A sequence $x_1, x_2, ..., x_n, ...$ consists of an initial block of $p$ positive distinct integers that then repeat periodically. This means that $\{x_1, x_2, \dots, x_p\}$ are $p$ distinct positive integers and $x_{n+p}=x_n$ for every positive integer $n$. The terms of the sequence are not known and the goal is to find the period $p$. To do this, at each move it possible to reveal the value of a term of the sequence at your choice.

(a) Knowing that $1 \le p \le 10$, find the least $n$ such that there is a strategy which allows to find $p$ revealing at most $n$ terms of the sequence.
(b) Knowing that $p$ is one of the first $k$ prime numbers, find for which values of $k$ there exist a strategy that allows to find $p$ revealing at most $5$ terms of the sequence.
(a) Knowing that \(1 \le p \le 10\), find the least \(n\) such that there is a strategy which allows to find \(p\) revealing at most \(n\) terms of the sequence.

1. If \(n \le 3\), then there are at most \(B_3 = 5\) ways to partition a set of \(n\) elements (where \(B_n\) is the \(n\)-th Bell number), so one can only get at most \(5 < 10\) distinct possible scenarios in any strategy. This is because each partition of a set with \(n\) elements corresponds with a possible scenario when \(n\) terms \(x_{a_1}, x_{a_2}, \dots, x_{a_n}\) of the sequence are revealed. For example, \(\{1, 2, 3, 4, 5\}\) can be partitioned into \(\{\{1, 2, 3\}, \{4\}, \{5\}\}\), which corresponds to \(x_{a_1} = x_{a_2} = x_{a_3}\) and \(x_{a_1} \ne x_{a_4} \ne x_{a_5} \ne x_{a_1}\). Thus, \(n \ge 4\).

2. For \(n = 4\), we can ask for the values of \(x_1, x_5, x_{11}, x_{46}\). Note that \(x_i = x_j\) for positive integers \(i < j\) if and only if \(p \mid j - i\) since the \(p\) repeated terms in the sequence are distinct. The following strategy works:
   - If \(x_1 = x_5 = x_{11} = x_{46}\), then \(p = 1\).
   - If \(x_1 = x_5 = x_{11}\) and \(x_{46} \ne x_1\), then \(p = 2\).
   - If \(x_1 = x_{46}\), \(x_{5} = x_{11}\), and \(x_1 \ne x_5\), then \(p = 3\).
   - If \(x_1 = x_5\) and \(x_1 \ne x_{11} \ne x_{46} \ne x_1\), then \(p = 4\).
   - If \(x_1 = x_{11} = x_{46}\) and \(x_1 \ne x_5\), then \(p = 5\).
   - If \(x_5 = x_{11}\) and \(x_1 \ne x_5 \ne x_{46} \ne x_1\), then \(p = 6\).
   - If \(x_{11} = x_{46}\) and \(x_1 \ne x_5 \ne x_{11} \ne x_1\), then \(p = 7\).
   - If \(x_1, x_5, x_{11}, x_{46}\) are distinct, then \(p = 8\).
   - If \(x_1 = x_{46}\) and \(x_1 \ne x_5 \ne x_{11} \ne x_1\), then \(p = 9\).
   - If \(x_1 = x_{11}\) and \(x_1 \ne x_5 \ne x_{46} \ne x_1\), then \(p = 10\).

Thus, the least \(n\) for which there exists a strategy is \(\boxed{4}\).

(b) Knowing that \(p\) is one of the first \(k\) prime numbers, find for which values of \(k\) there exists a strategy that allows to find \(p\) revealing at most \(5\) terms of the sequence.

1. We claim that there exists a strategy if and only if \(\boxed{k \le B_5 = 52}\).

2. If \(k \ge 53\), then by only asking at most \(5\) questions, one can only get at most \(B_5 = 52\) distinct possible scenarios, which is not enough to distinguish all \(k\) possible periods.

3. If \(k \le 52\), we will show there exists a strategy. Similar to (a), we will attempt to create a one-to-one correspondence between the \(52\) partitions of \(S = \{1, 2, 3, 4, 5\}\) and the distinct possible scenarios when revealing \(5\) terms \(x_{a_1}, x_{a_2}, x_{a_3}, x_{a_4}, x_{a_5}\) of the sequence, where \(x_{a_m} = x_{a_n}\) if and only if \(m\) and \(n\) are in the same subset for a unique partition of \(S\).

4. Let \(p_1 < p_2 < \dots < p_{52}\) be the first \(52\) prime numbers, and let \(P = p_1p_2 \dots p_{52}\). So the period of the sequence is \(p_t\) for some \(1 \le t \le 52\). We number the partitions of \(S\) from \(1\) to \(52\), with Partition \(1\) being \(\{\{1,2,3,4,5\}\}\) and Partition \(2\) being \(\{\{1\}, \{2,3,4,5\}\}\). The rest of the partitions can be numbered from \(3\) to \(52\) in any way.

5. Define the function \(f_i: S \to S\) such that for each \(x \in S\), \(f_i(x)\) is the smallest element of \(S\) in the same subset as \(x\) in Partition \(i\) (\(f_i(x)\) can be equal to \(x\)). Note that \(f_1(x) = 1\) for all \(x \in S\), \(f_2(1) = 1\), and \(f_2(x) = 2\) for all \(x \in \{2,3,4,5\}\). Since \(p_3 = 5\), we have \(1 \le f_i(x) \le p_i\) for all \(x \in S\) and \(1 \le i \le 52\).

6. Let \(a_n = \sum_{i = 1}^{52} f_i(n) \frac{P}{p_i}\) for \(n = 1,2,3,4,5\). Once again, \(x_{a_m} = x_{a_n}\) if and only if \(p_t \mid a_n - a_m\). Since \(p_t \mid \frac{P}{p_i}\) for all \(i \ne t\) and \(p_t \nmid \frac{P}{p_t}\), then \(a_n \equiv f_t(n)\frac{P}{p_t} \pmod{p_t}\). Hence, \(p_t \mid a_n - a_m\) if and only if \(p_t \mid \frac{P}{p_t}(f_t(n) - f_t(m)) \iff p_t \mid f_t(n) - f_t(m)\). Since \(1 \le f_t(m), f_t(n) \le p_t\), \(p_t \mid f_t(n) - f_t(m)\) if and only if \(f_t(n) = f_t(m)\), i.e. \(m\) and \(n\) are in the same subset in Partition \(t\).

7. Therefore, if \(p_t\) is the period of the sequence for some \(1 \le t \le 52\), then \(x_{a_m} = x_{a_n}\) if and only if \(m\) and \(n\) are in the same subset in Partition \(t\). This makes it possible for one to reveal \(x_{a_1}, x_{a_2}, x_{a_3}, x_{a_4}, x_{a_5}\) determine which of them are equal and unequal, match the scenario to a partition of \(S\), and find the value of \(t\) uniquely.

8. For \(k < 52\), a similar strategy can be made. Thus, for all \(k \le 52\), there exists a strategy to find the period \(p\) of the sequence.