A probability problem by Alejandro Castillo

The sum of the elements of the initial set $A$ , is $\frac { n(n+1) }{ 2 }$ , so we're looking for the least possible positive integer vale of $n$ , such that $\frac { n(n+1) }{ 2 }$ is divisible by $12$ . thus, $24|n(n+1)$ . Note that the sum of each subset must be equal or greater than $n$ (the greatest element of $A$ ), cause $n$ must be included in a subset.

As $n$ and $n+1$ are consecutive integers, they will have distinct parity, hence, $8|n$ or $8|(n+1)$ , and, $3|n$ or $3|(n+1)$ . From here, we can do trial and error in order to find the value of $n$ . Easily, we can discard $n=7,16$ . If $n=8$ or $n=15$ , then the sum of the elements of each subset is $3$ and $10$ , that are less than $8$ and $15$ , respectively. If $n=23$ , we can form the subsets as follows:

$\left\{ 1,22 \right\} ,\left\{ 2,21 \right\} ,...,\left\{ 11,12 \right\} ,\left\{ 23 \right\}$ .

$n=24$ is a possible value, but it's not the least possible value.

If you have the set {1,2, . . .n} and you group the elements in pairs starting from opposite ends----{1, n}, {2, n-1}, {3, n-2}, ect----you get sets with the same sum of their elements. So to get 12 subsets, I originally made n=24, but then realized that I could reduce that by 1 if I let the largest number be in a set by itself instead of a pair. Thus, I got n=23 with the set {1,22}, {2,21}, . . ., and {23}.