$3 \cdot 1$ If the numbers $p, p+10, p+14$ are all prime, find $p$.
(Kiev Mathematical Olympiad, 1981)
[Solution] If $p \equiv 1(\bmod 3)$, then
$$
p+14 \equiv 15 \equiv 0 \quad(\bmod 3) .
$$

In this case, $p+14$ is not a prime number.
If $p \equiv 2(\bmod 3)$, then
$$
p+10 \equiv 12 \equiv 0 \quad(\bmod 3) .
$$

In this case, $p+10$ is not a prime number.
Therefore,
$$
p \equiv 0 \quad(\bmod 3) .
$$

Since $p$ is a prime number, then $p=3$.
In this case, $p+10=13, p+14=17$ are both prime numbers.
Thus, the only solution to this problem is: $p=3$.