74th NMO Selection Tests for JBMO

a. If $s(n)=2$ and $n$ is odd, then $a_n=\frac{1}{2}$. The only solutions with these properties are of the form $n=10^k+1$, with $k\in {\mathbb{N}}^{\ast }$.

b. Let $n$ be a positive integer such that $a_n=\left\{\frac{n}{s(n)}\right\}=\frac{1}{6}$. Since $\frac{n}{s(n)}-\lfloor \frac{n}{s(n)}\rfloor =\frac{1}{6}$, we infer that $6n-6\cdot s(n)\cdot \lfloor \frac{n}{s(n)}\rfloor =s(n)$ (1). From here follows that $6\mid s(n)$, therefore $3\mid n$. Consider $n=3k$ and $s(n)=6m$, with $m,k$ positive integers. From (1) we deduce that $3k-6m\cdot \lfloor \frac{k}{2m}\rfloor =m$, hence $3\mid m$. Consequently $m=3u$, with $u\in {\mathbb{N}}^{\ast }$ and $s(n)=18u$, therefore $9\mid n$.

Consider $n=9v$, with $v$ a positive integer. From (1) we obtain $3v-6u\cdot \lfloor \frac{v}{2u}\rfloor =u$, hence $3\mid u$. Consider $u=3t$, with $t$ a positive integer. It follows that $m=9t$ and $s(n)=54t$, and the minimal sum of the digits of the natural number $n$ is 54.

The smallest positive integer with the sum of its digits 54 is $n=999999$.

But $a_{999999}=\left\{\frac{999999}{54}\right\}=\frac{1}{2}$, hence $n=999999$ is not a solution. The next positive integer with the sum of its digits 54 is $n=1899999$, for which we have $a_{1899999}=\left\{\frac{1899999}{54}\right\}=\{35185+\frac{1}{6}\}=\frac{1}{6}$, therefore $n_{\min }=1899999$.