Anička and Maruška each have their favorite natural number. If we multiply Anička's number by itself, we get a number that is a hundred times larger than when we multiply Maruška's number by itself. If we add Anička's and Maruška's favorite numbers, we get a number that is 18 more than half of Anička's number.

Determine Anička's and Maruška's favorite numbers.

(E. Semerádová)

Hint. How many times is Anička's number larger than Maruška's?
Aniččino číslo is ten times larger than Maruščino číslo - only in this case is the product of Aniččino číslo with itself equal to the hundredfold of Maruščino číslo $(10 \cdot 10=100)$. Half of Aniččino číslo is therefore five times larger than Maruščino číslo.

The sum of Aniččino číslo and Maruščino číslo is 18 greater than half of Aniččino číslo. Therefore, 18 is the sum of half of Aniččino číslo and Maruščino číslo. According to the conclusion of the previous paragraph, this sum is equal to six times Maruščino číslo.

Thus, Maruščino číslo is $3(18: 6=3)$ and Aniččino číslo is $30(10 \cdot 3=30)$.

Note. If we denote Aniččino číslo by $a$ and Maruščino číslo by $m$, then the previous considerations can be summarized as follows:

- $a \cdot a=100 \mathrm{~m} \cdot \mathrm{m}$, so $a=10 \mathrm{~m}$, so $\frac{1}{2} a=5 \mathrm{~m}$.
- $a+m=\frac{1}{2} a+18$, so $18=\frac{1}{2} a+m=6 m$.
- $m=18: 6=3$ and $a=10 m=30$.

After the initial observation $(a=10 m)$, it is also possible to successively substitute natural numbers for $m$, express the relevant difference $\left(a+m-\frac{1}{2} a=\frac{1}{2} a+m\right)$, and check whether it is or is not equal to 18:

| $m$ | 1 | 2 | 3 | 4 | $\ldots$ |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $a$ | 10 | 20 | 30 | 40 | $\ldots$ |
| $\frac{1}{2} a+m$ | 6 | 12 | 18 | 24 | $\ldots$ |