All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Twenty metal blocks (Posted on 2005-06-02)
Twenty metal blocks are of the same size and external appearance; some are aluminum, and the rest are duraluminum, which is heavier.

Using a pan balance to determine how many blocks are aluminum, what is the minimum number of weighings to be done?

 See The Solution Submitted by pcbouhid Rating: 3.5000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 between... | Comment 1 of 18

First of all, we have to assume that "some" means at least one, and that the "rest" also means at least one.  If not for these restrictions, there'd be 2^20 possibilities as to which are duraluminum and which not.  That leaves 2^20 - 2 -- not much different.

Ideally, each weighing would reduce the possibilities by a factor of 1/3: 1/3 of the cases would result in left pan heavy, 1/3 right pan heavy and 1/3 equal pan weights.  That would mean the process would take base-3-log ( 2^20 - 2 ) = 12.61859333528395.... Of course there are no fractional weighings, so ideally it might take as few as 13 weighings.  But possibly there'd be no way of arranging exact 1/3 situations for each weighing (or close enough to allow for 13 weighings).

There is certainly a way of doing this in 19 weighings.  Just weigh block 1 vs. block 2, block 2 vs. block 3, ..., block 18 vs. block 19, block 19 vs. block 20.

So it's somewhere between 13 and 19 weighings.

 Posted by Charlie on 2005-06-02 19:54:16

 Search: Search body:
Forums (0)