators to his annual party. As gifts, each senator brings the King a bottle of wine, for a grand total of 1000 bottles of wine. Each bottle is signed by the senator who gave it. At the end of the party, the Queen tells the King that one of the senators is trying to assassinate him, and has put deadly poison in the bottle of wine he gave as a gift. Unfortunately, the Queen doesn't know which senator is the traitor (and thus doesn't know which bottle of wine has the poison in it). The King has 10 servants. He views them as expendable, and does not care if they live or die. He decides to use them to figure out which bottle is poisoned, which will then indicate which senator is trying to assassinate him. His plan is to make each servant drink from zero or more of the bottles of wine. The King knows that the poison is such that if a servant drinks it, he will feel fine until noon on the next day, at which point he will instantly drop dead. The King must know for sure who the traitor is by noon on the day after the party, or else the traitor will try to find another way to assassinate him. This essentially means that he has one shot to make his servants drink the wine in order to figure out which is the poison wine. Note that the King can make any of the servants drink from any of the wine bottles. He does not need to make all of the servants drink wine if he doesn't want to. Any servant who drinks from the poisoned bottle will die the next day at noon. How can the King figure out for sure who the traitor is by noon on the following day? Critical Hint (Don't check if you want to solve everything on your own):+ Show Spoiler +Think of a grid, with 1000 rows representing the bottles of wine, and 10 columns representing the servants.
+ Show Spoiler +So this problem should be possible for up to 1024 bottles of wine. For each bottle, assign some combination of servants to drink it. (so bottle 1 can have no one drink it, bottle 2 just servant 1, etc. Since each of the 10 servants can either drink or not drink (2 possibilities), the number of unique bottle combinations is 2^10 = 1024. Upon seeing what combination of servants die, the correct bottle can be determined.
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Sanya12364 Posts
The Seat Snatcher Manager
A company bus with twenty seats picks up twenty employees at their homes and shuttles them to and from work everyday. The twenty employees each have an assigned seat from the back to the front in the order they are picked up. The first employee on the bus route is an obnoxious manager with unpredictable behavior. Every morning, the manager boards the bus randomly at any one of the twenty stops. The manager cuts in front other employees at the stop, boards the bus, sits in an unoccupied seats at random, and then yaks away on the cell phone.
All other employees roll their eyes at the behavior. When they board the bus, if they sit in their assigned seat if unoccupied. Otherwise, they pick an unoccupied seat at random and sit there. You are Frank, the last employee to be picked up along the bus route. How often do you sit in the frontmost seat?
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On May 10 2012 05:50 Aelfric wrote:
Here some riddles for you guys to kill some time...
The Clock:
Sam and John was trapped in a room with only a digital clock in front of them. A speaker from nowhere told them that they can get out only if they solved the "Clock Riddle". Here it is: The clock will beep every time it reads 3 consecutive same number(e.g. 01:11). How many beeps will the clock do 2 days from now? At the beginning the clock reads 12:00am.
+ Show Spoiler +I think it ends up being 34 each day for 68 over 2 days. times should include:
12:22
1:11
2:22
3:33
4:44
5:55
10:00
11:10
11:11
11:12
11:13
11:14
11:15
11:16
11:17
11:18
11:19
Once you hit 12pm it goes back to the top to repeat again.
If for whatever reason 11:11 beeps twice due to being able to be split into 2 sets of 3 1's then it would be 36 per day for 72 over the 2 days.
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Sanya12364 Posts
Mountaineers and Sherpas
A mountaineer is attempting a climb through a mountain gap from base camp A to base camp B. The route takes six days. At base camp A, the mountaineer finds sherpas for hire. He can get them to carry his supplies and equipment but must feed them and must not allow the sherpas to starve. The mountaineer and the sherpas can carry a maximum of four days supply while climbing. How does the mountaineer get from base camp to base camp in six days? How many days of supply does he buy?
The next year, the mountaineer returns to make the climb but no sherpas are available. How does the mountaineer make the climb on his own?
+ Show Spoiler +
The mountaineer hires two sherpas, who we will call Tarjee and Harjee, and buys 12 days worth of supplies. The mountaineer with his two companies set off with full packs. After the first day, Tarjee gives up two days' worth of supplies and returns home. Harjee and the mountaineer each receive one day's worth of supplies and continue the climb. After the climb on the second day, Harjee gives up one day's worth of supplies and return home. The mountain again receives one day's worth of supplies. He has four days' supply and four days' climb and makes it to Camp B.
The next year, when the mountaineer has to do it alone, the mountaineer figures out a way to cache and retrieve supplies in the wilderness after a day of climbing. The mountaineer buys 12 days worth of supplies and mimics, in series, the climb and supply exchanges of Tarjee, Harjee, and himself the year before. He climbs as Tarjee first, then as Harjee, then as himself, and makes it to Camp B after 12 days.
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On May 10 2012 12:44 TanGeng wrote:
The Seat Snatcher Manager
A company bus with twenty seats picks up twenty employees at their homes and shuttles them to and from work everyday. The twenty employees each have an assigned seat from the back to the front in the order they are picked up. The first employee on the bus route is an obnoxious manager with unpredictable behavior. Every morning, the manager boards the bus randomly at any one of the twenty stops. The manager cuts in front other employees at the stop, boards the bus, sits in an unoccupied seats at random, and then yaks away on the cell phone.
All other employees roll their eyes at the behavior. When they board the bus, if they sit in their assigned seat if unoccupied. Otherwise, they pick an unoccupied seat at random and sit there. You are Frank, the last employee to be picked up along the bus route. How often do you sit in the frontmost seat?
I like this one.
