Problem 1344

### Part (a)

1. Understanding the Problem: We need to arrange 14 students in a circle such that every two students sitting next to one another have different numbers. Each student has at least $k$ favorite numbers.

2. **Analyzing $k = 1$**: If $k = 1$, each student has only one favorite number. In the worst case, all students could have the same favorite number, making it impossible to arrange them in a circle with different numbers next to each other.

3. **Analyzing $k = 2$**: If $k = 2$, each student has at least two favorite numbers. In the worst case, all students have the same two favorite numbers, say 1 and 2. We can arrange the students in the sequence $1, 2, 1, 2, \ldots, 1, 2$, ensuring that no two adjacent students have the same number.

Thus, $k = 2$ is sufficient.

### Part (b)

1. Understanding the Problem: We have 14 students, 7 boys and 7 girls. Each student has at least $k$ favorite numbers, and no boy and girl share the same favorite number. We need to determine the least $k$ such that each student can be assigned a favorite number.

2. Graph Representation: Consider a bipartite graph $G(S, N)$ where $S = \{1, 2, \ldots, 14\}$ represents the students and $N = \{n_1, n_2, \ldots, n_k\}$ represents the favorite numbers. Each student $s \in S$ is connected to at least 4 numbers in $N$.

3. Partitioning: Partition the students into two groups $S_0$ (boys) and $S_1$ (girls). We need to partition $N$ into two disjoint sets $N_0$ and $N_1$ such that each boy is connected to a number in $N_0$ and each girl is connected to a number in $N_1$.

4. Bad Assignments: Define $\mathcal{A}$ as the set of all possible assignments (partitions) of $N$, which has $2^k$ elements. For each student $s \in S$, let $B_s \subset \mathcal{A}$ be the set of "bad" assignments for $s$. For boys, $B_s$ consists of assignments where all neighbors of $s$ are in $N_1$. For girls, $B_s$ consists of assignments where all neighbors of $s$ are in $N_0$.

5. Counting Bad Assignments: Since each student is connected to at least 4 numbers, $|B_s| = \frac{2^k}{2^d} \le 2^{k-4}$.

6. Union of Bad Assignments: Let $B = \bigcup_{s \in S} B_s$. Then,
$$|B| \le \sum_{s \in S} |B_s| \le 14 \cdot 2^{k-4}$$
For $k = 4$,
$$|B| \le 14 \cdot 2^{4-4} = 14 < 2^4 = 16$$
Therefore, $|B| < |\mathcal{A}|$, meaning there exists an assignment $A \in \mathcal{A} \setminus B$.

7. Conclusion: There exists an assignment that satisfies the conditions, proving that $k = 4$ is sufficient.

$\blacksquare$

The final answer is $\boxed{ k = 2 }$ for part (a) and $k = 4$ for part (b).