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Difference between revisions of "2015 AMC 12A Problems/Problem 18"

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==Problem==
 
==Problem==
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Given that the polynomial <math>x^2-kx+16</math> has only positive integer roots, find the average of all distinct possibilities for <math>k</math>.
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The zeros of the function <math>f(x) = x^2-ax+2a</math> are integers. What is the sum of the possible values of <math>a</math>?
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<math>\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18</math>
  
 
==Solution 1==
 
==Solution 1==
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There are a total of four possible values for <math>a</math>: <math>-1,0,8,</math> and <math>9</math>. Hence, the sum of all of the possible values of <math>a</math> is <math>\boxed{\textbf{(C) }16}</math>.
 
There are a total of four possible values for <math>a</math>: <math>-1,0,8,</math> and <math>9</math>. Hence, the sum of all of the possible values of <math>a</math> is <math>\boxed{\textbf{(C) }16}</math>.
  
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==Solution 2==
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 +
==Solution 2 ==
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By the quadratic formula, the roots <math>r</math> can be represented by
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<cmath>r=\frac{a\pm\sqrt{a^2-8a}}{2}</cmath>
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For <math>r\in\mathbb{Z}</math>, <math>a\in\mathbb{Z}</math>, since <math>\frac{\sqrt{a^2-8a}}{2}</math> and <math>\frac{a}{2}</math> will have different mantissas (mantissae?).
 +
 
 +
Now observe the discriminant <math>\sqrt{a^2-8a}=\sqrt{a(a-8)}</math> and have two cases.
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 +
 
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'''Positive <math>a</math>'''
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<math>a\geq8</math> and <math>a\leq0</math>, since <math>1\geq a \geq7</math> gives imaginary roots. Testing positive <math>a</math> values, quickly see that <math>a\leq9</math>. After <math>16</math> and <math>36</math>, the difference between the closest nonzero factor pairs of perfect squares exceeds <math>8</math>. For <math>8\geq a \geq9</math>, <math>a=8,9</math>. Checking both yields an integer.
 +
 
 +
'''Negative <math>a</math>'''
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 +
We can instead test with <math>\sqrt{-a(8-a)}</math>. If <math>b=8-a</math>, we have our original expression. Thus, for the same reasons, <math>b=8,9\implies 8,9=8-a</math>. <math>a=-1</math> (0 does not affect the answer).
 +
 
 +
 
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<math>-1+8+9=16\implies\boxed{\textbf{(C) }16}</math>
 +
 
 +
 
 +
(Solution by BJHHar)
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 +
==Solution 3==
  
 
Let <math>m</math> and <math>n</math> be the roots of <math>x^2-ax+2a</math>
 
Let <math>m</math> and <math>n</math> be the roots of <math>x^2-ax+2a</math>
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This means that the values for <math>(m,n)</math> are <math>(0,0),(4,4),(3,6),(1,-2)</math> giving us <math>a</math> values of <math>-1,0,8,</math> and <math>9</math>. Adding these up gets <math>\boxed{\textbf{(C) }16}</math>.
 
This means that the values for <math>(m,n)</math> are <math>(0,0),(4,4),(3,6),(1,-2)</math> giving us <math>a</math> values of <math>-1,0,8,</math> and <math>9</math>. Adding these up gets <math>\boxed{\textbf{(C) }16}</math>.
  
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==Solution 3==
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==Solution 4==
  
 
The quadratic formula gives <cmath>x = \frac{a \pm \sqrt{a(a-8)}}{2}</cmath>. For <math>x</math> to be an integer, it is necessary (and sufficient!) that <math>a(a-8)</math> to be a perfect square. So we have <math>a(a-8) = b^2</math>; this is a quadratic in itself and the quadratic formula gives <cmath>a = 4 \pm \sqrt{16 + b^2}</cmath>
 
The quadratic formula gives <cmath>x = \frac{a \pm \sqrt{a(a-8)}}{2}</cmath>. For <math>x</math> to be an integer, it is necessary (and sufficient!) that <math>a(a-8)</math> to be a perfect square. So we have <math>a(a-8) = b^2</math>; this is a quadratic in itself and the quadratic formula gives <cmath>a = 4 \pm \sqrt{16 + b^2}</cmath>
  
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We want <math>16 + b^2</math> to be a perfect square. From smartly trying small values of <math>b</math>, we find <math>b = 0, b = 3</math> as solutions, which correspond to <math>a = -1, 0, 8, 9</math>. These are the only ones; if we want to make sure then we must hand check up to <math>b=8</math>. Indeed, for <math>b \geq 9</math> we have that the differences between consecutive squares are greater than <math>16</math> so we can't have <math>b^2 + 16</math> be a perfect square. So summing our values for <math>a</math> we find '''16 (C)''' as the answer.
+
We want <math>16 + b^2</math> to be a perfect square. From smartly trying small values of <math>b</math>, we find <math>b = 0, b = 3</math> as solutions, which correspond to <math>a = -1, 0, 8, 9</math>. These are the only ones; if we want to make sure then we must hand check up to <math>b=8</math>. Indeed, for <math>b \geq 9</math> we have that the differences between consecutive squares are greater than <math>16</math> so we can't have <math>b^2 + 16</math> be a perfect square. So summing our values for <math>a</math> we find <math>\boxed{16 \textbf{ (C)}}</math>. as the answer.
 +
 
