2011 AIME I Problems/Problem 15

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:

$\|a\|$	$a^2 = 2011 + bc$
$45$	$14$
$46$	$14 + 91 =105$
$47$	$105 + 93 = 198$
$48$	$198 + 95 = 293$
$49$	$293 + 97 = 390$

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: 089

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 = 98

Answer: 089

2011 AIME I (Problems • Answer Key • Resources)
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2011 AIME I Problems/Problem 15

Contents

Problem

Solution

Solution 2

See also