Harvard-MIT Mathematics Tournament

Solution:
Number the stones $0, 1, \ldots, 41$, treating the numbers as values modulo $42$, and let $r_n$ be the length of your jump from stone $n$. If you jump from stone $n$ to $n+7$, then you cannot jump from stone $n+6$ to $n+7$ and so must jump from $n+6$ to $n+13$. That is, if $r_n = 7$, then $r_{n+6} = 7$ also. It follows that the 7 values $r_n, r_{n+6}, r_{n+12}, \ldots, r_{n+36}$ are all equal: if one of them is $7$, then by the preceding argument applied repeatedly, all of them must be $7$, and otherwise all of them are $1$.

Now, for $n = 0, 1, 2, \ldots, 42$, let $s_n$ be the stone you are on after $n$ jumps. Then $s_{n+1} = s_n + r_{s_n}$, and we have $s_{n+1} = s_n + r_{s_n} \equiv s_n + 1 \pmod{6}$. By induction, $s_{n+i} \equiv s_n + i \pmod{6}$; in particular $s_{n+6} \equiv s_n$, so $r_{s_n+6} = r_{s_n}$. That is, the sequence of jump lengths is periodic with period $6$ and so is uniquely determined by the first $6$ jumps. So this gives us at most $2^6 = 64$ possible sequences of jumps $r_{s_0}, r_{s_1}, \ldots, r_{s_{41}}$.

Now, the condition that you visit each stone exactly once before returning to the original stone just means that $s_0, s_1, \ldots, s_{41}$ are distinct and $s_{42} = s_0$. If all jumps are length $7$, then $s_6 = s_0$, so this cannot happen. On the other hand, if the jumps are not all of length $7$, then we claim $s_0, \ldots, s_{41}$ are indeed all distinct. Indeed, suppose $s_i = s_j$ for some $0 \leq i < j < 42$. Since $s_j \equiv s_i + (j-i) \pmod{6}$, we have $j \equiv i \pmod{6}$, so $j-i = 6k$ for some $k$. Moreover, since the sequence of jump lengths has period $6$, we have
$$s_{i+6} - s_i = s_{i+12} - s_{i+6} = \cdots = s_{i+6k} - s_{i+6(k-1)}$$
Calling this common value $l$, we have $k l \equiv 0 \pmod{42}$. But $l$ is divisible by $6$, and $j-i < 42 \Rightarrow k < 7$ means that $k$ is not divisible by $7$, so $l$ must be. So $l$, the sum of six successive jump lengths, is divisible by $42$. Hence the jumps must all be of length $7$, as claimed.

This shows that, for the $64-1=63$ sequences of jumps that have period $6$ and are not all of length $7$, you do indeed reach every stone once before returning to the starting point.