## Task A-4.1. (4 points)

The difference of the reciprocal values of two consecutive natural numbers is $0.0 a a a \ldots=0.0 \dot{a}$. What values can the digit $a$ take?
Solution. If the two consecutive natural numbers are $n$ and $n+1$, we get the equation

$$
\frac{1}{n}-\frac{1}{n+1}=0.0 a a a \ldots
$$

Since $0.0 a a a \ldots=0.0 \dot{a}=\frac{a}{90}$,

we solve the equation $\frac{1}{n}-\frac{1}{n+1}=\frac{a}{90}$, or $\frac{1}{n(n+1)}=\frac{a}{90}$

i.e.

$$
n(n+1)=\frac{90}{a} \quad \text { or } \quad a=\frac{90}{n(n+1)}
$$

We can proceed in several ways:

## First method.

The expression on the right side of the equation $n(n+1)=\frac{90}{a}$ must be an integer, so $a \in\{1,2,3,5,9\}$.

Therefore, $\frac{90}{a} \in\{90,45,30,18,10\}$.

Of these numbers, only $90=9 \cdot 10$ and $30=5 \cdot 6$ can be expressed as the product of two consecutive natural numbers. Thus, $a$ can be 1 or 3, since $\frac{1}{9}-\frac{1}{10}=\frac{1}{90}=0.0111 \ldots$ and $\frac{1}{5}-\frac{1}{6}=\frac{1}{30}=0.0333 \ldots$

## Second method.

In the right side of the equation $a=\frac{90}{n(n+1)}$, we substitute $n=1,2,3,4, \ldots$ sequentially.

For $n=1$ and $n=2$, we get $a \in \mathbb{N}$, but $a$ is not a digit.

For $n=5$, we get $a=\frac{90}{5 \cdot 6}=3$,

and for $n=9$, we get $a=\frac{90}{9 \cdot 10}=1$.

For $n>9$, the fraction $\frac{90}{n(n+1)}$ is less than 1,

so $a=1$ and $a=3$ are the only solutions to the problem.

(2 points)