Difference between revisions of "2014 AMC 10A Problems/Problem 21"
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<math> \textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8} </math> | <math> \textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8} </math> | ||
− | ==Solution== | + | ==Solution 1== |
− | Note that <math>y= | + | Note that when <math>y=0</math>, the <math>x</math> values of the equations should be equal by the problem statement. We have that |
+ | <cmath>0 = ax + 5 \implies x = -\dfrac{5}{a}</cmath> <cmath>0 = 3x+b \implies x= -\dfrac{b}{3}</cmath> | ||
+ | Which means that <cmath>-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15</cmath> | ||
+ | The only possible pairs <math>(a,b)</math> then are <math>(a,b) = (1,15), (3,5), (5,3), (15, 1)</math>. These pairs give respective <math>x</math>-values of <math>-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}</math> which have a sum of <math>\boxed{\textbf{(E)} \: -8}</math>. | ||
− | + | ==Solution 2== | |
+ | |||
+ | Going off of Solution 1, for the first equation, notice that the value of <math>x</math> cannot be less than <math>-5</math>. We also know for the first equation that the values of <math>x</math> have to be <math>5</math> divided by something. Also, for the second equation, the values of <math>x</math> can only be <math>-\frac13,-\frac23,-\frac33, \dots</math>. Therefore, we see that, the only values common between the two sequences are <math>-1, -5, -\frac13,-\frac53</math>, and adding them up, we get for our answer, <math>\boxed{\textbf{(E)} \: -8}</math>. | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://www.youtube.com/watch?v=-vZKwIazT08&list=PLyhPcpM8aMvKEM8u4Q-7Gi0rU5WU4WOb1&index=1 - AMBRIGGS | ||
==See Also== | ==See Also== | ||
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{{AMC10 box|year=2014|ab=A|num-b=20|num-a=22}} | {{AMC10 box|year=2014|ab=A|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category: Advanced Algebra Problems]] |
Latest revision as of 16:35, 30 July 2022
Problem
Positive integers and
are such that the graphs of
and
intersect the
-axis at the same point. What is the sum of all possible
-coordinates of these points of intersection?
Solution 1
Note that when , the
values of the equations should be equal by the problem statement. We have that
Which means that
The only possible pairs
then are
. These pairs give respective
-values of
which have a sum of
.
Solution 2
Going off of Solution 1, for the first equation, notice that the value of cannot be less than
. We also know for the first equation that the values of
have to be
divided by something. Also, for the second equation, the values of
can only be
. Therefore, we see that, the only values common between the two sequences are
, and adding them up, we get for our answer,
.
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=-vZKwIazT08&list=PLyhPcpM8aMvKEM8u4Q-7Gi0rU5WU4WOb1&index=1 - AMBRIGGS
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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.