Ireland_2017

We call a number eligible if its digits are all either $1$ or $2$. A number is divisible by $99=9\cdot 11$ if and only if it is divisible by both $9$ and $11$. Suppose $N\in \mathbb{N}$ has base-10 expansion $a_n{a}_{n-1}\dots a_2{a}_1$. We define three digit-sums (full, odd, even): $$S(N):=\sum _{1\le i\le n}{a}_i,o(N):=\sum _{\begin{array}{c}1\le i\le n\\ i\text{ odd}\end{array}}{a}_i,e(N):=\sum _{\begin{array}{c}1\le i\le n\\ i\text{ even}\end{array}}{a}_i.$$

As is well known (and easily established), $N$ is divisible by $9$ if and only if $S(N)$ is divisible by $9$, while $N$ is divisible by $11$ if and only if $d(N):=o(N)-e(N)$ is divisible by $11$.

Suppose first that $S(N)=9$. Then $N$ has at least five digits, $$3\le o(N)\le 9-2=7,2\le e(N)\le 9-3=6,$$ and $o(N)$, $e(N)$ are of opposite parity. Consequently, $0<|d(N)|\le 5$, and so $N$ cannot be divisible by $11$.

Thus, we must have $S(N)\ge 18$ for eligible $N$ to be divisible by $99$. Suppose next that $S(N)=18$. To minimize eligible $N$ with $S(N)=18$, we must certainly minimize the number of digits in $N$. We immediately rule out nine $2$s, since then $d(N)=10-8$ is not divisible by $11$.

The next smallest number of digits involves picking eight $2$s and two $1$s. The smallest such number is the one with $1$s in the leading positions, i.e. $N=1122222222$. Then $o(N)=e(N)=9$, and $N$ is divisible by $11$, and hence by $99$. Thus, this is the minimal example with $S(N)=18$.

It remains only to consider eligible numbers $N$ with $S(N)\ge 27$. Such numbers have at least $14$ digits, so are larger than the one we found above. Thus, the minimal number is indeed $1122222222$.