Pista received the following task. "Write the

$$
\frac{16 e-10}{10 e-3}
$$

expression, substituting integers for $e$, and simplify the resulting fraction if possible. Collect all the numbers with which you simplified!"

After several attempts, Pista found such a number and then proved that there are no more such numbers. Let's solve the problem too.
From the fraction $16 / 10=8 / 5$ subtracting, the numerator becomes independent of $e$:

$$
\frac{16 e-10}{10 e-3}-\frac{8}{5}=\frac{5(16 e-10)-8(10 e-3)}{5(10 e-3)}=\frac{26}{5(10 e-3)}
$$

It is clear from the second form that if the numerator and denominator of the original fraction are divisible by a number $d$, then the new fraction can also be simplified by $d$. However, this can only be done with the odd divisors of 26, which means it can only be simplified by 13, since both factors of the denominator and thus their product are odd numbers.

With this, we have solved the problem.

For $a=-1$, the fraction becomes -26, which can be simplified by -13, so it is indeed divisible by 13. It is easy to see that by substituting $a=13k-1$ for $e$, the fraction is always divisible by 13 $(k=0, \pm 1, \pm 2, \ldots)$, but for other values of $e$, this fraction cannot be simplified.