Difference between revisions of "2008 iTest Problems/Problem 84"

(Focus on problem)
 
(ACTUAL Solution to Problem 84)
 
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Easy problem, need to know basic
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==Problem==
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Let <math>S</math> be the sum of all integers <math>b</math> for which the polynomial <math>x^2+bx+2008b</math> can be factored over the integers. Compute <math>|S|</math>.
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==Solutions==
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===Solution 1 (credit to official solution)===
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Let the roots of the quadratic be <math>r</math> and <math>s</math>.  By [[Vieta's Formulas]], <math>r+s = -b</math> and <math>rs</math> = <math>2008b</math>.
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<br>
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We know that one of the possible values of <math>b</math> is 0 because <math>x^2</math> has integer roots.  However, adding or removing 0 does not affect the value of <math>S</math>, so we can divide both sides by <math>-b</math>.  Doing so results in
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<cmath>\begin{align*}
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\frac{rs}{r+s} &= -2008 \\
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rs &= -2008r - 2008s \\
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rs + 2008r + 2008s &= 0 \\
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(r+2008)(s+2008) &= 2008^2.
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\end{align*}</cmath>
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[[WLOG]], let <math>|a| \le 2008</math> be a factor of <math>2008^2</math>, so <math>r+2008 = a</math> and <math>s+2008 = \tfrac{2008^2}{a}</math>.  Thus,
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<cmath>-r-s = b = -a - \tfrac{2008^2}{a} + 4016.</cmath>
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Since <math>a</math> can be positive or negative, the positive values cancel with the negative values.  The prime factorization of <math>2008^2</math> is <math>2^6 \cdot 251^2</math>, so there are <math>\frac{21+2}{2} = 11</math> positive factors that are less than <math>2008</math>.  Thus, there are a total of <math>22</math> values of <math>a</math>, so the absolute value of the sum of all values of <math>b</math> equals <math>4016 \cdot 22 = \boxed{88352}</math>.
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===Solution 2===
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The discriminant of the function is <math>b^2 - 8032b</math>.  Since all roots are integers and leading term is 1, the discriminant must equal <math>n^2</math>, where <math>n</math> is an integer.
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<br>
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Thus, we know that
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<cmath>\begin{align*}
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b^2 - 8032b =& n^2 \\
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(b-4016)^2 - n^2 &= 4016^2 \\
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(b-4016+n)(b-4016-n) &= 4016^2.
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\end{align*}</cmath>
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Let <math>x</math> be a factor of <math>4016^2</math>, so
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<cmath>\begin{align*}
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b - 4016 + n &= x \\
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b - 4016 - n &= \frac{4016^2}{x} \\
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b &= 4016 + \frac{x}{2} + \frac{4016^2}{2x}
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\end{align*}</cmath>
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Note that if <math>x</math> is odd, then <math>\tfrac{4016^2}{x}</math> is even, so <math>b</math> can not be an integer.  Thus, <math>x</math> must be even.  Let <math>x = 2y</math>, so <math>b = 4016 + y + \frac{2008^2}{y}</math>.
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<br>
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If <math>y_0 = \frac{2008^2}{y}</math>, then <math>y + \tfrac{2008^2}{y} = y_0 + \tfrac{2008^2}{y_0}</math>.  Also, since <math>y</math> can be positive or negative, the positive values cancel with the negative values.  So [[WLOG]], let <math>|y| \le 2008</math>.
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<br>
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The prime factorization of <math>2008^2</math> is <math>2^6 \cdot 251^2</math>, so there are <math>\frac{21+2}{2} = 11</math> positive factors that are less than <math>2008</math>.  Thus, there are a total of <math>22</math> values of <math>y</math>, so the absolute value of the sum of all values of <math>b</math> equals <math>4016 \cdot 22 = \boxed{88352}</math>.
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==See Also==
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{{2008 iTest box|num-b=83|num-a=85}}
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Number Theory Problems]]

Latest revision as of 17:22, 22 May 2019

Problem

Let $S$ be the sum of all integers $b$ for which the polynomial $x^2+bx+2008b$ can be factored over the integers. Compute $|S|$.

Solutions

Solution 1 (credit to official solution)

Let the roots of the quadratic be $r$ and $s$. By Vieta's Formulas, $r+s = -b$ and $rs$ = $2008b$.


We know that one of the possible values of $b$ is 0 because $x^2$ has integer roots. However, adding or removing 0 does not affect the value of $S$, so we can divide both sides by $-b$. Doing so results in \begin{align*} \frac{rs}{r+s} &= -2008 \\ rs &= -2008r - 2008s \\ rs + 2008r + 2008s &= 0 \\ (r+2008)(s+2008) &= 2008^2. \end{align*} WLOG, let $|a| \le 2008$ be a factor of $2008^2$, so $r+2008 = a$ and $s+2008 = \tfrac{2008^2}{a}$. Thus, \[-r-s = b = -a - \tfrac{2008^2}{a} + 4016.\] Since $a$ can be positive or negative, the positive values cancel with the negative values. The prime factorization of $2008^2$ is $2^6 \cdot 251^2$, so there are $\frac{21+2}{2} = 11$ positive factors that are less than $2008$. Thus, there are a total of $22$ values of $a$, so the absolute value of the sum of all values of $b$ equals $4016 \cdot 22 = \boxed{88352}$.

Solution 2

The discriminant of the function is $b^2 - 8032b$. Since all roots are integers and leading term is 1, the discriminant must equal $n^2$, where $n$ is an integer.


Thus, we know that \begin{align*} b^2 - 8032b =& n^2 \\ (b-4016)^2 - n^2 &= 4016^2 \\ (b-4016+n)(b-4016-n) &= 4016^2. \end{align*} Let $x$ be a factor of $4016^2$, so \begin{align*} b - 4016 + n &= x \\ b - 4016 - n &= \frac{4016^2}{x} \\ b &= 4016 + \frac{x}{2} + \frac{4016^2}{2x} \end{align*} Note that if $x$ is odd, then $\tfrac{4016^2}{x}$ is even, so $b$ can not be an integer. Thus, $x$ must be even. Let $x = 2y$, so $b = 4016 + y + \frac{2008^2}{y}$.


If $y_0 = \frac{2008^2}{y}$, then $y + \tfrac{2008^2}{y} = y_0 + \tfrac{2008^2}{y_0}$. Also, since $y$ can be positive or negative, the positive values cancel with the negative values. So WLOG, let $|y| \le 2008$.


The prime factorization of $2008^2$ is $2^6 \cdot 251^2$, so there are $\frac{21+2}{2} = 11$ positive factors that are less than $2008$. Thus, there are a total of $22$ values of $y$, so the absolute value of the sum of all values of $b$ equals $4016 \cdot 22 = \boxed{88352}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 83
Followed by:
Problem 85
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