## Task A-1.1.

By deleting the first two digits of some natural number $n$, a number 73 times smaller is obtained. Determine the two smallest such numbers $n$.
## Solution.

Let $\overline{a b}$ be the two-digit beginning, and $x$ the remainder of the observed number $n$, and let $k$ be the number of digits in the number $x$.

The given condition can be written as:

$$
n=10^{k} \cdot \overline{a b}+x=73 x
$$

or

$$
10^{k} \cdot \overline{a b}=72 x
$$

(*) 1 point

Let's try to determine a solution for which $x$ is a single-digit number. Then $k=1$, and $x$ is a digit. The equation (*) in this case becomes $10 \cdot \overline{a b}=72 x$.

Since the left side of the obtained equality is positive and divisible by 5, it must be $x=5. \quad 1$ point

From $10 \cdot \overline{a b}=360$ we get $\overline{a b}=36$, and the desired number $n=365$. 2 points

Now let's assume that $x$ is a two-digit number and $k=2$. Then the equation (*) becomes $100 \cdot \overline{a b}=72 x$,

or $25 \cdot \overline{a b}=18 x$.

We see that $x$ must be divisible by 25, and $\overline{a b}$ by 18.

The smallest solution is obtained for $\overline{a b}=18$ - then $x=25$, and the number with the desired property is $n=1825$.