## Task B-1.2.

Let $n$ be a natural number and let

$$
A=0.1^{n} \cdot 0.01^{n} \cdot 0.001^{n} \cdot 0.0001^{n} \cdot \ldots \cdot 0 . \underbrace{00 \ldots 0}_{199 \text { zeros }} 1^{n} .
$$

The number $A^{-1}$ has 140701 digits. Determine the number $n$.
## Solution.

$A=10^{-n} \cdot 10^{-2 n} \cdot 10^{-3 n} \cdot 10^{-4 n} \cdot \ldots \cdot 10^{-200 n}=10^{-n(1+2+3+\ldots+200)} \quad 1 \text{ point}$

Then $A^{-1}=10^{n \cdot \frac{200 \cdot 201}{2}}=10^{20100 n}$. 2 points

This number has a leading digit of one and 20100n zeros in its decimal representation.

It follows that $20100 n+1=140701$.

Then $n=7$.