2. For an integer $x$, the following holds

$$
|\ldots||| x-1|-10|-10^{2}\left|-\ldots-10^{2006}\right|=10^{2007}
$$

Find the hundredth digit of the number $|x|$.
IV/2. Let

$$
x_{n}=|\ldots||| x-1|-10|-10^{2}\left|-\ldots-10^{n-1}\right|-10^{n}
$$

The equation in the problem thus states that $\left|x_{2006}\right|=10^{2007}$, so $x_{2006}= \pm 10^{2007}$. Since we can estimate

$$
x_{2006}=\left|x_{2005}\right|-10^{2006} \geq-10^{2006}
$$

it follows that $x_{2006}=10^{2007}$, hence $\left|x_{2005}\right|=10^{2006}+10^{2007}$. Further, from $x_{2006}=$ $\left|x_{2004}\right|-10^{2005} \geq 10^{2005}$ it follows that $x_{2006}=10^{2006}+10^{2007}$ and thus

$$
\left|x_{2004}\right|=10^{2005}+10^{2006}+10^{2007}
$$

We can reason similarly. If we already know that

$$
x_{n}=10^{n+1}+10^{n+2}+\ldots+10^{2007}
$$

it follows that $x_{n-1}= \pm\left(10^{n+1}+10^{2007-n+1}+\ldots+10^{2007}\right)$. Since $x_{n-1}=\left|x_{n}\right|-10^{n} \geq-10^{n}$, we get

$$
x_{n-1}=10^{n}+10^{n+1}+\ldots+10^{2007}
$$

Thus, $|x-1|=10+10^{2}+\ldots+10^{2007}$, from which it follows

$$
x=1+10+10^{2}+\ldots+10^{2007}=\underbrace{11 \ldots 1}_{2008}
$$

or

$$
x=1-10-10^{2}-\ldots-10^{2007}=-(\underbrace{11 \ldots 10}_{2007}-1)=-\underbrace{11 \ldots 1}_{2006} 09 \text {. }
$$

The 100th digit of the number $|x|$ is in both cases equal to 1.

Introduction of $x_{n}=|\ldots||| x-1|-10|-10^{2}\left|-\ldots-10^{n-1}\right|-10^{n}$. ............ 1 point

Proof that $x_{2006}=10^{2007}$ 2 points

Inductive conclusion $|x-1|=10+10^{2}+\ldots+10^{2007}$ 2 points

Solution $x=1+10+10^{2}+\ldots+10^{2007}=\underbrace{11 \ldots 1}_{2008}$

Solution $x=1-10-10^{2}-\ldots-10^{2007}=-\underbrace{11 \ldots 109}_{2006}$.