12. Given 98 distinct prime numbers $p_{1}, p_{2}, \cdots, p_{98}$, let $N=p_{1}^{2}+p_{2}^{2}+\cdots+p_{98}^{2}$, what is the remainder when $N$ is divided by 3?
【Answer】1 or 2
【Analysis】If the natural number $p$ is a multiple of 3, then $p^{2}$ is divisible by 3 with a remainder of 0; if the natural number is not a multiple of 3, then $p^{2}$ is divisible by 3 with a remainder of 1;
Therefore, if $p_{1}, p_{2}, \cdots, p_{98}$ contains 3, then $N \equiv 97 \equiv 1(\bmod 3)$;
if $p_{1}, p_{2}, \cdots, p_{98}$ does not contain 3, then $N \equiv 98 \equiv 2(\bmod 3)$;
In summary, the remainder when $N$ is divided by 3 is 1 or 2.