another logic

[beros87]

I know one answer to this quiz, but i just want to find out if there is more:

you have 9 stones in which one of them is heavier than the others, can you find that stone using the balance only twice?

thanks

[Acidic]

you weight 3 and 3
if one of the groups are heavier, throw away everything else (preferably through the window of who gave you this logic puzzle).
If they wieght the same throw both lots away and keep the 3 you didn't weight.

Out of the 3 left, weight 1 and 1. If one is heavier, that's the one. otherwise the one you didn't weight is the heavier one.

[NoteMe]

And that is the only solution.

[mendhak]

That was a Microsoft interview question.

There is another answer though: Through probability, you could pick up any one of the stones, and you have a 1/9 chance that it is the heavier one, and thus you don't even have to use the scale.

[NoteMe]

Originally posted by mendhak
That was a Microsoft interview question.

There is another answer though: Through probability, you could pick up any one of the stones, and you have a 1/9 chance that it is the heavier one, and thus you don't even have to use the scale.

Feeling lucky...

[beros87]

I already knew the first answer, but the answer with the probability is new to me.
but i think that the hard part in the quiz here is while using a scale.
thanks for all

[InfiniTeEn]

To add difficulty to the problem... you dont know if the odd stone is heavier or lighter... the answer won't be so trivial...

[alkatran]

That can't be solved (9 unknown in two)

Consider it this way:

Start:
1 and 1 - can't determine 7 remaining
2 and 2 - 5 remaining, same thing
3 and 3 - Can't even determine three!
4 and 4 - Can't determine if first is unequal

[InfiniTeEn]

Sorry, you can use the balance 3 times...

ok let me restate the problem:

You have 9 similar balls,
one of them weighs differently (heavier or lighter),
how do you isolate the odd ball in 3 steps using a balance?

[alkatran]

That's supposed to be a problem with 12 (or even 13) balls. but alright:

Step 1: 3 and 3: If they're different the remaining 3 are safe, otherwise the 6 you weighed are safe.

If 1 was unequal: Take 3 from one side, and a safe one. Put 2 on each side...


You know what? Here's a picture of the 13