# Tag Info

26

Part 1 The final solution (finally)

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These should do it: Just to show another example:

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If a rectangular piece of chessboard is $a\times b$ squares in size, then its diagonal squared is $a^2+b^2$ and its area squared is $a^2\cdot b^2$, and therefore the quantities $a^2,b^2$ are the roots of the quadratic equation $x^2-D^2x+A^2=0$ where $D,A$ are the diagonal and area. But this is enough to determine $\{a^2,b^2\}$ completely: these values are $\... 11 I think this is the answer Partial Reasoning 9 Solved grid: Solving path for right grid: 8 I think it can be done in: Method: 8 This can be done in Steps 8 Here is a new approach. The old solution was not perfect. Logician A can force the crates to balance. 8 I believe this is the solution :) Some information on how I solved it: 8 It seems rather straightforward to me, unless I've misunderstood some detail of the question. From what the audience heard, we have: This has the solution: The answers to the questions are therefore: 6 I think it's a safe assumption that the answer will be of the form: The reasoning is that Now, we want to get the longest sequence. To achieve this For instance, if we try Let's try another: Going even lower is not an alternative, since: Without a rigorous proof I'm going to say the combination is: 6 First of all, And now the point is that which means that 5 Because which means To be pedantic, however, 4 This can be done in Steps 4 When B is down to three coins left, they are left between the crates. This will happen before A runs out of coins. So the final steps are up to A to carry out. A will use the 3 coins left by B in the same way as A's own coins. Following exactly the same method as before, ensuring both crates have equal coins is simple. The only possible objection here is ... 4 An alternate approach: 4 The train is travelling at: Working: 4 This mistake lies: 3 First notes: Now, Therefore, That leaves only So the answer is 3 After looking at @hexomino's answer I came up with a slightly different proof. 3 This is only solvable in the sense that no solution is better than another. Since the judge has no prior knowledge, starting at any position is just as good, so the judge might as well toss a (38-sided) coin to decide, (treating the line as if its ends were connected to each other) and this is still optimal for the judge. If the judge does that, it becomes ... 3 Reasoning: 3 A and B have two logical choices. Let's say A choose strategy 1. Then, when it's A's turn again, either the crates will be the same weight, in which case A and B chose the same strategy. If they're not the same weight, that means the B chose strategy 2. If that happens, then A will now change their approach to strategy 2. After that it's straight ... 3 The (twice) given geometrical answer is very clever. Here's a much less clever solution, using very basic engineering only. It utilises the fact that the diagonal happens to be the longest dimension of the brick. Stand two of the bricks up on the ground with some room in between. Keeping one corner always touching one brick, Now you can easily measure ... 3 I think this is the answer. 2 I am not giving a new solution. But I'd like to propose a nicer expression of the solution given for the general case$n=2^k$. In fact my solution is the same as noedne's. And the explanation why it works is the same as OHO's. 2 Answer Steps If there is exactly one imbalance$a<b$, then coin$a$weighs 9 g and coin$b$weighs 11 g. If there are two imbalances$a<b$and$c<d$, then weigh only coins$a$and >!$c$(one extra weighing) Case 1:$a<c \implies$coin$a$weighs 9 g and coin$d$weighs 11 g Case 2:$c<a \implies$coin$c$weighs 9 g and coin$b$weighs 11 g ... 2 The answer is Unfortunately, this answer is neither elegant nor easy to explain since I found it via brute force. It's pretty disappointing to solve a puzzle this way, but I don't think anyone explained a correct answer yet (at least before I was sniped by Charles Gleason!). The General Approach Consider the case of 9 coins with one heavier than the rest.... 2 Instead of a cartesian plane, let's consider instead a... That means rotating$n$,$m$and$k$around, so that So armed with this nomenclature, let's run a few iterations starting with only the$(0,0)$point: ...we can conclude that the number of moves needed to clear the... ...row/diagonal is... 2 I claim that the long term average number of cars for$N\$ starting cars will be To prove this we can proceed by induction

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