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Count me in and weigh me

 
 
bobby47
 
Reply Mon 20 Oct, 2003 06:34 am
Here is the situation:
You are working for Fort Knox, and are responsible for storing the gold in large safes.
One day your boss walks up and has some bad news:
"Ok, someone screwed up. They delivered a package here that contains fake gold. The only way
to recognize the package is by its weight, so you will have to weigh every package and see
if it is lighter than the others. I already spoke with the folks at registry and narrowed it
down to 512000 packets, which is still a lot I guess. You have time until tomorrow to find the
quickest way to get the job done. I can get 4 scales from town, but every scale will only
hold 1000 packages max. Also you won't be able to get the weight from these scales, but you will
easily see the difference if the fake package is on one of these scales, of course only if all
the scales have the exact same number of packages on them. I want you to find the way that needs
the fewest scale attempts for the worst case scenario.
For every scaling you can put a stack of packages (not more than 1000 for every stack) on each of
the four scales, and see the difference between those four stacks of packages.
When you have the procedure please find the number of scale attempts needed for the best case and
the worst case and report with the numbers to me tomorrow. I hope it is better than mine, because
if not I may just fire you."
Now, thats your situation. Find the best answers fast and plug them in below.

Number of scaling attempts for the best case:

Number of scaling attempts for the worst case:
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Type: Discussion • Score: 1 • Views: 2,137 • Replies: 4
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Butrflynet
 
  1  
Reply Tue 21 Oct, 2003 10:50 pm
Hmm, I see this riddle posted on a few sites but no one has come up with an answer yet. This one might take one of our math wizards to answer.
0 Replies
 
bobby47
 
  1  
Reply Wed 22 Oct, 2003 07:42 pm
yes it's a very hard one....
dichotomy ?
0 Replies
 
Alt3rn4tiv3
 
  1  
Reply Thu 23 Oct, 2003 09:41 pm
heh... comes from HackQuest eh?
The question's rather ambiguous...
Not sure how to explain it tho... the answer's 5 and 133 respectively =P
0 Replies
 
bobby47
 
  1  
Reply Sat 25 Oct, 2003 11:51 am
Thank you very much!!!!
i'm at your service!
0 Replies
 
 

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