Difference between revisions of "2014 AMC 10A Problems/Problem 21"

Latest revision as of 16:41, 2 November 2024

Problem

Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$ -axis at the same point. What is the sum of all possible $x$ -coordinates of these points of intersection?

$\textbf{(A)}\ {-20}\qquad\textbf{(B)}\ {-18}\qquad\textbf{(C)}\ {-15}\qquad\textbf{(D)}\ {-12}\qquad\textbf{(E)}\ {-8}$

Solution 1

Note that when $y=0$ , the $x$ values of the equations should be equal by the problem statement. We have that $0 = ax + 5 \implies x = -\dfrac{5}{a}$ $0 = 3x+b \implies x= -\dfrac{b}{3}$ Which means that $-\dfrac{5}{a} = -\dfrac{b}{3} \implies ab = 15$ The only possible pairs $(a,b)$ then are $(a,b) = (1,15), (3,5), (5,3), (15, 1)$ . These pairs give respective $x$ -values of $-5, -\dfrac{5}{3}, -1, -\dfrac{1}{3}$ which have a sum of $\boxed{\textbf{(E)} \: -8}$ .

Solution 2

Using part of what we got from Solution 1, notice that the value of $x$ cannot be less than $-5$ . We also know for the first equation that the values of $x$ have to be $5$ divided by something. Also, for the second equation, the values of $x$ can only be $-\frac13,-\frac23,-\frac33, \dots$ . Therefore, we see that, the only values common between the two sequences are $-1, -5, -\frac13,-\frac53$ , and adding them up, we get for our answer, $\boxed{\textbf{(E)} \: -8}$ .

Video Solution by Pi Academy

https://youtu.be/uUxcMZVFSR8?si=zQWytkiPOVw1CRTG ~ Pi Academy

Video Solution by Richard Rusczyk

https://www.youtube.com/watch?v=-vZKwIazT08&list=PLyhPcpM8aMvKEM8u4Q-7Gi0rU5WU4WOb1&index=1 - AMBRIGGS

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.

@@ Line 12: / Line 12: @@
 ==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>.
+Using part of what we got from Solution 1, 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==
+==Video Solution by Pi Academy==
-Similar to the above solutions except we're using the equations
+https://youtu.be/uUxcMZVFSR8?si=zQWytkiPOVw1CRTG ~ Pi Academy
-<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>.
+==Video Solution by Richard Rusczyk==
+https://www.youtube.com/watch?v=-vZKwIazT08&list=PLyhPcpM8aMvKEM8u4Q-7Gi0rU5WU4WOb1&index=1 - AMBRIGGS
-~Starshooter11
-=== Video Solution by Richard Rusczyk ===
-https://artofproblemsolving.com/videos/amc/2014amc10a/375
-~ dolphin7
 ==See Also==
@@ Line 39: / Line 25: @@
 {{MAA Notice}}
-[[Category: Introductory Algebra Problems]]
+[[Category: Advanced Algebra Problems]]

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