1. Determine the digits $a, b$ so that the sum $\overline{323 a}+\overline{b 410}$ is divisible by 9?
Solution. Obviously, the sum will be a four-digit or five-digit number. If the sum is a four-digit number, then the digit $b$ is less than or equal to 6, and the digits of the sum are: the first $3+b$, the second 6, the third 4, and the fourth $a$. The given sum is divisible by 9 if and only if the sum of the digits $3+b+6+4+a=13+a+b$ is divisible by 9. We have

- $b=6$, the sum is $19+a$, so $a=8$
- $b=5$, the sum is $18+a$, so $a=0$ or $a=9$
- $b=4$, the sum is $17+a$, so $a=1$
- $b=3$, the sum is $16+a$, so $a=2$
- $b=2$, the sum is $15+a$, so $a=3$
- $b=1$, the sum is $14+a$, so $a=4$

In the case where the sum is a five-digit number, then its digits are the fifth $a$, the fourth 4, the third 6, the second $b+3-10$, and the first 1. Their sum is $a+4+6+b+3-10+1=a+b+4$ where the digit $b$ is equal to 7, 8, or 9. The following cases are possible:

- $b=7$, the sum is $11+a$, so $a=7$
- $b=8$, the sum is $12+a$, so $a=6$
- $b=9$, the sum is $13+a$, so $a=5$