247. Find the natural numbers $a$ and $b$, if out of the four statements:
1) $a-b$ is divisible by 3
2) $a+2b$ is a prime number;
3) $a=4b-1$
4) $a+7$ is divisible by $b$

three are true, and one is false. Find all solutions.
$\triangle$ We will look for a false statement among the given ones. For this, it would be good to find two incompatible statements.

The first and second statements cannot be true at the same time, because if the number $a-b$ is divisible by 3, then the number $a+2b = (a-b) + 3b$ is also divisible by 3, and therefore cannot be prime. Moreover, the first and third statements are incompatible, since if $a = 4b - 1$, then the number $a - b = (4b - 1) - b = 3b - 1$ cannot be divisible by 3.

Since in both cases of incompatible pairs of statements, the first statement appears, it is the false one. Therefore, the second, third, and fourth statements are true. We will use this to find $b$. Apply the third and fourth statements. The sum

$$
a + 7 = (4b - 1) + 7 = 4b + 6
$$

must be divisible by $b$, from which 6 must be divisible by $b$. In this case,

$$
b \in \{1; 2; 3; 6\}
$$

Let's consider all four cases, taking into account that the sum

$$
a + 2b = (4b - 1) + 2b = 6b - 1
$$

is a prime number. The values $b_1 = 1$, $b_2 = 2$, and $b_3 = 3$ fit. Let's find the corresponding values of $a$ using the formula $a = 4b - 1$.

Answer: $(3; 1), (7; 2), (11; 3)$.