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) | + | <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 | ||
+ | <cmath>\sum_a a = (5+4) + (-5+4) + (4+4) + (-4+4) = \boxed{\textbf{(C) }16}</cmath>. | ||
+ | |||
+ | |||
+ | |||
+ | === Video Solution by Richard Rusczyk === | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2015amc10a/397 | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Let the roots be <math>r</math> and <math>s</math>. By Vieta's we have that <math>r+s = a</math>, and <math>rs = 2a</math>. So by Simon's Favorite Factoring Trick: <cmath>2r+2s = rs \Longrightarrow rs-2r-2s = 0 \Longrightarrow (r-2)(s-2) = 4</cmath>. Now, we test out values of <math>r</math> and <math>s</math> to see which work. Lets say that <math>r-2 = 1</math>, and <math>s-2 = 4</math>. This implies that <math>(r,s) = (3,6)</math>, so <math>a = 9</math>. Now, we let <math>r-2 = 2</math>, and <math>s-2 = 2</math>, this gives us <math>(r,s) = (4,4)</math>, which gives us a value for <math>a</math> of <math>8</math>. Now, we circle an answer of <math>17</math> and move onto the next problem. Yike! They didnt say <math>a</math> cant be negative! This is where many people would go wrong. Lets let <math>r-2 = -1</math>, and <math>s-2 = -4</math>, this gives us that <math>(r,s) = (1,-2)</math>. This results in a value of <math>-1</math> for <math>a</math>. Now, we let <math>r-2 = -2</math>, and <math>s-2 = -2</math>. This gives <math>a = 0</math>. Now, we are done. Our total sum is <math>9+8-1 = \boxed{\textbf{(C) }16}</math>. | ||
+ | |||
+ | -jb2015007 - minor edit by mathlover1205 | ||
+ | |||
+ | ==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 16:59, 18 July 2025
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
Solution 3
Let the roots be and
. By Vieta's we have that
, and
. So by Simon's Favorite Factoring Trick:
. Now, we test out values of
and
to see which work. Lets say that
, and
. This implies that
, so
. Now, we let
, and
, this gives us
, which gives us a value for
of
. Now, we circle an answer of
and move onto the next problem. Yike! They didnt say
cant be negative! This is where many people would go wrong. Lets let
, and
, this gives us that
. This results in a value of
for
. Now, we let
, and
. This gives
. Now, we are done. Our total sum is
.
-jb2015007 - minor edit by mathlover1205
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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.