New answers tagged combinatorics
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summary of Gareth's solution:
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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,
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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
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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.
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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" ...
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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,...
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The answer is
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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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