# 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 $2 \leq \frac{N}{x} \leq 5$.

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If you want the least such integer $N$, the answer will be negative. Smallest is somewhat ambiguous when applied to integers that can be negative. –  Robert Israel Feb 13 '13 at 20:33
positive integer –  user62189 Feb 13 '13 at 20:36

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 $83$.
$x$ ranges from $N/5$ through $N/2$ (ignoring the breakage) so $N/2-N/5+1=25$ so $N=80$ and a check shows $x$ goes $16$ through $40$