Problem 195

From Vieta's formulas, we know that $a+b+c = 0$, and $ab+bc+ac = -2011$. Thus $a = -(b+c)$. All three of $a$, $b$, and $c$ are non-zero: say, if $a=0$, then $b=-c=\pm\sqrt{2011}$ (which is not an integer). $\textsc{wlog}$, let $|a| \ge |b| \ge |c|$. If $a > 0$, then $b,c 0$. We have $$-2011=ab+bc+ac = a(b+c)+bc = -a^2+bc$$
Thus $a^2 = 2011 + bc$. We know that $b$, $c$ have the same sign. So $|a| \ge 45 = \lceil \sqrt{2011} \rceil$.
Also, if we fix $a$, $b+c$ is fixed, so $bc$ is maximized when $b = c$ . Hence, \[2011 = a^2 - bc > \tfrac{3}{4}a^2 \qquad \Longrightarrow \qquad a ^2 4$,
$198$ is not divisible by $5$, $198/6 = 33$, which is too small to get $47$.
$293/48 > 6$, $293$ is not divisible by $7$ or $8$ or $9$, we can clearly tell that $10$ is too much.

Hence, $|a| = 49$, $a^2 -2011 = 390$. $b = 39$, $c = 10$.
Answer: $\boxed{098}$