Let $A={102^1, 102^2, 102^3, \cdots}$. How many primes $p$ are there such that $A$ has at least one element $a$ such that $a \equiv -1 \text{ (mod p)}$?

For example, one such prime is $103$, because $102^1 \equiv -1 \text{ (mod 103)}$.

Your answer seems reasonable.
Find out if you're right!