High I.

An irreducible fraction $\frac{a}{b}$ is such that

$$
\frac{a}{b}=\frac{999}{1999}+\frac{999}{1999} \cdot \frac{998}{1998}+\frac{999}{1999} \cdot \frac{998}{1998} \cdot \frac{997}{1997}+\ldots+\frac{999}{1999} \cdot \frac{998}{1998} \cdot \ldots \cdot \frac{1}{1001}
$$

Find $a$ and $b$.

#
Notice that in each term there is a fraction $\frac{999}{1999}$. Let's factor it out:

$$
\frac{a}{b}=\frac{999}{1999}\left(1+\frac{998}{1998}+\frac{998}{1998} \cdot \frac{997}{1997}+\ldots+\frac{998}{1998} \cdot \ldots \cdot \frac{1}{1001}\right)
$$

In each term inside the parentheses, except for the 1 left from $\frac{999}{1999}$, there will be a fraction $\frac{998}{1998}$, which can also be factored out:

$$
\frac{a}{b}=\frac{999}{1999}\left(1+\frac{998}{1998}\left(1+\frac{997}{1997} \cdot \frac{996}{1996}+\ldots+\frac{997}{1997} \cdot \ldots \cdot \frac{1}{1001}\right)\right)
$$

And in the parentheses, in each term except for the 1 left from $\frac{998}{1998}$, there will be a fraction $\frac{997}{1997}$, which can be factored out. This process can be continued until inside the parentheses there is only the sum $1+\frac{1}{1001}$:

$$
\frac{a}{b}=\frac{999}{1999}\left(1+\frac{998}{1998}\left(\ldots\left(1+\frac{3}{1003}\left(1+\frac{2}{1002}\left(1+\frac{1}{1001}\right)\right)\right) \ldots\right)\right)
$$

Now we will sequentially calculate the value of this expression from the inside out. Let's start with the calculation of the expression in the innermost parentheses:

$$
\left(1+\frac{2}{1002}\left(1+\frac{1}{1001}\right)\right)=\left(1+\frac{2}{1002} \cdot \frac{1002}{1001}\right)=\left(1+\frac{2}{1001}\right)
$$

$$
\left(1+\frac{3}{1003}\left(1+\frac{2}{1001}\right)\right)=\left(1+\frac{3}{1003} \cdot \frac{1003}{1001}\right)=\left(1+\frac{3}{1001}\right)
$$

Now we can make a hypothesis and verify it: at each step, we will get a fraction of the form $\frac{k}{1001}$. Indeed, if a fraction $\frac{k-1}{1001}$ has already been obtained, then on the next step, $\frac{k}{1001}$ will indeed be obtained:

$$
\begin{aligned}
\frac{a}{b}= & \frac{999}{1999}\left(1+\frac{998}{1998}\left(\ldots\left(1+\frac{k}{1000+k}\left(1+\frac{k-1}{1001}\right)\right) \ldots\right)\right)= \\
= & \frac{999}{1999}\left(1+\frac{998}{1998}\left(\ldots\left(1+\frac{k}{1000+k} \cdot \frac{1000+k}{1001}\right) \ldots\right)\right) \\
& =\frac{999}{1999}\left(1+\frac{998}{1998}\left(\ldots\left(1+\frac{k}{1001}\right) \ldots\right)\right)
\end{aligned}
$$

Continuing in this way, we will arrive at the answer:

$$
\frac{a}{b}=\ldots=\frac{999}{1999}\left(1+\frac{998}{1998}\left(1+\frac{997}{1001}\right)\right)=\frac{999}{1999}\left(1+\frac{998}{1001}\right)=\frac{999}{1001}
$$

## Answer

$a=999, b=1001$.