Symmetry forces $r=k$, and the pair-counting identity becomes $k(k-1)=2\cdot45=90$. 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. \b[No design with these parameters exists.]

Solution

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

 Solution (source code)