20160731, 16:37  #1 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2661_{16} Posts 
Avgvst 2016

20160731, 19:12  #2 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2×19×59 Posts 
That does not make sense. The wording implies that the scale is not a balance scale.
With a single Measurment reading for 10 bags any Measurment not involving any of the 10 bags will leave unmeasured bags unknown. If all 10 bags are measured at once then there would be no way to distinguish between them. The best you could do is to determine the number of counterfeit bags but not to know which specific bags are counterfeit. BTW There seem to be a typo (or two) in the title or am I missing something. Last fiddled with by a1call on 20160731 at 19:17 
20160731, 19:42  #3 
"Vincent"
Apr 2010
Over the rainbow
2·3·11·43 Posts 
The wording imply a scale with 2 plate.

20160731, 19:43  #4 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×5^{2}×131 Posts 

20160731, 19:47  #5  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
3×5^{2}×131 Posts 
Quote:


20160731, 20:09  #6 
"Vincent"
Apr 2010
Over the rainbow
2·3·11·43 Posts 
ah, this *accurate scale* has a limit?

20160731, 20:35  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
23141_{8} Posts 
None is implied by PT, but it is clear that within this problem a 20.470 kg limit will do fine, with precision of 1 g.
Such scales are quite routine. Our labs have scales that are doing a few grams but up to the fifth, I think, decimal digit behind dot, so they are more precise. 
20160801, 02:48  #8 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2242_{10} Posts 
The problem implies that for N<1024 it would take more than a single Measurment to solve. So the actual problem would require more than a single Measurment and no maximum is specified which makes the solution trivial. The problem is the N>=1024 being solved in a single Measurment does not make sense specially with an unknown number of fake coin bags.
I just don't understand the parameters of the problem. 
20160801, 03:20  #9 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2×17×293 Posts 
The 1024 is just an example, same as they use to give every month examples, for every challenge. It does not imply that you would need more measurements if N is smaller. It may be true or not, but the example itself implies nothing. Of course, we learned in elementary school how to solve the "example". In fact, the condition is very weak, even if N>=512, you could do it with a single scale very simply, you take 2^(n1) coins from the nth bag, and you have only 10 bags... So you don't need N >= 1024. When N>=512, you could take 1 coin from the first bag, 2 from the second, 4 from the third, etc, and 512 from the 10th bag, then you scale them together. At the end you have 1023 coins and expect to get 10230 grams if all coins are good, but you will get less. You get one gram less for each bad coin. Write the difference in binary and you know exactly how many bags have wrong coins, and which bags they are. If for example you get 29 grams less, which in binary is 11101, you know that there are 4 bags with bad coins, and those are the first, the third, forth and fifth.
Now, you have less coins in each bag. But you also know that there are at most 3 bags with bad coins. So, it should not be difficult to devise a scheme where you can point to the cheating bags, with only one scaling. This problem, same like the one last month, is actually very nice. Last fiddled with by LaurV on 20160801 at 03:49 
20160801, 03:28  #10 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4302_{8} Posts 
Thank you LaurV,
Somehow I was under the impression that bags where to be weighted and not the coins. The example does make sense now. Thanks again for the clarification. 
20160804, 19:15  #11 
"Robert Gerbicz"
Oct 2005
Hungary
11000100000_{2} Posts 
New update:
"Update (4/8): To get a '**' find a solution for the minimal N." 
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