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Difference between revisions of "2011 AIME I Problems/Problem 15"

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Hence, <math>|a| = 49</math>, <math>a^2 -2011 = 390</math>. <math>b = 39</math>, <math>c = 10</math>.
 
Hence, <math>|a| = 49</math>, <math>a^2 -2011 = 390</math>. <math>b = 39</math>, <math>c = 10</math>.
  
−
Answer: <math>098</math>
+
Answer: 089
 +
 
 +
== Solution 2==
 +
 
 +
 
 +
Starting off like the previous solution, we know that a + b + c = 0, and ab + bc + ac = -2011
 +
 
 +
Therefore, <math>c = -b-a</math>
 +
 
 +
Substituting, <math>ab + b(-b-a) + a(-b-a) = ab –b^2-ab-ab-a^2 = -2011</math>
 +
 
 +
Factoring the perfect square, we get: <math>ab-(b+a)^2=-2011 or (b+a)^2-ab=2011</math>
 +
 
 +
Therefore, a sum (a+b) squared minus a product (ab) gives 2011.
 +
 
 +
<br/>
 +
 
 +
We can guess and check different <math>a+b</math>’s starting with <math>45</math> since <math>44^2 < 2011</math>.
 +
 
 +
<math>45^2 = 2025</math> therefore <math>ab = 2025-2011 = 14</math>
 +
 
 +
Since no factors of <math>14</math> can sum to <math>45</math> (<math>1+14</math> being the largest sum), a + b cannot equal <math>45</math>.
 +
 
 +
<math>46^2 = 2116</math> making <math>ab = 105 = 3 * 5 * 7</math>
 +
 
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<math>5 * 7 + 3 < 46</math> and <math>3 * 5 * 7 > 46</math> so <math>46</math> cannot work either
 +
 
 +
<br/>
 +
 
 +
We can continue to do this until we reach <math>49</math>
 +
 
 +
<math>49^2 =  2401</math> making <math>ab = 390 = 2 * 3 * 5* 13</math>
 +
 
 +
<math>3 * 13 + 2* 5 = 49</math>, so one root is <math>10</math> and another is <math>39</math>. The roots sum to zero, so the last root must be <math>-49</math>.
 +
 
 +
<br/>
 +
 
 +
|-49|+10+39 = 98
 +
 
 +
Answer: 089
  
 
== See also ==
 
== See also ==

Revision as of 15:19, 20 June 2011

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

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
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1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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