Hmmm...a little mind problem for y'all...can YOU figure it out?

[eLiu]

hmmm...challenge question...this one is for how fast you can do it, and if you can get it right

You are given eight balls. One of them is heavier than the other seven. You are also given a double-pan balance. What is the minimum number of weighings required to find the heavy ball? And how?

I got the answer in about 40 seconds...and i figured out why 2 minutes later.

(dont mean to brag...well, yes i do hehe...im only 15 hehehe)

-eric

 

[Moonbeam]

Right just pick each one up and feel it.

 

[IJump]

4 weighings. All will be equal except the one that is heavier. Probably took me about 10 seconds.

 

[frizzlefry]

well the minimum would be 1. If you were to measure them with the scale. The maximum would be 4. about question

-edit-

took 5 seconds after reading your question a second time....

 

[Schlocemus]

Zero...

Just guess and you have a 12.5% of getting it correct; even more if you have any instinct...
You never stated that you *had* to use the balance, and you also never specified a limit for how many times you could try. Therefore, I stand by my answer


-Dave

...This took me 5 seconds to think of and type

 

[Chaotic42]

3

1. Balance 4 on one side, 4 on the other.
2. Take the heavier set of 4 and split them into 2 groups of 2. Weigh those
3. Take the heavier of the 2 and split them into 2 groups of 1. The heavier is
the odd man (er ball) out.

Took me 10 seconds and I'm 19

 

[eLiu]

arg...im an idiot...

need to add this to the problem:

You must find the MINIMUM number of weighings, and it must work EVERY time. You cannot feel the difference in weights, you MUST use the scale. Probability CANNOT be part of your answer.

(not so easy now, ehh?)

 

[IJump]

Throw them all at the wall, the one that leaves the biggest dent is the heaviest.

 

[Schlocemus]

ah, you realized...good update

 

[jaydee]

This is a repost so I already vaugely remember the answer. Three.

edit how:
Weigh any 3 on 3
If one side is heavier, weigh any 2, 1 on 1. If one is heavier, thats the ball, if not, its the one not weighed
If one side was not heavier, take the remaining 2 and wiegh them 1 on 1. Heavier side wins it.

Man, you guys are quick.

edit again
Duh, that makes 2. BTW, my first edit was 21:47:51 I wonder if I beat e-tech.

 

[Schlocemus]

D'oh!

Still thinking...

At least tell us *why* Chaotic is incorrect...Give us a situation where that method would not work

 

[eLiu]

hehehe

chaotic42, nice try...but no cigar

Three is incorrect hehe

 

[nicowju]

Good answer chaotic. Was thinking sort of something like that. There had to be a reason eLiu put "double-pan balance"
Just saw the post by eLiu... hmmm.. now I have to think hard

And now I just saw eTech's post. I feel dumb

 

[etech]

2.

Try starting with two groups of 3. If those are equal then you only weigh the other two balls. 2 weighs.

If the original 3 are not equal then you weigh two from the heavier side. Either they are equal in which case the 3rd ball is heavier or one of the two on the pans are heavier. 2 weighs total.

 

[Chaotic42]

<< 2.

Try starting with two groups of 3. If those are equal then you only weigh the other two balls. 2 weighs.

If the original 3 are not equal then you weigh two from the heavier side. Either they are equal in which case the 3rd ball is heavier or one of the two on the pans are heavier. 2 weighs total. >>



Yup. You're right. I concede

 

[Schlocemus]

Ah, etech is right ... I feel stupid

 

[eLiu]

Nice!

etech wins the grand prize! ONE MILLION US DOLLARS!

yea right hehe

but still, etech has the right answer there

 

[Schlocemus]

Got anymore puzzles for us? I like them

 

[eLiu]

hehe

ill post'em as i see them...or as a i remember old ones hehe

i had another one...a lot harder than this ball thing...but i forgot...hehehe

 

[jaydee]

Hey! I might have won (prolly not, odds are 51:9 against me). I should get a share of the pot Anyway, got more?

 

[Schlocemus]

How'd you get 51:9 ?

 

[jaydee]

Sorry did it too fast in my head. 51:60 against me. Never too good at probability.

 

[Schlocemus]

heh

 

[crazyjoekuta]

i heard big jobs like for microsoft they ask you a Q like this in the interview... one i heard was actually asked (i forget where) was the &quot;3 lightswitches 3 lightbulbs in the other room, which switch to which bulb&quot; puzzle.

EDIT: that was a couple years ago though, when the puzzle was new.

 

[Jothaxe]

I am not going to read all the posts, but here is the answer:

2 uses of the scale is all that is necessary

simply put three balls on each side of the scale:

i. if they are even, you know the ball is one of the remaining two, so simply compare them

ii. if one side is heavier, you know its one of those three balls. In this case compare two of the three left - if equal, you know the third ball is heaviest. If one side is heavier it is the one you want.