3.1. (12 points) The number

$$
\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}
$$

was written as an irreducible fraction with natural numerator and denominator. Find the last two digits of the numerator.
Answer: 99.

Solution. We have

$\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\ldots+\frac{2017}{2018!}=\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\ldots+\left(\frac{1}{2017!}-\frac{1}{2018!}\right)=1-\frac{1}{2018!}=\frac{2018!-1}{2018!}$.

In the end, we obtained an irreducible fraction, and the last two digits of the number 2018! - 1 are two nines.