Problem 4. Determine the natural numbers $a$ for which there exist exactly 2014 natural numbers $b$ that satisfy the relation $2 \leq \frac{a}{b} \leq 5$.
Solution. The relation $2 \leq \frac{a}{b} \leq 5$ is equivalent to $\frac{a}{5} \leq b \leq \frac{a}{2}$, which means $2 a \leq 10 b \leq 5 a$. ...1 $\mathbf{p}$

This means that in the sequence $2 a, 2 a+1, \ldots, 5 a$ there must be exactly 2014 multiples of 10, which means the sequence must contain at least 2013 decades of consecutive numbers and fewer than 2015 decades of consecutive numbers. We deduce that $2013 \cdot 10 \leq 5 a-2 a<2015 \cdot 10$. We obtain $a \in\{6710,6711, \ldots, 6716\}$. $3 \mathrm{p}$

The numbers that work are: 6710, 6712, and 6713.