One missing constraint is A<C. In other words, 2x<100-4x, or x<16.67

Another missing constraint is B+x=C. In other words, 2x+x<100-4x, or x<14.29

So, given this strategy, it seems that 14 is optimal.

]]>Following efimp’s solution in comments to knop’s blog:

split all coins to A,B,C; A=B in count; if A>B in weight, take x=y from A, if x>=y in wieght, x are geniune;

if A=B in weight, take x from A, (B+x)=C in count; (B+x)>C -> x genuine; (B+x)=C -> A-x genuine; (B+x) C genuine;

genuine outcomes are: x,y,A-x,C; constraints are x=y, 2x<=A, A=B, 100=C+A+B; this is simplified to maximization of x and 100-4x; thus x=100/5=20;

This of course only shows that 20 is possible; it’s not a proof that more is impossible.

]]>In the first message of this thread you cite Knop’s paper on Shapovalov coin problem (in Russian).

Does it go any further than the algorithm described above by Colorblind ?

If so, could you please write a short abstract/review stating the main result ?

That would be helpful for people like me who don’t understand Russian.

Thank you.

Jean Drabbe (Belgium) ]]>

For the proof of “no 15” hypothesis, I think a few things are to be proved:

1) All weighings must compare equal quantities of coins (Obvious, but even/odd will come into play later.)

2) Not all coins can be weighed on the first weighing (Kind of obvious, but allows us to partition the 100 coins into three groups for the first weighing, which is critical for the second weighing.)

3) If the first weighing results in equality, then all coins must be weighed at some point. (Not entirely obvious, but this forces any suitable algorithm to be very similar to the 13 coin solution mentioned above.)

4) If the first weighing results in equality, the unweighed coins from the first part must be in the same group when weighed during the second part. (A little work here, but no hard lifting.)

5) If the first weighing results in equality, the unweighed coins cannot be exactly one of the groups on the second weighing. (If that is so, the max coins that can be found is 13, from the generalization of the solution to the initial problem)

6) If the first weighing results in equality, adding 2 or more coins from one of the first two piles to the unweighed pile to create a group for the second weighing will result in no solution due to ambiguities.

Once those 6 points have been proved, we have what we need to finish both a uniqueness proof for the 14 solution and a “no 15” proof as desired. My main concern is 3). Intuitively, it makes sense; any unweighed coins will result in significantly less information destroying a possible solution. Thoughts?

]]>If you really want something to think about, consider a generalization of the problem:

Let “k” and “f” be given natural numbers with “f<=k”. Among “k” coins exactly “f” are fake. All genuine coins weigh the same; all fake coins, too. A fake coin is lighter than a genuine coin. For what values of “k” and “f” can we find at least one genuine coin using two weighings on a balance scale? How would you do it?

What is the biggest number “N” such that regardless of the results of each individual weighing, your strategy will allow you to identify at least “N” genuine coins?

What if we allow “w” weighings for some integer “w>=3”?

Of course, above “N” depends on the choice of “k,f,w”, so is a function. Find the function.

]]>Let “k” and “f” be given natural numbers with “f=3”?

Of course, above “N” depends on the choice of “k,f,w”, so is a function. Find the function.

]]>The mortgages were sold with the buyers having the belief that 1) the market wouldn’t collapse, 2) Interest rates wouldn’t increase, and 3) their personal income would improve. Without the benefit of hindsight, those are not horrible assumptions. Of course – and this is both the virtue and the vice of Americans (IMHO) – nobody bothered to hedge for the “unlikely” possibility something would go wrong.

With that in mind, I think “more education” is a red herring. There were educated people who understood the mortgages who could have used their knowledge, but they chose not to, in many cases because they wanted the largest personal gain and, here it comes again, did not hedge for the “unlikely” possibility something would go wrong.

I would certainly argue Americans have some innumeracy with regard to risk, but that it seems to be a matter of temperament rather than education.

(BTW, I really am working on the problem…the assumption being if there is a 15 coin solution, then it constitues a 14 coin solution. The known 14 coin solution is based on the solution to the original problem (I think…I’m trying to do this within down time that’s a function of seconds) and that does not extend to a 15 coin solution. Therefore, a 15 coin solution exists if and only if the known 14 coin solution is not unique. Luckily, the number of first and second weighings is finite and a large swath of first weighings possibilities can be eliminated easily, so a uniqueness proof seems doable.)

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