Daniel's path sums

If a vertex is not at the start or the end of the path, it connects to two edges in the path, and its label is added twice to the sum. If a vertex is at the start or the end of the path, it connects to one edge in the path, and its label is added once to the sum.

Therefore, if a path starts and ends at the vertices with the highest labels, it will have the minimum edge sum. Such a path will always exist since a complete graph has an edge between every pair of vertices.

Applying this, we have $\begin{aligned} f(G_n)&=2(1+2+\cdots+(n-2))+((n-1)+n) \\ &=(n-2)(n-1)+(n-1)+n \\ &=n^2-n+1 \end{aligned}$ Taking modulo $n$ , $f(G_n)\equiv 1\equiv 2013\pmod n$ Now, we must simply find the number of possible $n$ such that $2012\equiv 0\pmod n$ This is equal to the number of divisors of $2012=2^2\cdot 503$ We find that 2012 has 6 factors, but since $n>2$ , we must exclude 1 and 2. The answer is $\boxed{4}$ .

Let the vertices passed through in order be $v_1,v_2,\ldots v_n$ . Note that the edge sum would then be $v_1+v_n+2(v_2+v_3+\ldots + v_{n-1})$ .

To minimize $v_1+v_n+2(v_2+v_3+\ldots + v_{n-1})$ , we maximize $v_1$ and $v_n$ .

Therefore, $v_1=n$ and $v_n=n-1$ . This means that $2(v_2+v_3+\ldots +v_{n-1})=2(1+2+\cdots + n-2)=(n-2)(n-1)$ .

Therefore, $f(G_n)=n+n-1+(n-2)(n-1)=n^2-n+1$ .

We plug this into the bottom equation: $n^2-n+1\equiv 2013\pmod{n}$ .

Simplifying, we get that $1\equiv 2013\pmod{n}\implies 2012\equiv 0\pmod{n}\implies n|2012$ .

Factoring $2012$ , we see that $2012=2^2\cdot 503$ so $2012$ has $(2+1)(1+1)=6$ factors, and therefore there are $6$ values of $n$ that satisfy the equivalence.

However, since $n>2$ , we omit $n=1,2$ and arrive at a final answer of $6-2=\boxed{4}$ .

Daniel Chiu, great question! I had a lot of fun solving it.

Each path contains twice the number of each vertex, except the initial vertex and the final one that are counted only once. So we need to start and finish from vertex $n$ to vertex $n-1$ or vice versa and $f(G_n) = n(n-1)+1$ . So we have to solve $n(n-1) \equiv 2012 \pmod n \Leftrightarrow n | 2012$ and the solutions are $n=2012,1006,503,4$ .