Which are the three-digit numbers divisible by 45, whose digits in the order of writing form an arithmetic sequence?
Solution. The digits of the three-digit number are: $a+d, a, a-d$. Since 45 is a divisor of the number and $45=9 \cdot 5$, the sum of the digits is also divisible by 9. Therefore,

$$
a+d+a+a-d=3 a=9 k
$$

from which $a=3 k \leq 9$ so $k=1,2$ or 3. The possible values of $a$ are accordingly 3, 6, or 9. Since the three-digit number is also divisible by 5, its last digit is either 0 or 5.

If $a-d=0$, then $a=d$. In the case of $a=3$, the three-digit number is 630; and no more cases are possible.

If $a-d=5$, then $d=a-5$. In the case of $a=3$, $d=-2$ and the three-digit number is 135. Or $a=6, d=1$ and the number is 765; no more cases are possible in this scenario either.

The sought three-digit numbers are: $135,630,765$.