How many positive common divisors do $10^{100}$ and $10^{121}+10^{813}+10$ have?
The number of common divisors is equal to the number of divisors of the GCD of these two numbers. Indeed: a common divisor of the two numbers will be a divisor of the GCD.

$$
\begin{aligned}
10^{813}+10^{121}+10 & =\left(10^{713}+10^{21}\right) \times 10^{100}+10 \\
10^{100} & =10^{99} \times 10+0
\end{aligned}
$$

The GCD of these two numbers is then 10, which has 4 positive divisors: $1, 2, 5$, and 10.

The number of common divisors sought is therefore 4.