Chess

Here's a splendid problem for connoisseurs of the number 42. Composed by Bo Lindgren, it's a series self-mate in 42 - which means that White must play 42 consecutive moves to arrive at a position in which Black is forced to give mate in one.

You could try to solve it, but it may be more fun just to play through the answer.

The idea is that White's last move will be Qxe2+, with Black forced to reply Nxe2 mate; and that can work only if the squares round the white king are blocked by the appropriate pieces.

So here goes: 1.e8=N, 2.Nxf6, 3.Nd5, 4.f6, 5.f7, 6.f8=Q, 7.Qxf3, 8.Qg2 (lying in wait), 9.f4, 10.f5, 11.f6, 12.f7, 13.f8=B, 14.Bxh6, 15.Be3, 16.h6, 17.h7, 18.h8=R, 19.Rxh4, 20.Re4, 21.h4, 22.h5, 23.h6, 24.h7, 25.h8=Q, 26.Qxb8, 27.Qe5, 28.b8=B, 29.Bxa7, 30.Bc5, 31.a7, 32.a8=R, 33.Rxa4, 34.Rb4, 35.a4, 36.a5, 37.a6, 38.a7, 39.a8=N, 40.Nb6, 41.Nc4, 42.Qxe2+ and Black must play Nxe2 mate.

White has promoted all eight pawns, two to each type of piece. An extraordinary composition.