another riddle for you...

[DrCrap]

you have 8 coins, one of which is fake (i.e 7 original, and 1 fake).
you have scales which you can use to weigh the coins (two sides... duha) you can put as many (or as little...) as you want on each side.

what is the minimal amount of scaling comparisons you need to make, to figure out which coin is the fake one?

oh, and you don't know if the fake coin is lighter or heavier than the rest, all you know, is that it has a different weight...

 

[Fritzo]

Dude, do your own math homework

 

[Patt]

Originally posted by: Fritzo
Dude, do your own math homework

 

[DrCrap]

Originally posted by: Fritzo
Dude, do your own math homework

ok... Unfortunatly, I probably finished getting homework a little before you were born.

you know you could just admit you don't have the mental ability to solve this, rather than spam my thread. But thanx for the suggestion.

 

[DrCrap]

Originally posted by: Patt

Originally posted by: Fritzo
Dude, do your own math homework

read my previous reply, lamer.

 

[dighn]

3

 

[EngenZerO]

uhm divide the coins into sets of three, three and two.

place sets three and three on scale. see if it tips. if it tips you know one of those in that set of three is lighter. if it doesnt tip you know its in the set of two. now place they lighter set on the scale one coin, one coin and leave one on the floor (for set of three). if its not that lighter set just place one coin on one side and the other on the other side.

in two tries you will know the fake coin.

 

[DrCrap]

Originally posted by: dighn
3

duha... wanna elaborate?

 

[EngenZerO]

Originally posted by: DrCrap

Originally posted by: dighn
3

duha... wanna elaborate?

actually you can do it in two.

 

[flyboy84]

I'd say take four scales, and weigh two on each. one of the four weights will be different, then weigh both of the coins on that scale seperately, as well as one from the scales that had the other reading. Which ever of the two coins does not match that one is the fake.

 

[akubi]

Originally posted by: EngenZerO
uhm divide the coins into sets of three, three and two.

place sets three and three on scale. see if it tips. if it tips you know one of those in that set of three is lighter. if it doesnt tip you know its in the set of two. now place they lighter set on the scale one coin, one coin and leave one on the floor (for set of three). if its not that lighter set just place one coin on one side and the other on the other side.

in two tries you will know the fake coin.

bravo.
standard interview question with many variations.

but I remember the first time I saw it was on a rerun of columbo

 

[blackdogdeek]

Originally posted by: EngenZerO
uhm divide the coins into sets of three, three and two.

place sets three and three on scale. see if it tips. if it tips you know one of those in that set of three is lighter. if it doesnt tip you know its in the set of two. now place they lighter set on the scale one coin, one coin and leave one on the floor (for set of three). if its not that lighter set just place one coin on one side and the other on the other side.

in two tries you will know the fake coin.

but we don't know if the fake coin is heavier or lighter. therefore, from your explanation above:

let's assume that 3v3 tips the scales in one direction. how do we know which of the 2 groups has the fake coin?

let's assume that 3v3 does NOT tip the scales. how do we know which of the 2 is the fake coin without comparing one of them to one of the other from the 3v3?

 

[dighn]

Originally posted by: EngenZerO
uhm divide the coins into sets of three, three and two.

place sets three and three on scale. see if it tips. if it tips you know one of those in that set of three is lighter. if it doesnt tip you know its in the set of two. now place they lighter set on the scale one coin, one coin and leave one on the floor (for set of three). if its not that lighter set just place one coin on one side and the other on the other side.

in two tries you will know the fake coin.

problem is the coin could be either lighter or heavier

3 tries: weigh 2 vs 2, tips or not, 4 elimited. now weigh 1 vs 1 from remaining 4, another 2 eliminated. weigh 1 of the reimaining one vs one of the known real ones and you got the answer

 

[DrCrap]

Originally posted by: dighn

Originally posted by: EngenZerO
uhm divide the coins into sets of three, three and two.

place sets three and three on scale. see if it tips. if it tips you know one of those in that set of three is lighter. if it doesnt tip you know its in the set of two. now place they lighter set on the scale one coin, one coin and leave one on the floor (for set of three). if its not that lighter set just place one coin on one side and the other on the other side.

in two tries you will know the fake coin.

problem is the coin could be either lighter or heavier

3 tries: weigh 2 vs 2, tips or not, 4 elimited. now weigh 1 vs 1 from remaining 4, another 2 eliminated. weigh 1 of the reimaining one vs one of the known real ones and you got the answer

correct.
There are a couple of different solutions as well, but this one is also correct.

 

[AMCRambler]

Well, if you were lucky enough to pick up the fake one with the first two coins you compare, then it would only take two comparisons. The first comparison would show you that you had the fake one. For the second comparison take a coin from the pile and replace one of the ones on the scale. If the scale balances you know you took off the fake one. If it remains unbalanced, then you know the one you left on the scale is the fake one.

 

[Fritzo]

Originally posted by: DrCrap

Originally posted by: Fritzo
Dude, do your own math homework

ok... Unfortunatly, I probably finished getting homework a little before you were born.

you know you could just admit you don't have the mental ability to solve this, rather than spam my thread. But thanx for the suggestion.

It was a joke. You're getting cranky in your old age.