45.2. Find the natural number that is 40 times less than the sum of all preceding natural numbers.

$$
(9-10 \text { th } )
$$

10
45.2. We have the equation:

$$
\begin{gathered}
1+2+3+\ldots+(n-1)=40 n \text {, i.e., } \frac{1}{2} n(n-1)=40 n, \\
\text { hence } n=81 .
\end{gathered}
$$