Problem 96

Let $n$ be a positive integer for which the last 4 digits of $2011n$ is $9999$. Since the one's digit of $2011n$ is $9$, the one's digit of $n$ has to be $9$. Therefore, we can represent $n$ in the form $n = 10k + 9$ where $k$ is a non-negative integer. Then we must have $2011n = 10 \cdot 2011k + 18099$ and since the ten's digit of $2011n$ is $9$, we see that the one's digit of $k$ has to be $0$. Thus we conclude that $n = 100\ell + 9$ with some non-negative integer $\ell$. We then have $2011n = 100 \cdot 2011\ell + 18099$, and since the hundred's digit of this number is $9$ we must have $9$ for the one's digit of $2011\ell$, and therefore, the one's digit of $\ell$ must be $9$ as well. Consequently, we can represent $n$ as $n = 1000m + 909$ with a non-negative integer $m$, and we have $2011n = 1000 \cdot 2011m + 1827999$, and since the thousand's digit of this number must also be $9$, we have to have $2$ for the one's digit of $2011m$, which means that the one's digit of $m$ must be $2$. Thus we can conclude that the last 4 digits of $n$ must be $2909$ and since $2011 \cdot 2909 = 5849999$, we have $5849999$ for the desired answer.