Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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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>. | ||
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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>. | 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>. | ||
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+ | === Video Solution by Richard Rusczyk === | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2015amc10a/397 | ||
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+ | ==Video Solution== | ||
+ | https://youtu.be/RQ4ZCttwmA4 | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
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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