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3

To start, you should work out the sum of all the cubes, preferably using the formula $$1^3+2^3+\cdots+12^3=\frac{12^2\times13^2}{4}=6084\ .$$ So you need to find six of your twelve cubes which add up to $3042$. I should think it's trial and error from here, but would be glad to see if anyone has a smarter solution.


2

First, you've noticed that each side of the equation has to sum up to $3042$. Since $11^3+12^3=3059>3042$, each one of them has to be on a different group. Since $3042$ is even, each group must contain an even number of odd numbers. So each group must contain $0$ or $2$ or $4$ or $6$ odd numbers. With $0$ odd numbers on one group, it will definitely ...


2

The criteria demands that when you and your friends order a pizza and you each take a part of the pizza: Every one got some part of the pizza. No two people ate from the same slice. You ate the entire pizza. As for the proof, my best advice is once you get the intuitive idea, or even well before that, just work with the definitions carefully and slowly. ...


1

For the asymptotic, given $n$ we want to maximize $\frac {(n-1)^k}{k^{2k}}$. You probably know how to do that, take the derivative and set to zero. Alpha says this comes at $n=e^2k^2+1$, so $k=\frac{\sqrt{n-1}}e$


1

For each $a \in \{1,2,\ldots,n\}$, there are $k+1$ possibilities: $a$ can be element of $S_1$, of $S_2$, ..., of $S_n$ or of none of the $S_i$. It is not possible that $a$ is an element of more than one of the $S_i$, since that would contradict the condition $S_i \cap S_j$ for $i \neq j$. Multiplying these $k+1$ possibilities for all $a \in \{1,2,\ldots,n\}$ ...


1

Do I understand this correctly? The degrees $d_s$ are given, and you want to count the number of subsets $T$ of $S$ such that $\sum_{t \in T} d_t \ge N$ and $|T| \le N$. Note that given such a $T$, you can easily get the $p_t$ by starting with the $d_t$ and reducing one at a time. The number $c(m,n)$ of $T$ with each given cardinality $m = |T|$ and sum ...


1

As you started to say in your edit, this all comes down to plane partitions. An expression for the number of your partitions of $n$ is $$ \sum_{c=1}^n pp(n-c,c) $$ ($c$ for [core]) where $pp(i,j)$ is the number of plane partitions of $i$ with largest part $j$; this triangle of numbers is given in OEIS A242642 (with references back to MacMahon), which is ...


1

There are $2^n$ subsets of the first $n$ positive integers, and for each subset the sum is bounded below by zero and above by $\sum_{i=1}^n i^k = O(n^k/k)$. The problem of finding a partition with equal sums is equivalent to finding a subset with a particular sum in this range, specifically $\frac 12 \sum_{i=1}^n i^k$. If we think these subset sums are ...



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