The tenth question, set $\mathrm{X}=\{1,2, \ldots, 100\}$, function $\mathrm{f}: \mathrm{X} \rightarrow \mathrm{X}$ simultaneously satisfies: (1) For any $\mathrm{x} \in \mathrm{X}$, $f(x) \neq x$; (2) For any 40-element subset $A$ of $X$, there is $A \cap f(A) \neq \Phi$.

Find the smallest positive integer $\mathrm{k}$, such that for any function $\mathrm{f}$ satisfying the above conditions, there exists a $\mathrm{k}$-element subset $B$ of $\mathrm{X}$, such that $B \cup f(B)=X$. (Note: For a subset $T$ of $X$, define $f(T)=\{x \mid x=f(t), t \in T\}$.)
The tenth question,
Answer: If there exists $f\left(x_{1}\right)=y_{1}$, then remove $x_{1}, y_{1}$ from the set $x$, and continue this operation in the remaining set until it can no longer be performed. Suppose a total of $\mathrm{m}$ operations have been performed. At this point, there are $100-2 \mathrm{~m}$ numbers left in the set, and in the remaining set $x^{\prime}$, there no longer exist elements $x^{\prime}, y^{\prime}$ such that $f\left(x^{\prime}\right)=y^{\prime}$. This means $x^{\prime} \cap f\left(x^{\prime}\right)=\Phi$, so $\left|x^{\prime}\right|<40$, i.e., $100-2 \mathrm{~m}<40 \Rightarrow \mathrm{m} \geq 31$.

We first prove that for any function $\mathrm{f}$ that satisfies the given conditions, there exists a 69-element subset $\mathrm{B}$ of $\mathrm{x}$ such that $B \cup f(B)=X$. In fact, suppose the 31 pairs of numbers removed from the set $x$ are $x_{1} \rightarrow y_{1}, x_{2} \rightarrow y_{2}, \cdots, x_{31} \rightarrow y_{31}$, and the remaining 38 numbers are $x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{38}^{\prime}$. Let $B=\left\{x_{1}, x_{2}, \ldots, x_{31}, x_{1}^{\prime}, x_{2}^{\prime}, \ldots, x_{38}^{\prime}\right\}$, then $|B|=69$, and $B \cup f(B)=X$.

Next, we construct a function $\mathrm{f}$ that satisfies the conditions, such that for any 68-element subset $C$ of $\mathrm{X}$, $\mathrm{C} \cup f(\mathrm{C}) \neq \mathrm{X}$.
$$
\begin{array}{l}
\text { Let }\left\{\begin{array}{l}
f(3 k-2)=3 k-1 \\
f(3 k-1)=3 k \\
f(3 k)=3 k-2
\end{array} \text {, where } k=1, 2, \ldots, 30 \text {, and } f(91)=f(92)=\ldots=f(99)=100, \right. \text { } \\
f(100)=99 \text { . }
\end{array}
$$

Clearly, for any $x \in X, f(x) \neq x$. Also, for any 40-element subset $A$ of $X$, if $99, 100 \in A$, then $\{99,100\} \subseteq \mathrm{A} \cap \mathrm{f}(\mathrm{A})$; if $99, 100$ is not in $A$, then at least 31 numbers from $1, 2, \ldots, 90$ belong to $A$. Dividing $1, 2, \ldots, 90$ into 30 groups: $\{1,2,3\}, \{4,5,6\}, \ldots, \{88,89,90\}$, by the pigeonhole principle, $A$ contains at least two elements from the same group. According to the definition of $\mathrm{f}$, $\mathrm{A} \cap \mathrm{f}(\mathrm{A}) \neq \Phi$. Thus, the function $\mathrm{f}$ defined above satisfies the given conditions.
For this function $f$, to find a subset $B$ of $X$ such that $B \cup f(B)=X$, at least two numbers must be taken from each of the groups $\{1,2,3\}, \{4,5,6\}, \cdots, \{88,89,90\}$; all numbers from $\{91,92, \ldots, 98\}$ must be taken; and at least one number from $\{99,100\}$ must be taken. In this way, $|B| \geq 30 \times 2+8+1=69$. Therefore, for any 68-element subset $C$ of $X$, $C \cup f(C) \neq X$.
In summary, the smallest value of $\mathrm{k}$ that satisfies the conditions is 69.