1. Let $A$ be any $k$-element subset of the set $\{1,2,3,4, \cdots, 100\}$. Determine the minimum value of $k$ such that we can always guarantee the existence of two numbers $a$ and $b$ in $A$ such that $|a-b| \leq 4$.
1. Ans: 21
If $A$ is the set of multiples of 5 in $\{1,2,3, \cdots, 100\}$, then $|A|=20$ and $|a-b| \geq 5$ for every two numbers in $A$. Thus, if $|A|=20$, the existence such two numbers cannot be guaranteed. However, if $|A|=21$, by the pigeonhole principle, there must be two numbers $a, b$ in one of the following twenty sets:
$$
\{1,2,3,4,5\},\{6,7,8,9,10\}, \cdots,\{96,97,98,99,100\},
$$
and so $|a-b| \leq 4$. Thus the answer is 21 .