Solution SSLE.C50 Robert Beezer

Let $a$ be the hundreds digit, $b$ the tens digit, and $c$ the ones digit. Then the first condition says that $b+c=5$. The original number is $100a+10b+c$, while the reversed number is $100c+10b+a$. So the second condition is \begin{equation*} 792=\left(100a+10b+c\right)-\left(100c+10b+a\right)=99a-99c \end{equation*} So we arrive at the system of equations \begin{align*} b+c&=5\\ 99a-99c&=792 \end{align*} Using equation operations, we arrive at the equivalent system \begin{align*} a-c&=8\\ b+c&=5 \end{align*} We can vary $c$ and obtain infinitely many solutions. However, $c$ must be a digit, restricting us to ten values (0 – 9). Furthermore, if $c>1$, then the first equation forces $a>9$, an impossibility. Setting $c=0$, yields $850$ as a solution, and setting $c=1$ yields $941$ as another solution.