Finding the smallest positive integer $N$ such that there are $25$ integers $x$ with $2 \leq \frac{N}{x} \leq 5$ 
Find the smallest positive integer $N$ such that there are exactly $25$ integers $x$ satisfying $2 \leq \frac{N}{x} \leq 5$.


If $x > 0$, $2 \le N/x \le 5$ iff $\frac{N}{5}  \le x \le \frac{N}{2}$.  Thus you need
$\lfloor N/2 \rfloor - \lceil N/5 \rceil = 24$.  If $N = 10 q + r$, $r \in [0,1,\ldots,9]$, 
$$ \lfloor N/2 \rfloor -  \lceil N/5 \rceil  = 3 q +  \lfloor r/2 \rfloor -  \lceil r/5 \rceil$$
For $r = 0,1,\ldots,9$, $ \lfloor r/2 \rfloor -  \lceil r/5 \rceil = 0, -1, 0, 0, 1, 1, 1, 1, 2, 2$.  We need a value divisible by $3$ since $24$ and $3q$ are divisible by $3$, so $r = 0, 2$ or $3$; $q = 24/3 = 8$, and thus $N$ is either $80$, $82$ or $83$.
