[Combinatorics]
Problem: David builds a circular table; he then carves one or more positive integers into the table at points equally spaced around its circumference. He considers two tables to be the same if one can be rotated so that it has the same numbers in the same positions as the other. For example, a table with the numbers $8,4,5$ (in clockwise order) is considered the same as a table with the numbers 4, 5,8 (in clockwise order), but both tables are different from a table with the numbers 8, 5, 4 (in clockwise order). Given that the numbers he carves sum to 17 , compute the number of different tables he can make.

Solution: The problem calls for the number of ordered partitions of 17 , where two partitions are considered the same if they are cyclic permutations of each other. Because 17 is prime, each ordered partition of 17 into $n$ parts will be a cyclic permutation of exactly $n$ such partitions (including itself), unless $n=17$. (If $n=17$, then all the numbers are 1s, and there is exactly one table David can make.) By the sticks and stones method, the number of ordered partitions of 17 into $n$ nonzero parts is $\left(\begin{array}{c}16 \\ n-1\end{array}\right)$, and this overcounts the number of tables by a factor of $n$, except when $n=17$. Thus the number of possible tables is

$$
1+\sum_{n=1}^{16}\left(\begin{array}{c}
16 \\
n-1
\end{array}\right) \cdot \frac{1}{n}=1+\sum_{n=1}^{16}\left(\begin{array}{c}
17 \\
n
\end{array}\right) \cdot \frac{1}{17}=1+\frac{2^{17}-2}{17}=\mathbf{7 7 1 1}
$$

Final answer: 7711