# 2011 AIME I Problems/Problem 15

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## Problem

For some integer $m$, the polynomial $x^3 - 2011x + m$ has the three integer roots $a$, $b$, and $c$. Find $|a| + |b| + |c|$.

## Solution

With Vieta's formula, we know that $a+b+c = 0$, and $ab+bc+ac = -2011$. $a,b,c\neq 0$ since any one being zero will make the the other 2 $\pm sqrt{2011}$. $a = -(b+c)$. WLOG, let $|a| \ge |b| \ge |c|$.

Then if $a > 0$, then $b,c < 0$ and if $a < 0$, $b,c > 0$. $ab+bc+ac = -2011 = a(b+c)+bc = -a^2+bc$ $a^2 = 2011 + bc$

We know that $b$, $c$ have the same sign. So $|a| \ge 45$. ( $44^2<2011$ and $45^2 = 2025$)

Also, $bc$ maximize when $b = c$ if we fixed $a$. Hence, $2011 = a^2 - bc < \frac{3}{4}a^2$.

So $a ^2 < \frac{(4)2011}{3} = 2681+\frac{1}{3}$. $52^2 = 2704$ so $|a| \le 51$.

Now we have limited a to $45\le |a| \le 51$.

Let's us analyze $a^2 = 2011 + bc$.

Here is a table: $a$ $a^2 = 2011 + bc$ $45$ $14$ $5$ $46$ $14 + 91 =105$ $47$ $105 + 93 = 198$ $48$ $198 + 95 = 293$ $49$ $293 + 97 = 390$

We can tell we don't need to bother with $45$, $105 = (3)(5)(7)$, So $46$ won't work. $198/47 > 4$, $198 is not divisible by 5$, $198/6 = 33$, 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: $098$