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14

First of all, let's see why your brute-forcing fails. (This is the puzzle part, the rest is plain old math.) No matter which you chose, the number at the bottom right would have to be both odd and even at the same time, so there's no integer solution. However, there are four equations and four unknowns, so we should have at least one solution (unless the ...

7

If this is allowed you could ask him to evaluate the polynomial at $\pi$ or another transcendental number. The response will be something like $a_0 + a_1 \pi + a_2 \pi^2 + ...$, which is unique for all combinations of $\{a_i\}$, even for negative integers (by the definiton of transcendental numbers). The drawback of this method is just that depending on the ...

3

It's not possible to find it with a single query. Suppose you had one, e.g. asking for a single number $n$. Then you can't distinguish between the cases where your friend's secret polynomial is $P(x) = 2x$ and $P(x) = x + n$; in both cases, they will reply with $2n$. (I know it's unlikely that he chose the second option, especially if $n$ is a non-obvious ...

1

Preliminary answer. The work to get a definitive answer is more than I have time for. As OP points out: Also, as OP pointed out: As with all such problems, the goal is to try to have each test divide the remaining set of possibilities into 3 groups, as evenly as possible. The only possible option for the first test is to weigh N coins on the left versus ...

11

It can be done in flips. In general case of this puzzle is an open problem called the pancake sorting problem. It is known that with $n$ pancakes the worst case takes somewhere between $\frac{15}{14}n$ and $\frac{18}{11}n$ flips. It can of course easily be solved in no more than $2(n-1)$ flips by repeatedly flipping the largest unsolved pancake to the top ...

2

I can do it in Here is the stack, bottom to top, after each of the flips --

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