Just a spot to list some of my ideas, feelings, frustrations, etc.
The Problem: You have a set of Balance Scales like the one shown here:

You have 9 identical looking marbles. You can't tell the difference between any of the 9 marbles by looking at them.

One of the 9 marbles is ever so slightly denser than the others (thus it weighs more). The only way to find out which marble is the one that is denser is to use the balance scales.

One way to do this would be to put 1 marble on one side and 1 marble on the other. If one was heavier than the other, you would then know which of the nine was the heaviest. If they were the same, you could then take one of the marbles off the scales, and put another on. As you can see, this could take quite a few measurements. Worst case: too many to count.

Now, your task is to figure out how few measurements you can make to find the 1 dense marble.

Keep in mind; you can put more than 1 marble on each side of the scales at a time. Also, keep in mind the "worst case" for your solution. The solution I mentioned above could be solved in 1 measurement, but that's NOT worse case (it could take many more).

When you post your solution, don't just say I can find the dense marble with 6 measurements. You need to describe your solution. I have to be able to determine if the number of measurements you suggest is truly worst case or not.

By the way "6" is just used as an example here; it may or may not be the correct answer.

After this puzzle is solved, someone else could add another word puzzle to the thread.

Comments (Page 2)
2 Pages1 2 
on Jan 14, 2004
yes but you don't know if the marble is heavier or lighter so how would you know which group of four was wrong.


Well, either the first two groups of four are equal or their not. If they are, it's the third group.
on Jan 14, 2004
if you got lucky and they were equal on the scale then you would know which group of four (but this involves luck so it isn't applicable).


OK, I get your point.
on Jan 14, 2004
#11 by Ingui - 1/14/2004 8:19:44 PM
Divide $177 (in whole $ increments) into a number of bags so that I can ask for any amount between $1 and $177, and you can give me the proper amount by giving me a certain number of these bags without opening them. What is the minimum number of bags you will require?

177 bags.





Powered by SkinBrowser!
on Jan 14, 2004
for #11 by ingui:

8 bags: 88,44,22,11,5,4,2,1
on Jan 14, 2004
8 bags: 88,44,22,11,5,4,2,1


Travis, your good. Did you already know that one? You sure came up with an answer pretty fast. I guess we will wait for Ingui to see if it is the right one or not.
on Jan 14, 2004
nope didn't know it. thought to myself that the way to solve it would be to work backwards by halves. and once you got down to 11 you just had to make sure that the numbers below it allowed you to get up to 11 and hit every number. then all of them work. the other difficult part was getting it to add up to 177
on Jan 14, 2004
You better mark the bags with the amount contained in each or you'll have to open it to look!



Powered by SkinBrowser!
on Jan 15, 2004
yes you guys got it, sorry I didnt reply sooner. I got "called in" on Snow Duty. Your method seems to work but it's actually different, I think, than the answer I had. It is still 8 bags.
But the first even bags total $127 (1,2,4,8,16,32,64) the last bag is a $50 sac.

If the original total had been 12, you would end up with 4 bags(1,2,4 and 5)

the formula is : 2º, 2¹, 2², 2³, 2to the fourth, ... , 2 the sixth.
Anyway, seems we got a bunch o' einsteins here at WC
on Jan 15, 2004
Awww....Was away for that one. Well, nobody pays what I say much attention here anyway...



Powered by SkinBrowser!
2 Pages1 2