Chessbogglers Trivia Quiz

Answer: 50

Each rook has 7 legal moves (including one attack). Each knight has 3. Each bishop has 7. The king has 2 (as it cannot move to D2). The queen has 14 (including one attack). Totaling it up: 2×(7+3+7)+2+14=50. With the pawns in place, the number is 20.

Answer: King

The knight cannot perform the task. As it switches square colour on each of its moves, on even-numbered moves it lands on a black square. Thus, after the 62nd move, assuming that it has made it through all the squares besides H8, it is placed on a black square. This means that it cannot reach H8 on its 63rd and last move, since H8 is a black square as well.

The bishop cannot perform the task since it is limited to single-coloured squares.

There are numerous routes for the king to achieve the task. For example, covering the top six rows by N-like and inverted-N-like moves, next dashing from H7 to A7 and from A8 to H8.

Answer: 5

This is a variation of the classic "Queens Problem", which belongs to a branch called domination problems. There are 4860 solutions with 5 queens where the queens are allowed to attack each other. When the constraint is added that the 5 queens must be placed on a non-attacking formation, the number of solutions drops to 91. Here is one of them: A1, C7, D3, G8 and H4.

Answer: 1

Regardless of the result, 1/42 points are granted in each of the tournament's games.

The total number of games in the tournament is 7!/5!=42.

42 games, 1/42 points granted per game - 1 point in total is granted throughout the tournament.

Answer: 3136

No matter where you place a rook on the board, it poses an attack on 14 squares. Not forgetting to include the square the rook is placed on, the correct calculation is: 64×(64-14-1)=64×49=3136.

Answer: No

Each domino tile covers one black square and one white square. The chessboard in question has 32 black squares and 30 white squares, which means that the answer to my question is negative.

On the other hand, a generalized theorem exists (Gomory's Theorem) proving that any standard chessboard with one black square and one white square removed can be fully tiled by 31 domino pieces.

Answer: 712

A bishop placed on the any of the 14 black squares on an outer boundary of the chessboard poses an attack on 7 other black squares. As you move into the centre of the board the number increases to 9, 11 and finally 13 on the two centre black squares (D4 and E5). A summation on all options gives 14×(32-7-1)+10×(32-9-1)+6×(32-11-1)+2×(32-13-1)=712.

Answer: 32

A knight can pose an attack only on a square with a different colour than the one it stands on. Hence, when 32 knights are placed on 32 same-colour squares none of them attacks one another. Without formally proving it here, it is pretty intuitive to see that any formation which involves placing knights on different coloured squares results in a lower number.

Answer: Yes

The first elimination round has 2^(k-1) games, the second one 2^(k-2) and so forth till we reach the final.
The total number of games is always (2^k)-1 which is the sum of the geometric series 1,2,...,2^(k-2),2^(k-1).
There are infinitely many instances where (2^k)-1 takes a value evenly divisible by 3 (e.g. 3 for k=2 and 63 for k=6).

Answer: Yes, a queen

Symmetry considerations give away the queen while ruling out the other three options. Only a queen can roam freely between the unattacked squares of this formation (A3, A6, C1, C8, F1, F8, H3 and H6).