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 when <math>y=0</math>, the <math>x</math> values of the equations should be equal by the problem statement. We have that | 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> | <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> | 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>. | 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>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Similar to the above solutions except we're using the equations | ||
+ | <math>a = -\frac{-5}{x}</math> | ||
+ | and | ||
+ | <math>b = -3x</math> | ||
+ | With this, we know that c has to be negative. Doing some math, we find <math>x</math> to be | ||
+ | <math>-1, -5, -\frac{1}{3}, </math>and<math> -\frac{5}{3}</math> | ||
+ | |||
+ | Adding them up gives you our answer:<math>\boxed{\textbf{(E)} \: -8}</math>. | ||
+ | |||
+ | ~Starshooter11 | ||
+ | |||
+ | |||
+ | |||
+ | === Video Solution by Richard Rusczyk === | ||
+ | |||
+ | https://artofproblemsolving.com/videos/amc/2014amc10a/375 | ||
+ | |||
+ | ~ dolphin7 | ||
==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: Introductory Algebra Problems]] |
Revision as of 18:31, 30 August 2020
Contents
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, .
Solution 3
Similar to the above solutions except we're using the equations and With this, we know that c has to be negative. Doing some math, we find to be and
Adding them up gives you our answer:.
~Starshooter11
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2014amc10a/375
~ dolphin7
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.