75th Romanian Mathematical Olympiad

a) Denote $s(m)$ the sum of the digits of a positive integer $m$. Let $n=\overline{a_k{a}_{k-1}\dots a_2{a}_1}$ be a special number with $k\ge 1$ digits. The statement says that $11\mid s(n)$ and $11\mid s(n+1)$. If $a_1\le 8$, then $n+1=\overline{a_k{a}_{k-1}\dots a_2(a_1+1)}$, hence $s(n+1)=s(n)+1$. So, if $s(n)=M_{11}$, then $s(n+1)=M_{11}+1$, false. Consequently $a_1=9$. Suppose that the last $p\ge 1$ digits of $n$ are $9$ and the $(p+1{)}^{th}$ is different from $9$. Then $n=\overline{a_k{a}_{k-1}\dots a_{p+1}99\dots 9}$ and $n+1=\overline{a_k{a}_{k-1}\dots (a_{p+1}+1)00\dots 0}$, so $s(n)=a_k+a_{k-1}+\cdots +a_{p+1}+9p$ and $s(n+1)=a_k+a_{k-1}+\cdots +a_{p+1}+1$. Therefore $s(n)=s(n+1)+9p-1$, showing that $9p-1$ is a multiple of $11$. The smallest such $p$ is $p=5$, hence the last five digits of a special number are nines.

b) From a) follows that the special numbers are of the form $n=m\cdot 10^5+99999$, where $m$ is conveniently chosen. Since $s(n)=s(m)+45$ and $45=M_{11}+1$, it is enough to find $m$ with the last digit different from $9$ and with the property $s(m)=11\cdot t-1$, where $t$ is a positive integer. In particular, if $t=1$, the numbers of the form $m=2\cdot 10^q+8$, with $q\ge 1$, have the sum of their digits $10$. The above show that, for each $q\ge 1$, the number $n=2\cdot 10^{q+5}+899999=\underset{q-1\text{ digits}}{\underbrace{200\dots 0}}899999$ is special, and there are infinitely many such numbers.