## Task 1 - 210821

Determine all natural numbers $a$ for which $\frac{1}{4}<\frac{a}{a+12}<\frac{1}{3}$ holds!
I. If a natural number $a$ satisfies the required inequality, then:

$$
\frac{1}{4}<\frac{a}{a+12} \quad \text { and } \quad \frac{a}{a+12}<\frac{1}{3}
$$

From (1) it follows that $a+12<4a$, hence $12<3a$, thus $4<a$. From (2) it follows that $3a<a+12$, hence $2a<12$, thus $a<6$.

The only natural number that satisfies (3) and (4) is $a=5$. Therefore, only this number can satisfy the required inequality.

It satisfies this inequality, as a check shows.

2. Alternative solution:

If one forms $\frac{a}{a+12}$ for two consecutive natural numbers $a=n$ and $a=n+1$, one obtains the numbers $\frac{n}{n+12}$ and $\frac{n+1}{n+13}$.

Bringing them to the common denominator $(n+12)(n+13)$, they read

$$
\frac{n(n+13)}{(n+12)(n+13)}=\frac{n^{2}+13n}{(n+12)(n+13)} \quad \text { and } \quad \frac{(n+1)(n+13)}{(n+12)(n+13)}=\frac{n^{2}+13+12}{(n+12)(n+13)}
$$

Therefore, $\frac{n}{n+12}<\frac{n+1}{n+13}$ always holds, i.e., the numbers formed for $a=0,1,2, \ldots$ satisfy the inequalities

$$
0<\frac{1}{13}<\frac{2}{14}<\frac{3}{15}<\frac{4}{16}<\frac{5}{17}<\frac{6}{18}<\frac{7}{19}<\ldots
$$

Since $\frac{4}{16}=\frac{1}{4}$ and $\frac{6}{18}=\frac{1}{3}$, the natural number $a=5$ is the only one that satisfies the required inequality.