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Difference between revisions of "2015 AMC 10A Problems/Problem 23"

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Therefore <math>(a-4)^2 - k^2 = 16</math> and
 
Therefore <math>(a-4)^2 - k^2 = 16</math> and
 
<cmath>((a-4) - k)((a-4) + k) = 16.</cmath>
 
<cmath>((a-4) - k)((a-4) + k) = 16.</cmath>
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Let <math>(a-4) - k = u</math> and <math>(a-4) + k = v</math>; then, <math>a-4 = \dfrac{u+v}{2}</math> and so <math>a = \dfrac{u+v}{2} + 4</math>. Listing all possible <math>(u, v)</math> pairs (not counting transpositions because this does not affect (<math>u + v</math>), <math>(2, 8), (4, 4), (-2, -8), (-4, -4)</math>, yields <math>a = 9, 8, -1, 0</math>. These <math>a</math> sum to <math>16</math>, so our answer is <math>\boxed{\textbf{(C) }16}</math>.
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Let <math>(a-4) - k = u</math> and <math>(a-4) + k = v</math>; then, <math>a-4 = \dfrac{u+v}{2}</math> and so <math>a = \dfrac{u+v}{2} + 4</math>. Listing all possible <math>(u, v)</math> pairs (not counting transpositions because this does not affect (<math>u + v</math>)), <math>(2, 8), (4, 4), (-2, -8), (-4, -4)</math>, yields <math>a = 9, 8, -1, 0</math>. These <math>a</math> sum to <math>16</math>, so our answer is <math>\boxed{\textbf{(C) }16}</math>.
  
 
==Solution 2==
 
==Solution 2==
  
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Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic. ,
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Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic.  
 
By [[Vieta's Formulas]], <cmath>r_1 + r_2 = a\text{ and }r_1r_2 = 2a.</cmath>
 
By [[Vieta's Formulas]], <cmath>r_1 + r_2 = a\text{ and }r_1r_2 = 2a.</cmath>
  
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These factors <math>(f_1,f_2)</math> (ignoring order, because we want the sum of factors), can be <math>(1, 4), (-1, -4), (2, 2),</math> or <math>(-2, -2)</math>.
 
These factors <math>(f_1,f_2)</math> (ignoring order, because we want the sum of factors), can be <math>(1, 4), (-1, -4), (2, 2),</math> or <math>(-2, -2)</math>.
  
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The sum of distinct <math>a = r_1 + r_2 = (f_1+2) + (f_2+2)</math>, and these factors give <math>\sum_a a = (5+4) + (-5+4) + (4+4) + (0+4) = \boxed{\textbf{(C) }16}</math>.
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The sum of distinct <math>a = r_1 + r_2 = (f_1+2) + (f_2+2)</math>, and these factors give  
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<cmath>\sum_a a = (5+4) + (-5+4) + (4+4) + (-4+4) = \boxed{\textbf{(C) }16}</cmath>.
  
  
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https://artofproblemsolving.com/videos/amc/2015amc10a/397
 
https://artofproblemsolving.com/videos/amc/2015amc10a/397
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==Solution 3==
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Let the roots be <math>r</math> and <math>s</math>. By Vieta's we have that <math>r+s = a</math>, and <math>rs = 2a</math>. So by Simon's Favorite Factoring Trick: <cmath>2r+2s = rs \Longrightarrow rs-2r-2s = 0 \Longrightarrow (r-2)(s-2) = 4</cmath>. Now, we test out values of <math>r</math> and <math>s</math> to see which work. Lets say that <math>r-2 = 1</math>, and <math>s-2 = 4</math>. This implies that <math>(r,s) = (3,6)</math>, so <math>a = 9</math>. Now, we let <math>r-2 = 2</math>, and <math>s-2 = 2</math>, this gives us <math>(r,s) = (4,4)</math>, which gives us a value for <math>a</math> of <math>8</math>. Now, we circle an answer of <math>17</math> and move onto the next problem. Yike! They didnt say <math>a</math> cant be negative! This is where many people would go wrong. Lets let <math>r-2 = -1</math>, and <math>s-2 = -4</math>, this gives us that <math>(r,s) = (1,-2)</math>. This results in a value of <math>-1</math> for <math>a</math>. Now, we let <math>r-2 = -2</math>, and <math>s-2 = -2</math>. This gives <math>a = 0</math>. Now, we are done. Our total sum is <math>9+8-1 = \boxed{\textbf{(C) }16}</math>.
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-jb2015007 - minor edit by mathlover1205
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 16:59, 18 July 2025

Problem

The zeroes 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

By Vieta's Formula, $a$ is the sum of the integral zeros of the function, and so $a$ is integral.

Because the zeros are integral, the discriminant of the function, $a^2 - 8a$, is a perfect square, say $k^2$. Then adding 16 to both sides and completing the square yields \[(a - 4)^2 = k^2 + 16.\] Therefore $(a-4)^2 - k^2 = 16$ and \[((a-4) - k)((a-4) + k) = 16.\] Let $(a-4) - k = u$ and $(a-4) + k = v$; then, $a-4 = \dfrac{u+v}{2}$ and so $a = \dfrac{u+v}{2} + 4$. Listing all possible $(u, v)$ pairs (not counting transpositions because this does not affect ($u + v$)), $(2, 8), (4, 4), (-2, -8), (-4, -4)$, yields $a = 9, 8, -1, 0$. These $a$ sum to $16$, so our answer is $\boxed{\textbf{(C) }16}$.

Solution 2

Let $r_1$ and $r_2$ be the integer zeroes of the quadratic. By Vieta's Formulas, \[r_1 + r_2 = a\text{ and }r_1r_2 = 2a.\]

Plugging the first equation in the second, \[r_1r_2 = 2 (r_1 + r_2).\]

Rearranging gives \[r_1r_2 - 2r_1 - 2r_2 = 0 \implies (r_1 - 2)(r_2 - 2) = 4.\]

These factors $(f_1,f_2)$ (ignoring order, because we want the sum of factors), can be $(1, 4), (-1, -4), (2, 2),$ or $(-2, -2)$.

The sum of distinct $a = r_1 + r_2 = (f_1+2) + (f_2+2)$, and these factors give \[\sum_a a = (5+4) + (-5+4) + (4+4) + (-4+4) = \boxed{\textbf{(C) }16}\].


Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2015amc10a/397

Solution 3

Let the roots be $r$ and $s$. By Vieta's we have that $r+s = a$, and $rs = 2a$. So by Simon's Favorite Factoring Trick: \[2r+2s = rs \Longrightarrow rs-2r-2s = 0 \Longrightarrow (r-2)(s-2) = 4\]. Now, we test out values of $r$ and $s$ to see which work. Lets say that $r-2 = 1$, and $s-2 = 4$. This implies that $(r,s) = (3,6)$, so $a = 9$. Now, we let $r-2 = 2$, and $s-2 = 2$, this gives us $(r,s) = (4,4)$, which gives us a value for $a$ of $8$. Now, we circle an answer of $17$ and move onto the next problem. Yike! They didnt say $a$ cant be negative! This is where many people would go wrong. Lets let $r-2 = -1$, and $s-2 = -4$, this gives us that $(r,s) = (1,-2)$. This results in a value of $-1$ for $a$. Now, we let $r-2 = -2$, and $s-2 = -2$. This gives $a = 0$. Now, we are done. Our total sum is $9+8-1 = \boxed{\textbf{(C) }16}$.

-jb2015007 - minor edit by mathlover1205

Video Solution

https://youtu.be/RQ4ZCttwmA4

~savannahsolver

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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 10 Problems and Solutions

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