Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 30 / 4 / a / Solution

Symmetry forces $r=k$, and the pair-counting identity becomes $k(k-1)=2\cdot 45=90$. Its only positive integral solution is $k=r=10$. Since the number of blocks is even, the even-order symmetric design square obstruction requires $r-\lambda =8$ to be a square. It is not. No design with these parameters exists.