1. The sum of the digits of a three-digit number is 10. The digit in the tens place is equal to the sum of the digit in the hundreds place and the digit in the units place. If the digit in the units place and the digit in the tens place are swapped, a three-digit number is obtained which is 9 less than the original number. Find that number.
Solution. If $x$ is the unit digit, $y$ the tens digit, and $z$ the hundreds digit of the desired three-digit number, then it can be written in the form $x+10 y+100 z$. Since the sum of the digits of the desired number is 10, we have:

$$
x+y+z=10
$$

On the other hand, the tens digit is equal to the sum of the unit digit and the hundreds digit, i.e.

$$
y=x+z
$$

When the unit digit and the tens digit in the given three-digit number are swapped, we get a three-digit number $y+10 x+100 z$, which is 9 less than the original number $x+10 y+100 z$, so we have

$$
y+10 x+100 z+9=x+10 y+100 z
$$

Solving the system of equations (1), (2), and (3), we get $x=4, y=5$, and $z=1$, so the desired number is 154.