Difference between revisions of "2015 AMC 10A Problems/Problem 23"
m (fixed typos) |
m (→Solution 2) |
||
(32 intermediate revisions by 17 users not shown) | |||
Line 1: | Line 1: | ||
==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) | + | <math> \textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 18</math> |
+ | ==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. | ||
− | == | + | 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> | ||
+ | Therefore <math>(a-4)^2 - k^2 = 16</math> and | ||
+ | <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>. | ||
− | + | ==Solution 2== | |
+ | |||
+ | 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> | ||
+ | |||
+ | 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 <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>. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | |||
+ | |||
+ | === 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| | + | {{AMC10 box|year=2015|ab=A|num-b=22|num-a=24}} |
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Intermediate Number Theory Problems]] |
Latest revision as of 10:07, 10 July 2024
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. ,
By Vieta's Formulas,
Plugging the first equation in the second,
Rearranging gives
These factors (ignoring order, because we want the sum of factors), can be
or
.
The sum of distinct , and these factors give
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.