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summary of Gareth's solution:


There are If you seat $n$ such couples randomly, then where If we forbid couples to sit opposite one another, The increase in average distance is So,


It can be done without programming or mathematics. If the set of coins is compact that means So the maximum number of coins is The minimum value is


You don't need any code for that. No spoilers as solution is quite obvious and it has been a day. 3x20 simplifies to 50+10. 20+20+10 simplifies to 50 and 20+20 > 20+10. Therefore, we cannot have 2x20. The same for 200. Anything else is obviously optimal when you have 1 coin of each type because 2 already merge to the next one. So, 6 coins with 385 total. ...



The maximum number of coins is with a minimum total value of achieved by The integer linear programming solution approach I used might be of interest. Let nonnegative integer decision variable $x_c$ be the number of coins of type $c$. The first problem is to maximize $\sum_c x_c$ subject to $\sum_c c\cdot x_c \le 495$ and "compactness" ...


If the number of people per table is $p$, then to cover each pair we must have $$\frac{600}{p} \binom{p}{2} \ge \binom{30}{2},$$ which implies that $p \ge \lceil 49/20 \rceil = 3$. The following set of $200$ triples of years covers every pair and contains each year exactly $20$ times: {{1,2,22},{1,3,5},{1,3,7},{1,4,20},{1,6,8},{1,7,15},{1,9,10},{1,11,12},{1,...


The answer is


Here's a solution with incomplete logic (and could possibly be wrong). The largest $n$ is Consider the numbers $0$ to $10^m-1$. Now that we have a range where the largest $N$ lies, we can start to track down the exact value of the largest $n$.

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