At the vertices of a hexagon, the numbers $12,1,10,6,8,3$ (in that order) are written. In one move, it is allowed to choose two adjacent vertices and simultaneously add one to the numbers in these vertices or simultaneously subtract one from them. Can we end up with six numbers in the following order:

a) $14,6,13,4,5,2 ;$ b) $6,17,14,3,15,2 ?$
a) Yes, for example: $(12,1,10,6,8,3)$ - $(12,1,10,6,7,2)$ - $(14,3,10,6,7,2)$ - $(14,3,10,4,5,2)$ - $(14,6,13,4,5,2)$.

b) No, because the parity of the sum of all numbers cannot change, and at the beginning, the sum was 40, an even number. Therefore, it cannot become $6+17+14+3+15+2$ = 57, an odd number.

## Answer

a) yes; b) no.