## Task 1 - 100731

During the Peace Race, a leading group of cyclists passed by 6 Thälmann Pioneers in such a way that a clear order could be determined. To determine this order, each of the 6 Pioneers reported their observations, all of which are true:

(1) There were exactly 8 cyclists in the leading group, including exactly one Belgian and exactly two Poles.

(2) Among the cyclists who rode before the Belgian, there were at least two GDR cyclists.

(3) Among the cyclists who rode before the two Poles, there was at least one Soviet cyclist.

(4) Among the cyclists who rode after the Belgian, there was at least one Soviet cyclist.

(5) Two Soviet cyclists rode immediately one after the other.

(6) At the beginning and at the end of the leading group, there was a GDR cyclist each.

Determine the exact order of the cyclists in the leading group!
The Belgian drivers, GDR drivers, Polish and Soviet drivers are denoted in sequence as $B, D_{1}$, $D_{2}, \ldots, P_{1}, P_{2}, \ldots, S_{1}, S_{2}, \ldots$.

According to (5), at least two Soviet drivers were in the leading group. According to (2), at least one GDR driver neither drove at the beginning nor at the end, so due to (6), there were at least three GDR drivers in the leading group. Because of (1), these minimum numbers, 2 Soviet and 3 GDR drivers, must also be the exact numbers of Soviet and GDR drivers.

If $X, Y$ are designations of drivers, then $X<Y$ means that $X$ drove before $Y$. Then (2) $D_{1}<D_{2}<B$, (3) $S_{1}<P_{1}<P_{2}$, (4) $B<S_{2}$.

Since exactly one Belgian driver was in the leading group, it follows from (2) and (4)

(7) $D_{1}<D_{2}<B<S_{2}$.

Since exactly two Soviet drivers were in the leading group and, according to (5), drove immediately one after the other, it follows from this, as well as from (3) and (7),

(8) $D_{1}<D_{2}<B<S_{1}<S_{2}<P_{1}<P_{2}$.

From (6) and (8) it follows

(9) $D_{1}<D_{2}<B<S_{1}<S_{2}<P_{1}<P_{2}<D_{3}$.

This already accounts for 8 drivers, so (9) is the only possible sequence for the desired order.