Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2015 AMC 12A Problems/Problem 10"

(→‎Video Solution)
Line 5: Line 5:
 
<math> \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26</math>
 
<math> \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26</math>
  
−
==Solution==
+
==Solution 1==
  
 
Use [[SFFT]] to get <math>(x+1)(y+1)=81</math>. The terms <math>(x+1)</math> and <math>(y+1)</math> must be factors of <math>81</math>, which include <math>1, 3, 9, 27, 81</math>. Because <math>x > y</math>, <math>x+1</math> is equal to <math>27</math> or <math>81</math>. But if <math>x+1=81</math>, then <math>y=0</math> and so <math>x=\boxed{\textbf{(E)}\ 26}</math>.
 
Use [[SFFT]] to get <math>(x+1)(y+1)=81</math>. The terms <math>(x+1)</math> and <math>(y+1)</math> must be factors of <math>81</math>, which include <math>1, 3, 9, 27, 81</math>. Because <math>x > y</math>, <math>x+1</math> is equal to <math>27</math> or <math>81</math>. But if <math>x+1=81</math>, then <math>y=0</math> and so <math>x=\boxed{\textbf{(E)}\ 26}</math>.
  
 +
 +
==Solution 2==
 +
 +
Plug in values of <math>x</math> and solve for <math>y</math>, noting <math>x > y</math> and that y is an integer.
 +
 +
<math>x = 8</math>:
 +
 +
<math>
 +
8 + y + 8y = 80
 +
y = 8</math> (Does not work because <math>x > y</math>)
 +
 +
<math>x = 10</math>:
 +
 +
<math>
 +
10 + y + 10y = 80
 +
y = 70/11</math> (Does not work because y is not an integer)
 +
 +
 +
<math>x = 15</math>:
 +
 +
<math>
 +
15 + y + 15y = 80
 +
y = 65/16</math> (Does not work because y is not an integer)
 +
 +
 +
<math>x = 18</math>:
 +
 +
<math>
 +
18 + y + 18y = 80
 +
y = 62/19</math> (Does not work because y is not an integer)
 +
 +
Thus x = 26, <math>x=\boxed{\textbf{(E)}\ 26}</math>.
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==
 
https://youtu.be/ba6w1OhXqOQ?t=4512
 
https://youtu.be/ba6w1OhXqOQ?t=4512

Revision as of 12:42, 28 September 2025

Problem

Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$. What is $x$?

$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 15 \qquad\textbf{(D)}\ 18 \qquad\textbf{(E)}\ 26$

Solution 1

Use SFFT to get $(x+1)(y+1)=81$. The terms $(x+1)$ and $(y+1)$ must be factors of $81$, which include $1, 3, 9, 27, 81$. Because $x > y$, $x+1$ is equal to $27$ or $81$. But if $x+1=81$, then $y=0$ and so $x=\boxed{\textbf{(E)}\ 26}$.


Solution 2

Plug in values of $x$ and solve for $y$, noting $x > y$ and that y is an integer.

$x = 8$:

$8 + y + 8y = 80 y = 8$ (Does not work because $x > y$)

$x = 10$:

$10 + y + 10y = 80 y = 70/11$ (Does not work because y is not an integer)


$x = 15$:

$15 + y + 15y = 80 y = 65/16$ (Does not work because y is not an integer)


$x = 18$:

$18 + y + 18y = 80 y = 62/19$ (Does not work because y is not an integer)

Thus x = 26, $x=\boxed{\textbf{(E)}\ 26}$.

Video Solution by OmegaLearn

https://youtu.be/ba6w1OhXqOQ?t=4512

~ pi_is_3.14

See Also

2015 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 12 Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2015_AMC_12A_Problems/Problem_10&oldid=261697"