4A. Silvia chose three natural numbers $a, b$, and $c$. Petar found that the value of $a+\frac{b}{c}$ is 101, Lina found that the value of $\frac{a}{c}+b$ is 68, and Maria found that the value of $\frac{a+b}{c}$ is $k$. Determine $a, b, c$, and $k$.
Solution. From $a+\frac{b}{c}=101$ and $a$ is a natural number, it follows that $\frac{b}{c}$ is a natural number and therefore $b=c x$, where $x$ is a natural number. Similarly, from $\frac{a}{c}+b=68$ it follows that $\frac{a}{c}$ is a natural number and therefore $a=c y$, where $y$ is a natural number.

We have $a+x=101$ and $y+b=68$, and thus $a+x+y+b=169$. Since $\frac{a+b}{c}=k$ it follows that $k$ is a natural number and $k=x+y$. We have, $c y+x+y+c x=169$, so $(x+y)(c+1)=169$, i.e., $k(c+1)=169$. The divisors (natural) of 169 are 1, 13, and 169. Since $k=x+y \geq 1+1=2$, we have $k \neq 1$, and since $c+1 \geq 1+1=2$, it follows that $k \neq 169$. Therefore, $k=13$ and $c+1=13$, i.e., $c=12$. From $\frac{a}{c}+b=68$ it follows that $b \leq 68$, and from $a+\frac{b}{c}=101$ it follows that $b$ is divisible by 12. Therefore, $b \in\{12,24,36,48,60\}$. Since $a+k+b=169$, we get $a+b=156$, so $a=156-b \in\{144,132,120,108,96\}$. From $a+\frac{b}{c}=101$ it follows that $a \leq 101$, so $a=96$. Thus, $b=60$.

Finally, the required numbers are $a=96, b=60, c=12$ and $k=13$.

Note. For the linear functions

$$
f(x)=m x+m^{2}-1 \text { and } g(x)=(m-1) x+m(m+2), m \neq 0,1
$$

it is true that there exists a number $a$ such that $f(a)=g(a)$. For which $m$ is it true that $|a|<3$?