Say a player is in position H if H is the number she’s named.

Any player in position 90-99 is lost, because her opponent can advance immediately to 100.

That means that a player in position 89 will win, since her opponent will be forced to enter this losing territory.

But then position 78 is also winning, since the opponent must open a road to 89.

And so on backward: 67, 56, 45, 34, 23, 12, and 1 are all winning positions.

With correct play, then, the first player always wins: She chooses 1 and then aims for each of these milestones.

The key is to notice that the 1 × 1 × 1 blocks must fall along a diagonal of the finished cube, to ensure that each 3 × 3 × 1 “slice” of the cube (in every orientation) can contain an odd number of blocks:

White’s knights began the game on squares of opposite color, but now they’re both on white squares. That means that, between them, they’ve made an odd number of moves in this game. By a similar argument White’s rooks and king have made an even number of moves in total, and we can see that White has made a single pawn move. White’s bishops haven’t moved, and his queen must have been captured on her home square. So White has made an even number of moves.

The same considerations apply to Black’s position, but he’s made an extra pawn move, so altogether he’s made an odd number of moves. Since White moved first, this means it’s Black turn to move, and he can mate with 1. Nxc2#.

(From Schach, 1957, via Noam D. Elkies and Richard P. Stanley, “The Mathematical Knight,” Mathematical Intelligencer 25:1 [December 2003], 22-34.)

Henry Dudeney presented this solution in his 1917 Amusements in Mathematics; it’s now known as the “method of buttons and strings.” No knight will ever visit the center square, so we can ignore that; regard each of the remaining squares as a button and connect it by a string to each square that’s a knight’s move away. That produces diagram B. Now if we untangle the strings we get diagram C, a unique cycle that any knight must follow in making its way around this board.

The four knights must follow this cycle, and no two knights may ever occupy the same square. The white knights start on squares 1 and 3, the black on 6 and 8. So to exchange the white and black knights, we must advance all four knights four steps along the circle, either clockwise or counterclockwise. Since the four knights make four moves each, the task is done in 16 moves. That’s the minimum. (Somewhat paradoxically, if three white knights oppose three black ones in the starting position, the minimum number of moves drops to 8.)

Yes. Put the coin in one pan, and put all 100 remaining coins in the other. Now suppose a true coin weighs t grams and a fake coin weighs f grams. If the coin is fake, then the difference between the weights will be

51t + 49f – f = 51t + 48f = 51t + 48(t ± 1) = 99t ± 48,

which is always divisible by 3. If the coin is true, then the difference will be

50t + 50f – t = 49t + 50f = 49t + 50(t ± 1) = 99t ± 50,

which is not divisible by 3.

Another solution is to put aside the chosen coin and divide the remaining 100 coins into two piles. If the difference in weight between these piles is even, then the chosen coin is genuine; if it’s odd then it’s false.

(From Sergei Aleksandrovich Genkin, Dmitrii Vladimirovich Fomin, and Ilia Itenberg, Mathematical Circles: (Russian Experience), 1996.)

They’re touching. If the rope is 150 feet long then we have to double it to reach 75 feet.

1. Qa8!

If the king goes to c4 then 2. Qe4 is mate.

If he moves to the a-file then Bc5 traps him there.

From the American Chess Bulletin, 1920.

(Thanks, Lee.)

This solution is by Clayton W. Dodge of the University of Maine at Orono:

Number the boxes 1-9 from top left to lower right, so that the first column is 1-4-7. Now:

1. The middle box must be empty. If it were marked, say with an X, then the board’s second X would have to be adjacent to it somewhere, and an O must complete that row, column, or diagonal. But now wherever we place the remaining O, the O player can immediately force a win by making a double threat if she has the move. X would be guilty of malpractice for allowing this disaster to come about, and we know she’s an expert player. So it must be X’s turn to move. But now we’ve deduced whose turn it is, and we’ve been told this is impossible. So this whole scenario cannot be the case — the center box must be empty.

2. No two similar marks can appear symmetrically around box 5, since then (if that player has the move) it leads to an immediate win or (if that player doesn’t have the move) we’re violating observation 1 above.

3. The same mark, say X, can’t appear in two adjacent corners, because then an O must appear between them, as well as in another box that will allow O to make two simultaneous threats if she has the move.

4. The same mark, say X, can’t appear in two middle-of-an-edge squares. If Xs appeared in, say, 2 and 8 we’d be violating observation 2. If they appeared in boxes 2 and 4, then one O must go in box 1 (to prevent the double threat) and another in box 6 or 8 to prevent X from winning with box 5. But then O can win with box 9.

5. The same mark, say X, can’t occupy two adjacent squares (say 1 and 2), since then an O must go in box 3 and another box that would allow O to make a double threat if she has the move.

6. Putting all this together, each player has marked one corner and one box opposite to it in the middle of an edge. There are four possibilities:

• If X has played to 1 and 6 and O to 2 and 9, then X can play to 4 and force a win.
• If X has played to 1 and 6 and O to 2 and 7, then X can play to 5 and force a win.
• If X has played to 1 and 6 and O to 4 and either 3 or 9, then expert play will always lead to a draw.

So the envelope must bear one of these two diagrams (or its symmetrical counterpart):

(From Pi Mu Epsilon Journal 5:10 [Spring 1974], 534.)