19. For any positive integer $M$, let
$$
M!=M \cdot(M-1) \cdots \cdot 2 \cdot 1 \text {, }
$$
$x^{n} \| y$ denotes $x^{n} \mid y$, and $x^{n+1} \nmid y\left(x, y \in \mathbf{Z}_{+}\right)$.
Then the largest integer $n$ such that $5^{n} \|(98!+99!+100!)$ is ( ).
(A) 23
(B) 24
(C) 25
(D) 26
(E) 27
19. D.

Notice,
$$
\begin{array}{l}
98!+99!+100!=98!(1+99+100 \times 99) \\
=98! \times 2^{4} \times 5^{4}.
\end{array}
$$

Let $[x]$ denote the greatest integer not exceeding the real number $x$.
Also, $\left[\frac{98}{5}\right]+\left[\frac{98}{5^{2}}\right]=19+3=22$, so the largest integer $n=22+4=26$.