2. For any natural number $n$, we denote by $p(n)$ the product of the digits of this natural number, written in decimal. Calculate the sum

$$
p(1)+p(2)+\cdots+p(2001)
$$
II/2. We calculate $p(1)+p(2)+\cdots+p(9)=45$. If we denote hundreds by $k$ and tens by $j$, then

$$
\begin{aligned}
p(j 0)+p(j 1)+\cdots+p(j 9) & =j \cdot 45 \\
p(k j 0)+p(k j 1)+\cdots+p(k j 9) & =k \cdot j \cdot 45 \text { and } \\
p(1 k j 0)+p(1 k j 1)+\cdots+p(1 k j 9) & =k \cdot j \cdot 45
\end{aligned}
$$

It follows that $p(1)+\cdots+p(99)+p(100)+\cdots+p(999)+p(1000)+\cdots+p(1999)+p(2000)+p(2001)=45(1+$ $1+2+\cdots+9)+45(1+2+\cdots+9)(1+2+\cdots+9)+45(1+2+\cdots+9)(1+2+\cdots+9)+0+0=184320$.