We know that among the following four statements about the positive integers $a$ and $b$, three are true and one is false:

1) $(a+1)$ is divisible by $b$;
2) $a=2b+5$
3) $a+b$ is divisible by $3$
4) $a+7b$ is a prime number.

Determine the numbers $a$ and $b$.
If (3) is true, then neither (2) nor (4) can be true. From (2), $a+b=3b+5$ follows, and the right side is not divisible by 3; furthermore, according to (3), $a+7b=(a+b)+6b$ would be divisible by 3, and since it is greater than 3, it could not be prime.

Therefore, the three true statements are: (1), (2), and (4). From (2), $a+1=2b+6$; this number, according to (1), is divisible by $b$, so $b$ must be a divisor of 6, hence its possible values are 1, 2, 3, and 6. Since (2) implies that $b$ is odd, for $a+7b$ not to be even, $b$ must be even.

If $b=2$, then $a=9$, $a+b=11$ is not divisible by 3, and $a+7b=23$ is a prime number.

If $b=6$, then $a=17$, $a+b=23$ is not divisible by 3, and $a+7b=59$ is also a prime number. Therefore, the two solutions are:

$$
a=9, \quad b=2
$$

and

$$
a=17, \quad b=6
$$

Based on the work of Weisz Csaba (Kaposvár, Táncsics M. Gymnasium, 2nd grade)