[Guest]

you have comitted a crime and you have one option to keep from getting the death penalty. What you have to do is to make sure 11 extremely expensive drinks get to an important person.You were given 12 drinks so that you have a spare one incase something happens to one of the bottles.turns out that your rival pours a couple of drops of a poison that takes 24 hours to kill some body. you have exactly 24 hours to give the man the drinks and you only have 4 lab rats to find out wich of the drinks is poisoned. how do you find out wich is the poisoned drink?

[peace*out]

divide the bottles in half, poar some wine from each of the bottles into a drink to give the lab rat. if it dies, then you know the poisoned wine was in that group, if not, it was in the other group. take the group the poisioned wine was in. split it in half, with 3 drinks in each. test it. same method deciding which group it was in. then, divide the wine again, with testing 2 in a group. if the rat dies, then its in that group, and you test it on the last rat. if not, then you keep a rat!

I don't like this cruetly to animals though...

[gvg]

First, give six drinks to one rat and the other six to another. Then, whichever group kills the rat, give three of them to one rat and the other three to the remaining. Then, from whichever group kills the rat, guess and hope for the best =)

[Guest]

Spoiler for ummm...

divide the bottles in half, poar some wine from each of the bottles into a drink to give the lab rat. if it dies, then you know the poisoned wine was in that group, if not, it was in the other group. take the group the poisioned wine was in. split it in half, with 3 drinks in each. test it. same method deciding which group it was in. then, divide the wine again, with testing 2 in a group. if the rat dies, then its in that group, and you test it on the last rat. if not, then you keep a rat!

I don't like this cruetly to animals though...

that cant be done becaus you only have 24 hours to test which bottle is poisoned and it takes 24 hours to kill

[Guest]

the poison takes 24 hours to kill and you only have 24 hours to test. also if the man dies, you get the death penalty

Edited May 13, 2010 by G3HR1G

[Glycereine]

divide the bottles in half, poar some wine from each of the bottles into a drink to give the lab rat. if it dies, then you know the poisoned wine was in that group, if not, it was in the other group. take the group the poisioned wine was in. split it in half, with 3 drinks in each. test it. same method deciding which group it was in. then, divide the wine again, with testing 2 in a group. if the rat dies, then its in that group, and you test it on the last rat. if not, then you keep a rat!

I don't like this cruetly to animals though...

If it takes 24 hours for the rat to die though you will only be able to perform the first stage of your test in time . I don't see a readily available solution though because I was going this route as well.

[Glycereine]

I can see how to narrow it down to 3 of the drinks in 24 hours perhaps someone else can run with my logic and get the answer:

Rat 1: ABCDEFGHI

Rat 2: DEFGHIJKL

Rat 3: GHIJKLABC

Rat 4: JKLABCDEF

This will result in only one surviving rat. You will know the poison was in one of the 3 drinks that you didn't give that rat

[Guest]

I can see how to narrow it down to 3 of the drinks in 24 hours perhaps someone else can run with my logic and get the answer:

Rat 1: ABCDEFGHI

Rat 2: DEFGHIJKL

Rat 3: GHIJKLABC

Rat 4: JKLABCDEF

This will result in only one surviving rat. You will know the poison was in one of the 3 drinks that you didn't give that rat

you are onto something

[Glycereine]

I can see how to narrow it down to 3 of the drinks in 24 hours perhaps someone else can run with my logic and get the answer:

Rat 1: ABCDEFGHI

Rat 2: DEFGHIJKL

Rat 3: GHIJKLABC

Rat 4: JKLABCDEF

This will result in only one surviving rat. You will know the poison was in one of the 3 drinks that you didn't give that rat

Alternatively you can give yourself a 33% chance of survival by

Drink from 8 of the drinks (A-H) and give each rat the following:

Rat 1: IJK

Rat 2: JKL

Rat 3: KLI

Rat 4: LIJ

If you die, you die (2/3 chance)

If you live, 1 rat will also live and you can see the poisoned drink was the one the other 3 rats drank.

As mentioned only 1/3 chance of survival

[Guest]

there is a way to do it without risking your life

[Glycereine]

Rat 1: ABCDEF

Rat 2: GHIJKL

Rat 3: ABCGHI

Rat 4: ADGJ

You: BEHK

In this method you have a 2/3 chance of survival.

Rat 1 or 2 will die. This narrows it down to half the drinks. Then if rat 3 lives or dies you are narrowed down to 3 drinks.

If rat 4 dies you know exactly which one it is (and you live). If he doesnt die, you assume the poison is in C, F, I or L (as decided based on which of the first 3 rats lives and dies). You then give the man the 11 drinks other than those. If you are wrong and the poison is in B, E, H or K, you are dead anyways.

