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Since $x$ and $y$ are positive, $x+y$ is always greater than $x$ and $y$ - this means that undoing a move always involves decreasing the greater coordinate by some multiple of the lesser.


From (6,21) the robot can either go to (27,21) or (6,27). Another turn after that, the robot will be at either (48,21), (27, 48), (33, 27) or (6, 33). Because these numbers have no way of ever decreasing, the robot can go to any point (6a+21b,6c+21d) where a, b, c and d are all positive integers, b and c being larger than 0. Hope this answers your question!


Gareth's answer is correct. Here's the proof I found before posting: Now, Is this optimal? Therefore, yes, the solution found above is optimal, and the final answer is


(After I saw Rand's solution, it seemed to me that there ought to be a way of streamlining mine a bit. So I've done that. The original version of mine is preserved below in case anyone feels that the improved one is "polluted" by my having seen OP's own answer.) The smallest possible $a+b$ is First of all, notice that Let's Now There is one ...


I have found the following solutions:


Other than n = 1, it is easy to see that too small n's do not have such sets: The first n that has a solution in both parts is which can be solved as follows: Given this, I conjecture that because


Find the prime factorization of the modulus: 3^1*5^1 For each prime number in the factorization, take the prime number reduced by one power times one less than the prime number, and then multiply all of those numbers together. In this case, both prime numbers are raised to the first power, so reducing the power leaves just p^0, or 1. So we have (3-1)(5-1) = ...


Essentially the same problem as FizzBuzz. But this is just looking for a closed-form rather than a heuristic solution.


This is a special case of a more general problem: How to find the $N$th positive integer coprime to $k$? The important thing to know is that Therefore, to find the $N$th positive integer coprime to $k$, we first So basically, to solve this general problem you only need to know and then the rest is just basic arithmetic.


The trick is this:

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