Thai Mathematical Olympiad

We will use the well-known fact that $\nu (n)={\sum }_{k=1}^{\infty }\lfloor \frac{n}{2^k}\rfloor$. If the base-2 form of the number $n$ is $n=\overline{d_l{d}_{l-1}\dots d_1{d}_0}={\sum }_{i=0}^{\infty }{d}_i\cdot 2^i$, where the digits $d_0,\dots ,d_l$ are all $0$ or $1$, then $$\begin{array}{rl}\nu (n)& =\sum _{k=1}^{\infty }\lfloor \frac{n}{2^k}\rfloor =\sum _{k=1}^l\lfloor \frac{\overline{d_l{d}_{l-1}\dots d_1{d}_0}}{2^k}\rfloor =\sum _{k=1}^l\overline{d_l{d}_{l-1}\dots d_k}\\ & =\sum _{k=1}^l\left(\sum _{i=k}^l{d}_i\cdot 2^{i-k}\right)=\sum _{i=1}^l{d}_i\left(\sum _{k=1}^i{2}^{i-k}\right)=\sum _{i=1}^l{d}_i\cdot (2^i-1).\end{array}$$ Now we will show that there exists an integer $r$ which is relatively prime to $m$, and an infinite sequence $i_1<i_2<i_3<\dots$ of positive integers such that $$2^{i_1}-1\equiv 2^{i_2}-1\equiv 2^{i_3}-1\equiv \cdots \equiv r\mathrm{mod}m.$$ Let $m=2^{t}u$ where $u$ is odd, and consider an arbitrary positive integer $i\ge t$ for which $\phi (u)$ divides $i-1$. By the Euler-Fermat theorem, $$u\mid 2^{\phi (u)}-1\mid 2^{i-1}-1,$$ $$2^i-1=2(2^{i-1}-1)+1\equiv 1\text{mod }u,$$ and, due to $i\ge t$, $$2^i-1\equiv -1\text{mod }{2}^t.$$ The relations (1) and (2) determine the residue class of $2^i-1$ modulo $m$, and it must be relatively prime to $m$. Since $m$ and $r$ are relatively prime, there is a positive integer $u$ such that $a\equiv ur\mathrm{mod}m$. For $n=2^{i_1}+\cdots +2^{i_u}$ we achieve $$\nu (n)=\nu (2^{i_1}+\cdots +2^{i_u})=(2^{i_1}-1)+\cdots +(2^{i_u}-1)\equiv u\cdot r\equiv a\text{mod }m.$$