[Glycereine]

Hmm I'll have to look some more if there's a way to do it without risking your life.

Edited May 13, 2010 by Glycereine

[Guest]

This was on car talk a few weeks ago, and I solved it. Here's how you pinpoint the poisoned bottle:

You have drinks 1-12. Write each number in binary, with 4 digits. You get:

1 = 0001

2 = 0010

3 = 0011

4 = 0100

5 = 0101

6 = 0110

7 = 0111

8 = 1000

9 = 1001

10 = 1010

11 = 1011

12 = 1100

Now, number the rats 1-4. If the Nth digit (counting from the left) of the binary representation of drink D is a 1, then you give that drink to rat N. Otherwise, rat N does not receive a sample of drink D. For example, drink #7 would be given to rats 2, 3, and 4, but not to 1, and drink #8 would be given only to rat #1.

Now, look at the rats that die. The rats that die are the digits that are 1's. For example, if rats 1 and 3 die, then the 1st and 3rd digit of the binary representation of the poisoned bottle were 1's, and the other digits were 0, so bottle 1010 = 10 was poisoned, and if rats 2 and 3 die, then bottle 0110 = 6 was the poisoned bottle.

Bazinga.

[Glycereine]

Rat 1: ABCD

Rat 2: EFGH

Rat 3: ABEFIJ

Rat 4: AEICGK

A: 1,3,4 Die

B: 1,3 Die

C: 1,4 Die

D: 1 Dies

E: 2,3,4 Die

F: 2,3 Die

G: 2,4 Dies

H: 2 Dies

I: 3,4 Die

J 3 Dies

K 4 Dies

L None Die

Edited May 13, 2010 by Glycereine

[Guest]

Rat 1: ABCD

Rat 2: EFGH

Rat 3: ABEFIJ

Rat 4: AEICGK

A: 1,3,4 Die

B: 1,3 Die

C: 1,4 Die

D: 1 Dies

E: 2,3,4 Die

F: 2,3 Die

G: 2,4 Dies

H: 2 Dies

I: 3,4 Die

J 3 Dies

K 4 Dies

L None Die

correct

[Glycereine]

This was on car talk a few weeks ago, and I solved it. Here's how you pinpoint the poisoned bottle:

You have drinks 1-12. Write each number in binary, with 4 digits. You get:

1 = 0001

2 = 0010

3 = 0011

4 = 0100

5 = 0101

6 = 0110

7 = 0111

8 = 1000

9 = 1001

10 = 1010

11 = 1011

12 = 1100

Now, number the rats 1-4. If the Nth digit (counting from the left) of the binary representation of drink D is a 1, then you give that drink to rat N. Otherwise, rat N does not receive a sample of drink D. For example, drink #7 would be given to rats 2, 3, and 4, but not to 1, and drink #8 would be given only to rat #1.

Now, look at the rats that die. The rats that die are the digits that are 1's. For example, if rats 1 and 3 die, then the 1st and 3rd digit of the binary representation of the poisoned bottle were 1's, and the other digits were 0, so bottle 1010 = 10 was poisoned, and if rats 2 and 3 die, then bottle 0110 = 6 was the poisoned bottle.

Bazinga.

Much more elegant, additionally with your method you can actually determine 16 bottles with 4 rats Since using binary you can count from 0-15 with 4 digits.

[Guest]

This was on car talk a few weeks ago, and I solved it. Here's how you pinpoint the poisoned bottle:

You have drinks 1-12. Write each number in binary, with 4 digits. You get:

1 = 0001

2 = 0010

3 = 0011

4 = 0100

5 = 0101

6 = 0110

7 = 0111

8 = 1000

9 = 1001

10 = 1010

11 = 1011

12 = 1100

Now, number the rats 1-4. If the Nth digit (counting from the left) of the binary representation of drink D is a 1, then you give that drink to rat N. Otherwise, rat N does not receive a sample of drink D. For example, drink #7 would be given to rats 2, 3, and 4, but not to 1, and drink #8 would be given only to rat #1.

Now, look at the rats that die. The rats that die are the digits that are 1's. For example, if rats 1 and 3 die, then the 1st and 3rd digit of the binary representation of the poisoned bottle were 1's, and the other digits were 0, so bottle 1010 = 10 was poisoned, and if rats 2 and 3 die, then bottle 0110 = 6 was the poisoned bottle.

Bazinga.

