My riddle...From Columbo - Page 2

#11 10-12-2002, 01:51 AM

You place all ten bags on the scale. Insert the penny in the scale then remove the bags one at a time. Whichever bag lowers the weight by a greater amount than any other bag is the bag of gold bars. This is assuming that there an equal number of bars in each bag but there is no reason to assume otherwise given the parameters of the problem.

For those of you not old enough to remember the penny scales you were able to slowly remove weight without resetting the coin mechanism but if you tried to add additional weight they would immediately reset.

Jimbo

#12 10-12-2002, 02:01 AM

irchans,

My version of the solution (see my post above) saves the 2nd penny 100% of the time. Remember a penny saved is a penny earned. [img]/forums/images/icons/smile.gif[/img] This might be a good place to mention that it is no less likely that each bag has the same numer of bars than to assume you have at least nine bars in each bag which is required for the other 1 penny solutions above. My solution requires no minimum number in each bag which I think is important.

Jimbo

baggins · #13 10-12-2002, 05:15 AM

very nice thought process, Jimbo.

#14 10-12-2002, 09:19 AM

This would require the scale to be accurate over 100 oz. The 2 penny solution is for the case where we can only weigh 100 oz at a time. Also, we don't know if the bags have an equal numbers of bars.

BruceZ · #15 10-12-2002, 09:55 AM

We could save a penny 100% of the time if we place just 1 bar from bags 1-9 on the scale, and take them off one at a time. Then we don't care how many are in each bag.

Just my 2 cents...

irchans · #16 10-12-2002, 11:53 AM

#17 10-12-2002, 12:05 PM

Very true BruceZ,

I was also avoiding bumping up against the 10 minute time limit.

Jimbo

#18 10-12-2002, 12:06 PM