Example 6. Given a three-digit integer that is a multiple of 5, the sum of its digits is 20, and the sum of one digit and the hundreds digit is a multiple of 3, find this integer.
Solution: Let the required three-digit integer be
$$
N=100 x+10 y+z \text {. }
$$

Since $N$ is a multiple of 5, it must be that $z=0$ or $z=5$. It is also given that the sum of the digits of $N$ is 20, so
$$
x+y+z=20 \text { . }
$$

If $z=0$, then $x+y=20$.
Given $1 \leqslant x \leqslant 9,0 \leqslant y \leqslant 9$, we have $1 \leqslant x+y \leqslant 18$,
which contradicts (1). Therefore, only $z=5$, in which case $x+y=15, y=15$ $-x \leqslant 9,6 \leqslant x \leqslant 9$. It is also given that $x+z=x+5$ is a multiple of 3, so $x=7$, and thus $y=8$.
Therefore, the required three-digit integer is 785.