Riddles, brain puzzles and mathematical problems

[eliasfrost]

Can't you do it on all of the ten rats at the same time?

[BAndrew]

Can't you do it on all of the ten rats at the same time?

Yes you can, but it won't work with exactly the same method Neptune used.




PS. The solution is really simple.

[MyRedNeptune]

I don't get it. The last rat still receives "512" bottles so it is impossible to find the poison in only 24 hours, wasn't that the point of the puzzle? You said it takes 10-20 hours for the poison to run its course and the scientist only has 24 hours to find the poison, implying that in a worst case scenario it must be done in one go (though it is possible to do it in two in if the poison decides to kill quickly.) Therefore, to satisfy the conditions of the puzzle it must be done in one go. With this method, ruling out 512 bottles is only the first step, you must then divide 512 bottles the same way, and then 256, then 128, etc … until you have only two. Therefore, way more than 24 hours and way more than 10 rats are required.

Time limit. Ooops. I completely forgot. You are right. The method Neptune used can be done with 10 rats but it will take much more than 24 hours because you must wait 10-20 hours for the first rat to die. Sorry for that. It was late when I checked his solution and it slipped my mind.

The riddle remains open. There is a way to do it ON time. Neptune is close.

Hint:The answer is 10 rats (as Neptune predicted), but you have to modify Neptune's method to make it faster

Why is that so? My method involves neither waiting nor knowledge of prior results. Just do all the steps at the same time. It is the pattern of deaths after 20 hours that will show where the poison is located. I was worried that this wasn't clear with how I worded it in some parts, but it was indeed the intention.

[Bridge]

I don't get it. The last rat still receives "512" bottles so it is impossible to find the poison in only 24 hours, wasn't that the point of the puzzle? You said it takes 10-20 hours for the poison to run its course and the scientist only has 24 hours to find the poison, implying that in a worst case scenario it must be done in one go (though it is possible to do it in two in if the poison decides to kill quickly.) Therefore, to satisfy the conditions of the puzzle it must be done in one go. With this method, ruling out 512 bottles is only the first step, you must then divide 512 bottles the same way, and then 256, then 128, etc … until you have only two. Therefore, way more than 24 hours and way more than 10 rats are required.

Time limit. Ooops. I completely forgot. You are right. The method Neptune used can be done with 10 rats but it will take much more than 24 hours because you must wait 10-20 hours for the first rat to die. Sorry for that. It was late when I checked his solution and it slipped my mind.

The riddle remains open. There is a way to do it ON time. Neptune is close.

Hint:The answer is 10 rats (as Neptune predicted), but you have to modify Neptune's method to make it faster

Why is that so? My method involves neither waiting nor knowledge of prior results. Just do all the steps at the same time. I was worried that this wasn't clear with how I worded it in some parts, but it was indeed the intention.

The point is that you have only eliminated half of the bottles by the end of it. You have to repeat the process 9 more times which means you must wait in a worst-case scenario 9*20 = 180 hours/7.5 days and you only have one to find the poison. That's assuming you can do all of the steps at the same time, but then you need to wait worst-case scenario 20 hours. You of course need more than 10 rats to do that too.

EDIT: Either that or I really do not understand the method and how the results can possibly be interpreted in such a way that it singles out one bottle.

[BAndrew]

Why is that so? My method involves neither waiting nor knowledge of prior results. Just do all the steps at the same time. It is the pattern of deaths after 20 hours that will show where the poison is located. I was worried that this wasn't clear with how I worded it in some parts, but it was indeed the intention.

In that case your answer is correct. I misread your answer. There has been a misconception as to whether you need or not to know the outcome of the previous rat before going to the next. Thanks to clearing this. Your answer is correct.

@Bridge
Actually he didn't meant to do this.
Here's what you do:
You split 1024 into Group A of 512 bottles and Group B of 512 bottles (actually 488 and the rest are imaginary)

You give rat 1 to drink group A

You split group A and B to group A1,A2,B1,B2 each of 256 bottles
Give rat 2 to drink A1 and B1 groups.

Continue to do this for 10 steps.

The pattern of the death of the rats determines where the poison is.
So his answer doesn't need more rats or more time.

---

My answer to the riddle is the following (maybe it helps you)
We make an array (Rats/Bottles) like this (where X means that the X rat will drink the Y bottle):



So for instance after 20 hours if all of the 10 rats are dead this means that the bottle 1 has the poison. If only rat 10 is dead then the 1000th bottle has the poison. There are 2^10 = 1024 possible combinations so you have 24 that will not be used.

However Neptune's answer is equally correct.

[Bridge]

No, I don't get it. if only rat 10 dies, you still have 511 (or 487) more bottles to check, right?

[BAndrew]

No, I don't get it. if only rat 10 dies, you still have 511 (or 487) more bottles to check.

You don't get my solution or Neptune's (or neither)?

[Bridge]

Neither. The 10th rat drinks from 512 bottles just like the rest of them. At the end of the process, all you know is which 512 bottles contain the poison (e.g. you can rule out the other 512.) You still need to repeat the process until only two possibilities remain.

[BAndrew]

Neither. The 10th rat drinks from 512 bottles just like the rest of them. At the end of the process, all you know is which 512 bottles contain the poison (e.g. you can rule out the other 512.) You still need to repeat the process until only two possibilities remain.

In which solution are you reffering to?

[Bridge]

MyRedNeptune's, but the solutions are pretty similar.