GREAT Gold Bags DEALS BELOW THE ARTICLES
Part 2 - Twelve bags of gold in three or four dollars?
Then we're back with the king to leave. Thanks for your help with 8 bags of gold, King was able to accumulate more gold. Now you need 12 bags to hold all of this gold. Each bag weighs exactly the same or may be not! The King has discovered that one of his guards of confidence was counterfeiting of their gold. They found, but did not tell them who had the gold pirates! All the king's men could not know that a bag was false and that the weight of the bag was different in the king had to know: 1 Which bag weighted an amount different from the rest? 2 is that the bag heavier or lighter? The king wanted to determine that was false, and because it is cheap, I wanted to do two weight as before. The mathematician said he could not make three. Can you imagine how? No, no I. Can you help please guys? Please explain in detail how you got your solution.
It is delicate ... First, we divided into three bags of Groups A, B, C and A and B weigh There are three possible outcomes ... 1st I. weight A = B II. A> B III. An identical, so we will try together. Because I: If A = B, this means that the counterfeits are in C and A and B contain real money. Let's sign as the stock of CA, B, C, D and actual pieces weigh as R a, b, C v R, R, R 3 results back - 2 weighing 1. ABC = rrr is clear that 2 d is false. ABC> rr then Counterfeiting is heavier than real and is one of ABC. Weighing b vs identifies. (If a> b then if one is real and between ABC, with a weight b vs identifies (if you have a> b-> b, a = b-> c) II. Suppose that A B> This is even more delicate. For those who are going to be the number of A1, A2, A3, A4 and for B, B1, B2, B3, B4. One of them is false. Us Alco's C, which we call r because we are now convinced they are real. We have implemented a second weighing: A1A2A3B1vsA4rrr we have 3 options 1. A1A2A3B1 A4rrr = 2. A1A2A3B1> A4rrr 3. A1A2A3B1
Take them one by one...
1. A1A2A3B1=A4rrr
We savoir ici que la contrefaçon est parmi B2B3B4 et il est plus léger la cosa real. (porque A> B) so we can weigh B2vsB3. Weighing 3: B2> B3-> B3, B2 , B2, B3 = B2-B4. 2. A1A2A3B1> A4rrr This tells us that the B2, B3, B4 are real. This makes that A1A2A3B1> A4B2B3B4 Thus, A4 and B1 must be real. Otherwise, they would have tipped the scales A> B on the other side. This leaves us with the certainty that is false A1, A2, A3 and is heavier than the real thing. Thus, a weight of No. 3 in A1vsA2 readily reveals. 3.A1A2A3B1
once de plus, B2, B3, B4 sont réels (ils ne sont pas sur la balance et il there is unbalanced) Aussi A1, A2, A3 are real (they were previously in the heaviest of the group in the 1st and then weighed on a barge in the second, so they do not affect the imbalance. if false is A4 or B1. The third is a weight r vs. A4 If there is an imbalance, then A4 is false, if equal, B1 is false. It merely II. III is the same as II but with A and B switching. Ugh. enjoy
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