2015 AMC 12A Problems/Problem 10

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

~ Solution by CYB3RFLARE7408

Video Solution by OmegaLearn

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

~ pi_is_3.14

2015 AMC 12A (Problems • Answer Key • Resources)
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2015 AMC 12A Problems/Problem 10

Contents

Problem

Solution 1

Solution 2

Video Solution by OmegaLearn

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