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Difference between revisions of "2015 AMC 12A Problems/Problem 10"
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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>.
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 +
== Video Solution ==
 +
https://youtu.be/ba6w1OhXqOQ?t=4512
 +
 +
~ pi_is_3.14
  
 
== See Also ==
 
== See Also ==
 
{{AMC12 box|year=2015|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2015|ab=A|num-b=9|num-a=11}}

Revision as of 19:11, 15 January 2021

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

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

Video Solution

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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