CEMC Cayley

Suppose that, when the $n$ students are put in groups of 2, there are $g$ complete groups and 1 incomplete group. Since the students are being put in groups of 2, an incomplete group must have exactly 1 student in it. Therefore, $n=2g+1$. Since the number of complete groups of 2 is 5 more than the number of complete groups of 3, then there were $g-5$ complete groups of 3. Since there was still an incomplete group, this incomplete group must have had exactly 1 or 2 students in it. Therefore, $n=3(g-5)+1$ or $n=3(g-5)+2$. If $n=2g+1$ and $n=3(g-5)+1$, then $2g+1=3(g-5)+1$ or $2g+1=3g-14$ and so $g=15$. In this case, $n=2g+1=31$ and there were 15 complete groups of 2 and 10 complete groups of 3. If $n=2g+1$ and $n=3(g-5)+2$, then $2g+1=3(g-5)+2$ or $2g+1=3g-13$ and so $g=14$. In this case, $n=2g+1=29$ and there were 14 complete groups of 2 and 9 complete groups of 3. If $n=31$, dividing the students into groups of 4 would give 7 complete groups of 4 and 1 incomplete group. If $n=29$, dividing the students into groups of 4 would give 7 complete groups of 4 and 1 incomplete group. Since the difference between the number of complete groups of 3 and the number of complete groups of 4 is given to be 3, then it must be the case that $n=31$. In this case, $n^{2}-n=31^{2}-31=930$; the sum of the digits of $n^{2}-n$ is 12.