 +
Additional note: You can use the quadratic and plug in squares for a (since for <math>b^2</math> to be an integer <math>a</math> would have to be some square), and eventually you can notice a limit to get the answer~
 +
 
 +
 
 +
==Solution 5==
 +
 
 +
First of all, we know that <math>a</math> is the sum of the quadratic's two roots, by Vieta's formulas. Thus, <math>a</math> must be an integer. Then, we notice that the discriminant <math>a^2-8a</math> must be equal to a perfect square so that the roots are integers. Thus, <math>a(a-8)=b^2</math> where <math>b</math> is an integer.
 +
 
 +
We can complete the square and rearrange to get <math>(a-4)^2-b^2=16</math>. Let's define <math>m=a-4</math>, just to make things a little easier to write, so now we have <math>(m+b)(m-b)=16</math>. We can now list out the integer factor pairs of 16 and the resulting values of <math>m</math> and <math>b</math>. (Note that <math>m</math> and <math>b</math> must both be integers)
 +
 
 +
 
 +
<math>(m+b)(m-b)=16</math>
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<math>16*1=16</math> <math>\Rightarrow</math> Doesn't work
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<math>8*2=16</math> <math>\Rightarrow</math> <math>m=5, b=3</math>
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<math>4*4=16</math> <math>\Rightarrow</math> <math>m=4, b=0</math>
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<math>2*8=16</math> <math>\Rightarrow</math> <math>m=5, b=-3</math>
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<math>1*16=16</math> <math>\Rightarrow</math> Doesn't work
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<math>-16*-1=16</math> <math>\Rightarrow</math> Doesn't work
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<math>-8*-2=16</math> <math>\Rightarrow</math> <math>m=-5, b=-3</math>
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<math>-4*-4=16</math> <math>\Rightarrow</math> <math>m=-4, b=0</math>
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<math>-2*-8=16</math> <math>\Rightarrow</math> <math>m=-5, b=-3</math>
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<math>-1*-16=16</math> <math>\Rightarrow</math> Doesn't work
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We want the possible values of <math>m</math>, which are <math>-5,-4,4,</math> and <math>5</math>. As <math>m+4=a</math>, <math>a</math> can equal <math>-1,0,8,</math> or <math>9.</math> Adding all of that up gets us our answer, <math>\boxed{\textbf{(C) }16}</math>.
 +
 
 +
 
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(Solution by Curious_crow)
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2015|ab=A|num-b=17|num-a=19}}
 
{{AMC12 box|year=2015|ab=A|num-b=17|num-a=19}}

Revision as of 13:25, 25 July 2022

Problem

The zeros of the function $f(x) = x^2-ax+2a$ are integers. What is the sum of the possible values of $a$?

$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 8 \qquad\textbf{(C)}\ 16 \qquad\textbf{(D)}\ 17 \qquad\textbf{(E)}\ 18$

Solution 1

The problem asks us to find the sum of every integer value of $a$ such that the roots of $x^2 - ax + 2a = 0$ are both integers.

The quadratic formula gives the roots of the quadratic equation: $x=\frac{a\pm\sqrt{a^2-8a}}{2}$

As long as the numerator is an even integer, the roots are both integers. But first of all, the radical term in the numerator needs to be an integer; that is, the discriminant $a^2 - 8a$ equals $k^2$, for some nonnegative integer $k$.

$a^2-8a=k^2$

$a(a-8)=k^2$

$((a-4)+4)((a-4)-4)=k^2$

$(a-4)^2-4^2=k^2$

$(a-4)^2=k^2+4^2$

From this last equation, we are given a hint of the Pythagorean theorem. Thus, $(k,4,|a-4|)$ must be a Pythagorean triple unless $k = 0$.

In the case $k=0$, the equation simplifies to $|a-4|=4$. From this equation, we have $a=0,8$. For both $a=0$ and $a=8$, $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers, so these values satisfy the constraints from the original problem statement. (Note: the two zero roots count as "two integers.")

If $k$ is a positive integer, then only one Pythagorean triple could match the triple $(k,4,|a - 4|)$ because the only Pythagorean triple with a $4$ as one of the values is the classic $(3,4,5)$ triple. Here, $k=3$ and $|a-4|=5$. Hence, $a=-1,9$. Again, $\frac{a\pm\sqrt{a^2-8a}}{2}$ yields two integers for both $a=-1$ and $a=9$, so these two values also satisfy the original constraints.

