jmc

Factoring out $98!+99!+100!$, we have $98!(1+99+99\ast 100)$, which is $98!(10000)$. Next, $98!$ has $\lfloor \frac{98}{5}\rfloor +\lfloor \frac{98}{25}\rfloor =19+3=22$ factors of $5$. The $19$ is because of all the multiples of $5$.The $3$ is because of all the multiples of $25$. Now, $10,000$ has $4$ factors of $5$, so there are a total of $22+4=\boxed{26}$ factors of $5$. Note: Can you say what formula this uses? most AMC 8 test takers won't know it.