Example 15 Given a set $M$, which consists of 7 consecutive natural numbers, and there exists a 5th-degree polynomial $p(x)$, such that: (1) each coefficient of the polynomial $p(x)$ is an integer; (2) there are 5 numbers $k$ in $M$, including the largest and smallest numbers in $M$, that satisfy $p(k)=k$; (3) there exists $k \in M$ such that $p(k)=0$.
(1980 British Olympiad Problem)
Let the 7 consecutive natural numbers be $a, a+1, a+2, \cdots, a+6$.
From condition (2), we can construct the polynomial
$$
p(x)=x+(x-a)(x-a-b)\left(x-b_{1}\right)\left(x-b_{2}\right)\left(x-b_{3}\right),
$$

where $b_{1}, b_{2}, b_{3} \in\{a+1, a+2, \cdots, a+5\}$.
From condition (3), we know that if $p(k)=0$, then $k \in\{a+1, \cdots, a+5\}$.
If $k=a+1$, then by (*), we have $a+1+1 \cdot(-5) \cdot c_{1} \cdot c_{2} \cdot c_{3}=0$, where $c_{1}, c_{2}, c_{3}$ are three of $-1,-2,-3,-4$.
Let $c_{1}=-1, c_{2}=-2, c_{3}=-3$, then by $a+1+1 \cdot(-5) \cdot(-1) \cdot(-2) \cdot(-3)=0$, we get $a<0$, which is a contradiction.
Similarly, if $\left\{c_{1}, c_{2}, c_{3}\right\} \varsubsetneqq\{-1,-2,-3,-4\}$, we always get $a<0$, which is a contradiction.
If $k=a+2$, then by (*), we have $a+2+2 \cdot(-4) \cdot c_{1} \cdot c_{2} \cdot c_{3}=0$, where $\left\{c_{1}, c_{2}, c_{3}\right\} \varsubsetneqq\{1,-1,-2,-3\}$.
Taking $c_{1}=1, c_{2}=-1, c_{3}=-2$, then by
$$
a+2+2 \cdot(-4) \cdot 1 \cdot(-1) \cdot(-2)=0 \text {, }
$$

we get $a=14$. Thus, the 7 numbers are $14,15,16,17,18,19,20$.
Therefore, $p(x)=x+(x-14)(x-20)(x-15)(x-17)(x-18)$ satisfies
$$
\begin{array}{l}
p(16)=16+2 \cdot(-4) \cdot 1 \cdot(-1) \cdot(-2)=0 \\
p(14)=14, p(20)=20, p(15)=15, p(17)=17, p(18)=18 .
\end{array}
$$