8. (1981 Kyiv Mathematical Olympiad) If the numbers $p, p+10, p+14$ are all prime, find $p$.

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8. If $p \equiv 1(\bmod 3)$, then
$$
p+14 \equiv 15 \equiv 0(\bmod 3) \text {. }
$$

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

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