16

If the person can stay in those 2 minutes, the problem will be unsolvable because he can only move if and only if we check his room. Otherwise, here is a solution with exactly 15 14 (thanks @JaapScherphuis!) checkings, just in time!


8

Assumption: Shifty coin only switches if weighed. I believe this is optimal solution; if (case 1) otherwise (case 2) If it is Case 1; if (Case 1.1 - They are equal) if (Case 1.2 - They are not equal) then if Case 2 happens (no balance at first weighing); if there is balance (Case 2.1), (also meaning it was heavier at first) if no balance again as ...


5

They can guarantee at least two of them guess correctly with the following strategy


4

This is not a new solution but an easy way to understand how it works:


4

I think this is more of a Mathematical problem than a puzzle, and one possible solution (for a single refuel) is: More about the solution:


4

I think the answer is Reasoning:


3

I think the following would work (8 trips)


3

Proof that 51 moves is optimal


3

Solution explanation: Here is a solution for strategies of a certain form: Before departing, declare a threshold price $T_i$ for each gas station $i \in \{1,2,3,4,5\}$. Upon arriving at station $i$, fill up 100% of the tank if the station's price is less than or equal to $T_i$. What are the optimal thresholds? I conjecture that a strategy of this form is ...


3

Because


2

The answer is: Explanation: Tip:


2



2

I have managed to lock the grid in As shown:


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