Difference between revisions of "2015 AMC 10A Problems/Problem 23"
(→Solution) |
|||
Line 5: | Line 5: | ||
− | ==Solution== | + | ==Solution 1== |
By Vieta's Formula, <math>a</math> is the sum of the integral zeros of the function, and so <math>a</math> is integral. | By Vieta's Formula, <math>a</math> is the sum of the integral zeros of the function, and so <math>a</math> is integral. | ||
Line 13: | Line 13: | ||
<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 16, so our answer is <math>\textbf{(C)}</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 16, so our answer is <math>\textbf{(C)}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>r_1</math> and <math>r_2</math> be the integer zeroes of the quadratic. | ||
+ | |||
+ | Since the coefficent of the <math>x^2</math> term is <math>1</math>, the quadratic can be written as <math>(x - r_1)(x - r_2)</math> or <math>x^2 - (r_1 + r_2)x + r_1r_2)</math>. | ||
+ | |||
+ | By comparing this with <math>x^2 - ax + a^2</math>, <math>r_1 + r_2 = a</math> and <math>r_1r_2 = 2a</math>. | ||
+ | |||
+ | Plugging the first equation in the second, <math>r_1r_2 = 2(r_1 + r_2). Rearranging gives </math>r_1r_2 - 2r_1 - 2r_2 = 0<math>. | ||
+ | |||
+ | This can be factored as </math>(r_1 - 2)(r_2 - 2) = 4<math>. | ||
+ | |||
+ | These factors can be: </math>(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)<math>. | ||
+ | |||
+ | We want the number of distinct </math>a = r_1 + r_2<math>, and these factors gives </math>a = {-1, 8, 9}. | ||
+ | |||
+ | So the answer is <math>-1 + 8 + 9 = \boxed{\textbf{(C) }16}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}} | {{AMC10 box|year=2015|ab=A|before=First Problem|num-a=2}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:48, 4 February 2015
Contents
Problem
The zeroes of the function are integers .What is the sum of the possible values of a?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}}\ 17\qquad\textbf{(E)}\ 18$ (Error compiling LaTeX. ! Extra }, or forgotten $.)
Solution 1
By Vieta's Formula, is the sum of the integral zeros of the function, and so is integral.
Because the zeros are integral, the discriminant of the function, , is a perfect square, say . Then adding 16 to both sides and completing the square yields Hence and Let and ; then, and so . Listing all possible pairs (not counting transpositions because this does not affect ), , yields . These sum to 16, so our answer is .
Solution 2
Let and be the integer zeroes of the quadratic.
Since the coefficent of the term is , the quadratic can be written as or .
By comparing this with , and .
Plugging the first equation in the second, r_1r_2 - 2r_1 - 2r_2 = 0$.
This can be factored as$ (Error compiling LaTeX. ! Missing $ inserted.)(r_1 - 2)(r_2 - 2) = 4$.
These factors can be:$ (Error compiling LaTeX. ! Missing $ inserted.)(1, 4), (-1, -4), (4, 1), (-4, -1), (2, 2), (-2, -2)$.
We want the number of distinct$ (Error compiling LaTeX. ! Missing $ inserted.)a = r_1 + r_2a = {-1, 8, 9}.
So the answer is .
See Also
2015 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.