B1. What is the maximum possible greatest common divisor of the numbers $a-2b+3, 2a-3b-1$, and $3a+b-2$, if $a$ and $b$ are natural numbers?
B1. Let $d$ be the greatest common divisor of the given numbers. Then $d$ also divides the numbers

$$
\begin{aligned}
& (2 a-3 b-1)-2(a-2 b+3)=b-7 \\
& (a-2 b+3)+2(b-7)=a-11 \\
& (3 a+b-2)-(b-7)=3 a+5 \\
& (3 a+5)-3(a-11)=38
\end{aligned}
$$

Thus, $d$ can be at most 38. If we choose $a=11$ and $b=7$, then $a-2 b+3=0$, $2 a-3 b-1=0$, and $3 a+b-2=38$, so the greatest common divisor can indeed be 38.

Mentioning or using the fact that for any integers $e$ and $f$, their greatest common divisor $\operatorname{gcd}(e, f)$ divides any linear combination of $e$ and $f \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point Writing the equation $(2 a-3 b-1)-2(a-2 b+3)=b-7$ (or equivalent elimination of one variable from the equations) 1 point Writing the equation $(a-2 b+3)+2(b-7)=a-11$ (or equivalent elimination of the second variable from the equations) 1 point Writing the equation $(3 a+b-2)-(b-7)=3 a+5$ (or equivalent elimination of the second variable from the equations in a different way than before) 1 point Writing the equation $(3 a+5)-3(a-11)=38$ (or equivalent elimination of all variables from the equations) and concluding that $d$ can be at most 38. 1 point

Constructing an example with $\mathbf{z} a=11$ and $b=7$ (or equivalent), which gives $d=38$. 2 points

Note: If the same variable is eliminated twice, the contestant also gets the point for eliminating the second variable.