8. Given a two-digit prime number $n$ such that $n +$ 20180500 is also a prime number. Then the value of $n$ that satisfies the condition is $\qquad$
8. 61 .

From $n$ being a two-digit prime number, we know
$$
n \equiv 1(\bmod 2) \text {. }
$$

Noting that, $20180500 \equiv 1(\bmod 3)$, and $n + 20180500$ is a prime number.

Thus, $n \equiv 1(\bmod 3)$, meaning $n$ is a two-digit prime number of the form $6k+1$.
Therefore, the possible values of $n$ are
$$
13, 19, 31, 37, 43, 61, 67, 73, 79, 97 \text {. }
$$

Checking each one using a TI calculator, we find that only 20180561 is a prime number.
Hence, $n=61$.