1. A three-digit number has its middle digit three times smaller than the sum of the other two, and the sum of the last two digits is half of the first digit. If the digits in the tens and units places are swapped, the resulting number is 18 less than the given number. What is that number?
Solution. Let $x, y$ and $z$ denote the units, tens, and hundreds digits, respectively, of the desired three-digit number. According to the conditions of the problem,

$$
\left\{\begin{array}{l}
3 y=x+z \\
x=2(y+z) \\
100 x+10 y+z=100 x+10 z+y+18
\end{array}\right.
$$

From which we get $x=8, y=3$ and $z=1$. Therefore, the desired number is 831.