Puzzle time!
Jan. 24th, 2005 02:27 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
fluffymarkInsomnia makes me do odd things. Last night I decided to twist a well known puzzle to make it more evil, and then solve it, which I did. And this stopped me thinking of monsters under the bed and other scaries (I read the Nightmares and Fairy tales graphic novel yesterday, and it may have gotten to me...ooops), so was actually a good thing, I'm trying to tell myself.
Anyway, lets see which of you clever people solves this first: (all comments initially screened)
You have a balance scale and a big bag of marbles. The marbles are all identical in weight. You take 39 of the identical marbles and mix in another special marble (so 40 marbles now), which may be heavier or lighter than the others, or may not, but looks identical, so can be identified by weight alone. Allowing only 4 weighings, how can you identify whether the special marble is heavier or lighter than the others, and if so, which marble it is?
Update: Congratulations to the clever![[livejournal.com profile]](https://www.dreamwidth.org/img/external/lj-userinfo.gif)
ixwin who solved it by adding more marbles from the bag. So, lets make the problem HARDER. Assume the bag only has one more marble left in it. You have the original 40 marbles (39 identical and one special) and a bag with one more marble in it. Now solve it.
Anyway, lets see which of you clever people solves this first: (all comments initially screened)
You have a balance scale and a big bag of marbles. The marbles are all identical in weight. You take 39 of the identical marbles and mix in another special marble (so 40 marbles now), which may be heavier or lighter than the others, or may not, but looks identical, so can be identified by weight alone. Allowing only 4 weighings, how can you identify whether the special marble is heavier or lighter than the others, and if so, which marble it is?
Update: Congratulations to the clever
ixwin who solved it by adding more marbles from the bag. So, lets make the problem HARDER. Assume the bag only has one more marble left in it. You have the original 40 marbles (39 identical and one special) and a bag with one more marble in it. Now solve it.
![[identity profile]](https://www.dreamwidth.org/img/silk/identity/openid.png){kind=small}
no subject
Date: 2005-01-27 08:54 pm (UTC)Is there a formula for the maximum number of marbles you can sort out for N uses of the scales?
no subject
Date: 2005-01-28 12:22 am (UTC)Formula is easy to work out - each use of the scale has 3 possible results (balanced, left side down or right side down), so N uses of the scales gives 3^N possible results. Now if there are M marbles, there are M results where a marble is heavier, and M results where a marble is lighter, and also the single possible case where the special marble is the same weight as all the others. Hence we must have that 3^N = 2M+1, which we can rearrange to get the formula for the number of marbles M = (3^N - 1)/2 (which correctly gives M=40 when N=4).
I must sound like a maths teacher. Forgive me.
Anyway, I'm wondering who you are. Have I met you on my wanderings in Cambridge or not?