Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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==Problem== | ==Problem== | ||
− | The zeroes of the function <math>f(x)=x^2-ax+2a</math> are integers .What is the sum of the possible values of a? | + | The zeroes 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> |
<math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18</math> | <math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18</math> | ||
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Because the zeros are integral, the discriminant of the function, <math>a^2 - 8a</math>, is a perfect square, say <math>k^2</math>. Then adding 16 to both sides and completing the square yields | Because the zeros are integral, the discriminant of the function, <math>a^2 - 8a</math>, is a perfect square, say <math>k^2</math>. Then adding 16 to both sides and completing the square yields | ||
<cmath>(a - 4)^2 = k^2 + 16.</cmath> | <cmath>(a - 4)^2 = k^2 + 16.</cmath> | ||
− | + | 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> | ||
− | 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>. | + | 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== | ||
− | Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic. | + | Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic. Since the coefficient of the <math>x^2</math> term is <math>1</math>, the quadratic can be written as <cmath>(x - r_1)(x - r_2)=x^2 - (r_1 + r_2)x + r_1r_2</cmath> |
− | + | By comparing this with <math>x^2 - ax + 2a</math>, <cmath>r_1 + r_2 = a\text{ and }r_1r_2 = 2a.</cmath> | |
− | + | Plugging the first equation in the second, <cmath>r_1r_2 = 2 (r_1 + r_2).</cmath> Rearranging gives <cmath>r_1r_2 - 2r_1 - 2r_2 = 0\implies (r_1 - 2)(r_2 - 2) = 4.</cmath> These factors can be<math>(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2),</math> or <math>(-2, -2).</math> | |
− | + | We want the number of distinct <math>a = r_1 + r_2</math>, and these factors gives <math>a = -1, 0, 8, 9</math>. So the answer is <math>-1 + 0 + 8 + 9 = \boxed{\textbf{(C) }16}</math>. | |
− | |||
− | |||
− | + | === Video Solution by Richard Rusczyk === | |
− | + | https://artofproblemsolving.com/videos/amc/2015amc10a/397 | |
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/RQ4ZCttwmA4 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|num-b=22|num-a=24}} | {{AMC10 box|year=2015|ab=A|num-b=22|num-a=24}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Intermediate Number Theory Problems]] |
Revision as of 00:28, 27 September 2021
Contents
Problem
The zeroes of the function are integers. What is the sum of the possible values of
Solution 1
By Vieta's Formula, is the sum of the integral zeros of the function, and so
is integral.
Because the zeros are integral, the discriminant of the function, , is a perfect square, say
. Then adding 16 to both sides and completing the square yields
Therefore
and
Let
and
; then,
and so
. Listing all possible
pairs (not counting transpositions because this does not affect (
),
, yields
. These
sum to
, so our answer is
.
Solution 2
Let and
be the integer zeroes of the quadratic. Since the coefficient of the
term is
, the quadratic can be written as
By comparing this with ,
Plugging the first equation in the second, Rearranging gives
These factors can be
or
We want the number of distinct , and these factors gives
. So the answer is
.
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2015amc10a/397
Video Solution
~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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