The elements of the set $\{1,2,3,4,5,6,7,8,9,10,11\}$ can be separated into the sets $\{3,9,10,11\}$ and $\{1,2,4,5,6,7,8\}$ such that the sum of the elements in each set is equal to 33.

(a) Display a way to separate the elements of the set $\{1,2,3,4,5,6,7,8\}$ into three sets such that the sum of the elements in each set is the same.
(b) Explain why it is not possible to separate the numbers of the set $\{1,2,3,4,5,6,7,8,9,10\}$ into two sets of the same sum.

(c) For each positive integer $n \geq 2$, determine the smallest positive integer $N$ such that the set $\{1,2, \ldots, N\}$ can be separated into exactly $n$ sets of the same sum.
Solution

(a) The sum of the numbers from 1 to 8 is equal to 36, so each of the three sets must have a sum of 12. It is sufficient to separate them into the sets: $\{4,8\},\{5,7\}$ and $\{1,2,3,6\}$.

(b) The sum of the numbers from 1 to 10 is equal to 55. To separate them into two sets of the same sum, each would have positive integer elements and a sum of $\frac{55}{2}$. Since $\frac{55}{2}$ is not an integer, such a separation is not possible.

(c) If we form $n$ sets of the same sum, at most one of them can have one element, and the other $n-1$ sets would have at least two elements. This allows us to conclude that $N \geq 1+2 \cdot(n-1)=2 n-1$. To verify that $N=2 n-1$ is the smallest, we need to separate the set $\{1,2, \ldots, 2 n-1\}$ into $n$ sets of the same sum. Consider the sets $\{2 n-1\},\{2 n-2,1\},\{2 n-3,2\}, \ldots,\{n+1, n-2\}$ and $\{n, n-1\}$. They are exactly $n$ sets, and each has a sum of elements equal to $2 n-1$. Thus, we conclude that $N=2 n-1$ is the smallest positive integer that satisfies the conditions of the problem.