Integers $x$ and $y$ with $x>y>0$ satisfy $x+y+xy=80$. What is $x$?
I notice that the equation $x+y+xy=80$ can be rewritten as $(x+1)(y+1)=81$ by adding $1$ to both sides and factoring. This means that $x+1$ and $y+1$ are factors of $81$ that are greater than $1$. I also know that $x>y$, so $x+1>y+1$. I can list the factors of $81$ that are greater than $1$ as follows: $3, 9, 27, 81$. Since $x+1$ is the larger factor, it must be either $27$ or $81$. If $x+1=27$, then $y+1=3$, which implies that $x=26$ and $y=2$. If $x+1=81$, then $y+1=1$, which implies that $x=80$ and $y=0$. However, the problem states that $x>y>0$, so the second case is invalid. Therefore, the only possible value of $x$ is $26$. 