Difference between revisions of "2015 AMC 10A Problems/Problem 23"
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<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>. However, <math>a = 8</math> and <math>a=0</math> do not work because the problem states that there are "zeros" of the function that are "integers", which clearly signifies more than one root. Thus the only <math>a</math> that work are <math>a=9</math> and <math>a=-1</math>. These <math>a</math> sum to <math>8</math>, so our answer is <math>\textbf{(B)}</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>. However, <math>a = 8</math> and <math>a=0</math> do not work because the problem states that there are "zeros" of the function that are "integers", which clearly signifies more than one root. Thus the only <math>a</math> that work are <math>a=9</math> and <math>a=-1</math>. These <math>a</math> sum to <math>8</math>, so our answer is <math>\textbf{(B)}</math>. | ||
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+ | FOR THE WRITER OF THIS SOLUTION: Generally, roots that are double roots are still counted as "two" roots by multiplicity. See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3735612&#p3735612 (the forum discussion for this problem). The answer should be C as shown below. I won't edit your solution; instead I'll let you edit your own solution how you want it. | ||
==Solution 2== | ==Solution 2== |
Revision as of 13:00, 5 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. Unknown error_msg)
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
. However,
and
do not work because the problem states that there are "zeros" of the function that are "integers", which clearly signifies more than one root. Thus the only
that work are
and
. These
sum to
, so our answer is
.
FOR THE WRITER OF THIS SOLUTION: Generally, roots that are double roots are still counted as "two" roots by multiplicity. See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3735612&#p3735612 (the forum discussion for this problem). The answer should be C as shown below. I won't edit your solution; instead I'll let you edit your own solution how you want it.
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, . Rearranging gives
.
This can be factored as .
These factors can be: .
We want the number of distinct , and these factors gives
.
So the answer is .
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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