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

Revision as of 13:22, 29 March 2015

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

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

@@ Line 3: / Line 3: @@
 Integers <math>x</math> and <math>y</math> with <math>x>y>0</math> satisfy <math>x+y+xy=80</math>. What is <math>x</math>?
-<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==

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

Revision as of 13:22, 29 March 2015

Problem

Solution

See Also