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

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==Solution 2==
 
==Solution 2==
  
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Plug in values of <math>x</math> and solve for <math>y</math>, noting <math>x > y</math> and that y is an integer.  
+
Plug in values of <math>x</math> and solve for <math>y</math>, noting <math>x > y</math> and that <math>y</math> is an integer.  
  
 
<math>x = 8</math>:
 
<math>x = 8</math>:
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<math>
 
<math>
 
8 + y + 8y = 80  
 
8 + y + 8y = 80  
−
y = 8</math> (Does not work because <math>x > y</math>)
+
</math>
 +
 
 +
<math>y = 8</math> (Does not work because <math>x > y</math>)
 +
 
  
 
<math>x = 10</math>:
 
<math>x = 10</math>:
  
 
<math>
 
<math>
−
10 + y + 10y = 80  
+
10 + y + 10y = 80
−
y = 70/11</math> (Does not work because y is not an integer)
+
</math>
 +
 +
<math>y = 70/11</math> (Does not work because <math>y</math> is not an integer)
  
  
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<math>
 
<math>
 
15 + y + 15y = 80  
 
15 + y + 15y = 80  
−
y = 65/16</math> (Does not work because y is not an integer)
+
</math>
 +
 
 +
<math>y = 65/16</math> (Does not work because <math>y</math> is not an integer)
  
  
Line 38: Line 45:
 
<math>
 
<math>
 
18 + y + 18y = 80  
 
18 + y + 18y = 80  
−
y = 62/19</math> (Does not work because y is not an integer)
+
</math>
 +
 
 +
<math>y = 62/19</math> (Does not work because <math>y</math> is not an integer)
 +
 
 +
 
 +
Thus <math>x = 26</math>, <math>\boxed{\textbf{(E)}\ 26}</math>.
 +
 
  
−
Thus <math>x = 26</math>, <math>x=\boxed{\textbf{(E)}\ 26}</math>.
+
~ Solution by CYB3RFLARE7408
  
 
== Video Solution by OmegaLearn ==
 
== Video Solution by OmegaLearn ==

Latest revision as of 15:47, 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 $x = 26$, $\boxed{\textbf{(E)}\ 26}$.


~ Solution by CYB3RFLARE7408

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
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