There are a total of four possible values for $a$: $-1,0,8,$ and $9$. Hence, the sum of all of the possible values of $a$ is $\boxed{\textbf{(C) }16}$.


Solution 2

By the quadratic formula, the roots $r$ can be represented by \[r=\frac{a\pm\sqrt{a^2-8a}}{2}\] For $r\in\mathbb{Z}$, $a\in\mathbb{Z}$, since $\frac{\sqrt{a^2-8a}}{2}$ and $\frac{a}{2}$ will have different mantissas (mantissae?).

Now observe the discriminant $\sqrt{a^2-8a}=\sqrt{a(a-8)}$ and have two cases.


Positive $a$

$a\geq8$ and $a\leq0$, since $1\geq a \geq7$ gives imaginary roots. Testing positive $a$ values, quickly see that $a\leq9$. After $16$ and $36$, the difference between the closest nonzero factor pairs of perfect squares exceeds $8$. For $8\geq a \geq9$, $a=8,9$. Checking both yields an integer.

Negative $a$

We can instead test with $\sqrt{-a(8-a)}$. If $b=8-a$, we have our original expression. Thus, for the same reasons, $b=8,9\implies 8,9=8-a$. $a=-1$ (0 does not affect the answer).


$-1+8+9=16\implies\boxed{\textbf{(C) }16}$


(Solution by BJHHar)

Solution 3

Let $m$ and $n$ be the roots of $x^2-ax+2a$

By Vieta's Formulas, $n+m=a$ and $mn=2a$

Substituting gets us $n+m=\frac{mn}{2}$

$2n-mn+2m=0$

Using Simon's Favorite Factoring Trick:

$n(2-m)+2m=0$

$-n(2-m)-2m=0$

$-n(2-m)-2m+4=4$

$(2-n)(2-m)=4$

This means that the values for $(m,n)$ are $(0,0),(4,4),(3,6),(1,-2)$ giving us $a$ values of $-1,0,8,$ and $9$. Adding these up gets $\boxed{\textbf{(C) }16}$.

Solution 4

The quadratic formula gives \[x = \frac{a \pm \sqrt{a(a-8)}}{2}\]. For $x$ to be an integer, it is necessary (and sufficient!) that $a(a-8)$ to be a perfect square. So we have $a(a-8) = b^2$; this is a quadratic in itself and the quadratic formula gives \[a = 4 \pm \sqrt{16 + b^2}\]

We want $16 + b^2$ to be a perfect square. From smartly trying small values of $b$, we find $b = 0, b = 3$ as solutions, which correspond to $a = -1, 0, 8, 9$. These are the only ones; if we want to make sure then we must hand check up to $b=8$. Indeed, for $b \geq 9$ we have that the differences between consecutive squares are greater than $16$ so we can't have $b^2 + 16$ be a perfect square. So summing our values for $a$ we find $\boxed{16 \textbf{ (C)}}$. as the answer.

Additional note: You can use the quadratic and plug in squares for a (since for $b^2$ to be an integer $a$ would have to be some square), and eventually you can notice a limit to get the answer~


Solution 5

First of all, we know that $a$ is the sum of the quadratic's two roots, by Vieta's formulas. Thus, $a$ must be an integer. Then, we notice that the discriminant $a^2-8a$ must be equal to a perfect square so that the roots are integers. Thus, $a(a-8)=b^2$ where $b$ is an integer.

We can complete the square and rearrange to get $(a-4)^2-b^2=16$. Let's define $m=a-4$, just to make things a little easier to write, so now we have $(m+b)(m-b)=16$. We can now list out the integer factor pairs of 16 and the resulting values of $m$ and $b$. (Note that $m$ and $b$ must both be integers)


$(m+b)(m-b)=16$

$16*1=16$ $\Rightarrow$ Doesn't work

$8*2=16$ $\Rightarrow$ $m=5, b=3$

$4*4=16$ $\Rightarrow$ $m=4, b=0$

$2*8=16$ $\Rightarrow$ $m=5, b=-3$

$1*16=16$ $\Rightarrow$ Doesn't work

$-16*-1=16$ $\Rightarrow$ Doesn't work

$-8*-2=16$ $\Rightarrow$ $m=-5, b=-3$

$-4*-4=16$ $\Rightarrow$ $m=-4, b=0$

$-2*-8=16$ $\Rightarrow$ $m=-5, b=-3$

$-1*-16=16$ $\Rightarrow$ Doesn't work


We want the possible values of $m$, which are $-5,-4,4,$ and $5$. As $m+4=a$, $a$ can equal $-1,0,8,$ or $9.$ Adding all of that up gets us our answer, $\boxed{\textbf{(C) }16}$.


(Solution by Curious_crow)

See Also

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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