2. For natural numbers $a$ and $b$, it is known that $5a-1$ is divisible by $b$, $a-10$ is divisible by $b$, but $3a+5$ is not divisible by $b$. What values can the number $b$ take?
Answer: 49.

Solution. The number $(5 a-1)-5 \cdot(a-10)=49$ is divisible by $b$, so either $b=1$, or $b=7$, or $b=49$. If the number $(3 a+5)-3 \cdot(a-10)=35$ were divisible by $b$, then $3 a+5$ would be divisible by $b$, which is not the case. Therefore, the options $b=1$ and $b=7$ are impossible. The numbers $a=10$ and $b=49$ show that $b=49$ is achievable.

Comment. (a) Incorrect answer -1 point.

(b) Proven that $b$ is a divisor of $49-3$ points.

(c) Proven that $b$ is not equal to 7 or a stronger statement from which it follows that $b \neq 7-1$ point.

Points for the above items are summed except in the following cases:

Proven items (b) and (c) -6 points.

Correct solution -7 points.

Note. Checking the existence of such numbers $a$ and $b$ is not mandatory, as it is assumed in the problem statement.