Problem 9.3. Natural numbers $a$ and $b$ are such that $a$ is divisible by $b+1$ and 43 is divisible by $a+b$.

(a) (1 point) Indicate any possible value of $a$.

(b) (3 points) What can $b$ be? Indicate all possible options.
Answer: (a) any of the numbers $22,33,40,42$. (b) $1,3,10,21$.

Solution. Since 43 is a prime number and $a+b$ is a natural number greater than 1, we get that $a+b=43$. Therefore, $b \leqslant 42$ and $a=43-b$.

From the condition, it follows that the number $b+1$ is a divisor of the number $a+(b+1)=44$. Since $b+1$ is a natural number greater than 1 and less than 44, we get that $b+1$ can equal one of the values $2,4,11,22$.

- Let $b+1=2$. Then $b=1$ and $a=42$.
- Let $b+1=4$. Then $b=3$ and $a=40$.
- Let $b+1=11$. Then $b=10$ and $a=33$.
- Let $b+1=22$. Then $b=21$ and $a=22$.

It is easy to verify that all pairs $(42,1),(40,3),(33,10),(22,21)$ satisfy the condition.