# 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 formulas, 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 other two $\pm \sqrt{2011}$.

$a = -(b+c)$. WLOG, let $|a| \ge |b| \ge |c|$.

Then if $a > 0$, then $b,c < 0$ and if $a < 0$, then $b,c > 0$.

$ab+bc+ac = -2011 = a(b+c)+bc = -a^2+bc$

$a^2 = 2011 + bc$

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

Also, if we fix $a$, $bc$ is maximized when $b = c$ . 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 analyze $a^2 = 2011 + bc$.

Here is a table:

$|a|$$a^2 = 2011 + bc$
$45$$14$
$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: $\boxed{098}$

## Solution 2

Starting off like the previous solution, we know that $a + b + c = 0$, and $ab + bc + ac = -2011$.

Therefore, $c = -b-a$.

Substituting, $ab + b(-b-a) + a(-b-a) = ab-b^2-ab-ab-a^2 = -2011$.

Factoring the perfect square, we get: $ab-(b+a)^2=-2011$ or $(b+a)^2-ab=2011$.

Therefore, a sum ($a+b$) squared minus a product ($ab$) gives $2011$..

We can guess and check different $a+b$’s starting with $45$ since $44^2 < 2011$.

$45^2 = 2025$ therefore $ab = 2025-2011 = 14$.

Since no factors of $14$ can sum to $45$ ($1+14$ being the largest sum), a + b cannot equal $45$.

$46^2 = 2116$ making $ab = 105 = 3 * 5 * 7$.

$5 * 7 + 3 < 46$ and $3 * 5 * 7 > 46$ so $46$ cannot work either.

We can continue to do this until we reach $49$.

$49^2 = 2401$ making $ab = 390 = 2 * 3 * 5* 13$.

$3 * 13 + 2* 5 = 49$, so one root is $10$ and another is $39$. The roots sum to zero, so the last root must be $-49$.

$|-49|+10+39 = \boxed{098}$.

## Solution 3

Let us first note the obvious that is derived from Vieta's formulas: $a+b+c=0, ab+bc+ac=-2011$. Now, due to the first equation, let us say that $a+b=-c$, meaning that $a,b>0$ and $c<0$. Now, since both $a$ and $b$ are greater than 0, their absolute values are both equal to $a$ and $b$, respectively. Since $c$ is less than 0, it equals $-a-b$. Therefore, $|c|=|-a-b|=a+b$, meaning $|a|+|b|+|c|=2(a+b)$. We now apply Newton's sums to get that $a^2+b^2+ab=2011$,or $(a+b)^2-ab=2011$. Solving, we find that $49^2-390$ satisfies this, meaning $a+b=49$, so $2(a+b)=\boxed{098}$.

## Solution 4

We have

$(x-a)\cdot (x-b)\cdot (x-c)=x^3-(a+b+c)x+(ab+ac+bc)x-abc$


As a result, we have

$a+b+c=0$

$ab+bc+ac=-2011$

$abc=-m$

So, $a=-b-c$

As a result, $ab+bc+ac=(-b-c)b+(-b-c)c+bc=-b^2-c^2-bc=-2011$

Solve $b=\frac {-c+\sqrt{c^2-4(c^2-2011)}}{2}$ and $\Delta =8044-3c^2=k^2$, where $k$ is an integer

Cause $89<\sqrt{8044}<90$

So, after we tried for $2$ times, we get $k=88$ and $c=10$

then $b=39$, $a=-b-c=-49$

As a result, $|a|+|b|+|c|=10+39+49=\boxed{098}$

## Note

This is a misplaced number theory problem.