+ Show Spoiler +50%. + Show Spoiler +Suppose there were only 2 seats. Then clearly the manager either sits in your seat or his own, and you get your own seat 50% of the time. Increase the number of seats by one at a time and watch what happens. 3 seats. We have three equally likely cases. If the manager takes his own seat, everyone (including you) gets their own seat. If the manager takes your seat, then regardless of where the second guy goes, you're not getting your seat. In the third case, the manager takes the second guy's seat, and then the second guy either takes yours (50%) or leaves yours for you (50%). That evens out to a 50-50 overall shot you get your own seat.
Four seats (and after this I trust the pattern will be obvious enough that you can see it holds for any number of seats). We've got four equally likely cases. If the manager takes his own seat, you get your own seat, and so does everyone else. If the manager takes your seat, you obviously get someone else's seat. In the other two cases, the manager takes either the 2nd or 3rd guy's seat. If he had taken the 3rd guy's seat, then the 2nd guy takes his own seat, and the third guy either goes in the manager's seat or in yours (50% each way). If he had taken the 2nd guy's seat, then we're back in the scenario with 3 seats total described above -- we have three empty seats, you're going last, and the next person to get on is taking a random seat, which we already know comes out 50-50 that you get your seat.
This continues. With N seats, there's N equally likely scenarios -- one in which everyone gets their own seat, one in which you definitely get someone else's seat, and N-2 more which all reduce to previous cases that all have a 50% chance of you getting your own seat. Basically, if the manager takes his seat we know what happens, if the manager takes you seat we know what happens, and if he takes anyone else's seat we don't care -- all that does is delay the inevitable arrival of someone who has two seats to choose randomly between; yours and the one corresponding to whoever took his seat.
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On May 10 2012 07:05 JeeJee wrote:Show nested quote +On May 10 2012 05:50 Aelfric wrote:
Numbers:
How can you place the numbers 1 through 9 in a 3x3 grid such that every row, column, and the two diagonals all add up to 15?
In addition to this answer (as there are many if you count rotational symmetry), here's a quick and easy method to answer that for any odd-numbered square. + Show Spoiler +start at the top row, in the center with the number 1. Assume rows and columns wrap around (e.g. going "up" from the first row ==> last row, and so on). Go up and right, and place the next number there. If a number exists in that position already, then instead of going up and right, go down once instead. Continue until square is filled.
In this case, this would result in the following square
816
357
492
If you'll notice, that's the same square as frog's except rotated 90* ccw and flipped horizontally!
+ Show Spoiler +
Interesting, I did it simply by finding the central number first. If you put, for example, the nine in the centre, you'll find that wherever you put the eight/seven/six in, that line will come to over fifteen. The highest number that you can put in the centre is five, and it makes sense that you want the highest number possible in the centre since you're going to have to make use of one and two etc, and they need 'counter balancing' so to speak.
For the rest you can pretty much fill it in quite easily because of the rotational symmetry. Just put a random number elsewhere in the box, complete that line, and continue from there
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On May 10 2012 07:49 xxpack09 wrote:Show nested quote +On May 10 2012 05:50 Aelfric wrote:Here some riddles for you guys to kill some time... Assassination Attempt (This one is a bit harder):The King of a small country invites 1000 senators to his annual party. As gifts, each senator brings the King a bottle of wine, for a grand total of 1000 bottles of wine. Each bottle is signed by the senator who gave it. At the end of the party, the Queen tells the King that one of the senators is trying to assassinate him, and has put deadly poison in the bottle of wine he gave as a gift. Unfortunately, the Queen doesn't know which senator is the traitor (and thus doesn't know which bottle of wine has the poison in it). The King has 10 servants. He views them as expendable, and does not care if they live or die. He decides to use them to figure out which bottle is poisoned, which will then indicate which senator is trying to assassinate him. His plan is to make each servant drink from zero or more of the bottles of wine. The King knows that the poison is such that if a servant drinks it, he will feel fine until noon on the next day, at which point he will instantly drop dead. The King must know for sure who the traitor is by noon on the day after the party, or else the traitor will try to find another way to assassinate him. This essentially means that he has one shot to make his servants drink the wine in order to figure out which is the poison wine. Note that the King can make any of the servants drink from any of the wine bottles. He does not need to make all of the servants drink wine if he doesn't want to. Any servant who drinks from the poisoned bottle will die the next day at noon. How can the King figure out for sure who the traitor is by noon on the following day? Critical Hint (Don't check if you want to solve everything on your own):+ Show Spoiler +Think of a grid, with 1000 rows representing the bottles of wine, and 10 columns representing the servants. + Show Spoiler +So this problem should be possible for up to 1024 bottles of wine. For each bottle, assign some combination of servants to drink it. (so bottle 1 can have no one drink it, bottle 2 just servant 1, etc. Since each of the 10 servants can either drink or not drink (2 possibilities), the number of unique bottle combinations is 2^10 = 1024. Upon seeing what combination of servants die, the correct bottle can be determined.
+ Show Spoiler +
An alternative explanation that illuminates the coding principle behind the solution:
Number the bottles from 1 to 1000 and each servant from 1 to 10. For each bottle i, let b be i in binary (base 2) with digits b1...b10 (since 2^10 = 1024, you need 10 binary digits (bits) to represent the numbers from 1 to 1000), and let the nth servant drink from the bottle if bn is 1. For example, 75 in binary is 0001001011 so servants 1, 2, 4, and 7 should drink from the 75th bottle.
The ith bottle is poisoned if and only if all the servants that drank from that bottle die. Because there is a unique representation in binary for each digit from 1 to 1000 using 10 bits, there is a unique combination of servants that drink from each bottle.
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EPT