Rat 1: ABCD

Rat 2: EFGH

Rat 3: ABEFIJ

Rat 4: AEICGK

A: 1,3,4 Die

B: 1,3 Die

C: 1,4 Die

D: 1 Dies

E: 2,3,4 Die

F: 2,3 Die

G: 2,4 Dies

H: 2 Dies

I: 3,4 Die

J 3 Dies

K 4 Dies

L None Die

both are correct

[Guest]

remember to rate the puzzle

[Glycereine]

Good puzzle .

[Guest]

Good puzzle .

thank you

[Guest]

there is a way to do it without risking your life

Answer....Lengthy, sorry. I wouldve drawn a diagram but it wouldve taken me too long to figure out how...

Feed drinks 7-12 to Rat A....Feed drinks 5,6,7,8 to Rat B….Feed drinks 3,4,5,8,9,10 to Rat C….Feed drinks 2,3,10,11 to Rat D.

If none of the rats die, then discard drink 1.

If rat A dies and Rat B dies but Rat C lives, then discard drink 7.

If Rat A Dies and Rat B Dies and Rat C Dies then discard drink 8.

If Rat A Dies and Rat B lives and Rat C dies and Rat D lives then discard drink 9.

If Rat A dies and Rat B lives and Rat C dies and Rat D dies then discard drink 10.

If Rat A dies and Rat B lives and Rat C lives and Rat D dies then discard drink 11.

If Rat A dies and Rat B lives and Rat C lives and Rat D lives then discard drink 12.

If Rat A lives and Rat B dies and Rat C lives, discard drink 6.

If Rat A lives and Rat B dies and Rat C Dies and Rat D lives discard drink 5.

If Rat A lives and Rat B lives and Rat C dies and Rat D lives discard drink 4.

If Rat A lives and Rat B lives and Rat C dies and Rat D dies discard drink 3.

If Rat A lives and Rat B lives and Rat C lives and Rat D dies, discard drink 2.

[Guest]

You use base two. One, then two, then one and two. So if you have four rats or four digits, using base two you can create 16 distinct combinations. Here's how you do it. Drink number one goes only to rat one. Drink number two goes to rat number two. Here's the interesting part. Drink number three goes to rats one and two. So if rats one and two are belly up the next day, it can only be because drink number three had the poison in it. If you continue using base two, drink four goes to rat three and rat three only. drink five goes to rats one and three, etc. etc. etc. And each combination corresponds to a different drink, which you of course, keep track of. Make sense?

[Guest]

You can use this way to solve it with up to 16 drinks. You give the drink to all rats listed then see which, if any, rats die.

Rats A,B,C,D

Drink 1-16

1 - A

2 - B

3 - C

4 - D

5 - AB

6 - BC

7 - CD

8 - DA

9 - AC

10 - BD

11 - ABC

12 - BCD

13 - ABD

14 - ACD

15 - ABCD

16 - none

[Guest]

Using the same alpha code for the 12 drinks (A-L)

Rat 1 drinks A E G I K

Rat 2 drinks B E H I J K

Rat 3 drinks C F G I J K

Rat 4 drinks D F H J K

drink L is not given to any rat. Now depending on which rats and how many determins which bottle was poisoned (Left for you to Prove that).

I can see how to narrow it down to 3 of the drinks in 24 hours perhaps someone else can run with my logic and get the answer:

Rat 1: ABCDEFGHI

Rat 2: DEFGHIJKL

Rat 3: GHIJKLABC

Rat 4: JKLABCDEF

This will result in only one surviving rat. You will know the poison was in one of the 3 drinks that you didn't give that rat

[Guest]

divide the 12 bottles by for, assign one rat to each batch of 3 bottles, one of the 4 batches will kill 1 rat, assign the remaining 3 rats to the poisoned batch and you'll find out immediately with bottle was poisoned, take it out, serve the remaining 11 to the person...

you have comitted a crime and you have one option to keep from getting the death penalty. What you have to do is to make sure 11 extremely expensive drinks get to an important person.You were given 12 drinks so that you have a spare one incase something happens to one of the bottles.turns out that your rival pours a couple of drops of a poison that takes 24 hours to kill some body. you have exactly 24 hours to give the man the drinks and you only have 4 lab rats to find out wich of the drinks is poisoned. how do you find out wich is the poisoned drink?

[Guest]

I cant figure out the actual demonstration to it on paper, but i know the logistics.

You cant divide drinks up by rat. Instead each drink needs to have been drunk by at least two rats, so when one of them dies, you can tell by which other rats die along with it that any combination of two or three rats narrows it down to one drink.

like if each bottle was drank by 3 rats each, then there would be only one combination of possible rats to have died from any given permutation. Like you'd be able to tell by the code of which three rats died which of the 12 drinks it was because only one of the twelve drinks was drunk by specifically and only those three rats.

The permutations would take forever to write out