1. Determine the number of all infinite arithmetic sequences of integers that have both numbers 1 and 2005 among their first ten terms.
1. Let's consider the question for which integer arithmetic sequences $\left(a_{i}\right)_{i=1}^{\infty}$ there exist indices $i, j \in\{1,2, \ldots, 10\}$ such that $a_{i}=1$ and $a_{j}=2005$. Emphasize that such a pair of indices $(i, j)$, if it exists at all, is unique, because in a non-constant arithmetic sequence, each number appears at most once.

Assume that the mentioned indices $i$ and $j$ are known, and use them to express the first term $a_{1}$ and the common difference $d$ of the sequence. Since the general term of an arithmetic sequence is given by $a_{k}=a_{1}+(k-1) d$, we get the system of equations

$$
a_{i}=a_{1}+(i-1) d=1 \quad \text { and } \quad a_{j}=a_{1}+(j-1) d=2005
$$

which we can easily solve for the unknowns $a_{1}, d$:

$$
d=\frac{2004}{j-i} \quad \text { and } \quad a_{1}=1-\frac{2004(i-1)}{j-i}.
$$

Such values of $a_{1}, d$ are integers if and only if the natural number $|j-i|$ is a divisor of 2004, so $|j-i|$ must be one of the numbers $1,2,3,4$ or 6 (from the condition $i, j \in\{1,2, \ldots, 10\}$ it follows that $|j-i|<10$ and the number 2004 has no other single-digit divisors). The number of sequences we are looking for is therefore equal to the number of pairs of indices $(i, j)$ chosen from the set $\{1,2, \ldots, 10\}$ for which $|j-i| \in\{1,2,3,4,6\}$. There are $2 \cdot 9, 2 \cdot 8, 2 \cdot 7, 2 \cdot 6$, and $2 \cdot 4$ such pairs $(i, j)$, so the total number of sequences is $18+16+14+12+8=68$.

For a complete solution, award 6 points. Consider a solution complete even if it does not explicitly state that in the given arithmetic sequence, the indices $i, j$ are uniquely determined by the conditions $a_{i}=1, a_{j}=2005$. If the solver correctly determines the number of 34 sequences with a positive common difference $d$, but neglects the possibility $d<0$, deduct 2 points.