No, it can’t be done. On a given move, suppose that t pennies change from tails to heads. That means that 4-t pennies change from heads to tails. So this move has changed the total number of tails by (4-t) – t = 4-2t = 2(2-t). Since this number is a multiple of 2, it will always be even. If we started with 0 tails, and each move changes the total number of tails by an even number, then we can never reach 7 tails.

From Paul Vaderlind, Richard K. Guy, and Loren C. Larson, The Inquisitive Problem Solver, 2002.

White wins with the preposterous-looking 1. Qh7!, threatening 2. Qxd7#. If Black captures the queen then White mates with the knight.

From A Collection of Two Hundred Chess Problems, 1866.

Call two opposing points A and B, and suppose that the temperature at A is higher than B. So A – B is positive. Now rotate both points around the equator, maintaining their opposition. Their difference can’t remain positive, because we know that when they’ve traded places it will be negative (the opposite of its initial value). Because temperature varies continuously, there must be some position in between where their difference is zero.

1. Qg6 and White will mate with 2. Qb1, 2. Rb1, 2. Na7, or 2. Rxc5.

From J. Paul Taylor, Chess Chips, 1878.

The smallest possible number of men is seven. They could be accounted for in three different ways. (1) Two with both arms sound, one with broken right arm, and four with both arms broken. (2) One with both arms sound, one with broken left arm, two with broken right arm, and three with both arms broken. (3) Two with left arm broken, three with right arm broken, and two with both arms broken. But if every man was injured, the last case is the only one that would apply.

Since the final state matches the initial state, we can imagine this process going on continuously. Consider the dwarf whose cup contains the smallest amount of milk just before he begins pouring. Call him D, and call this quantity of milk a ounces. Since D’s cup contains the least milk before pouring, then each other dwarf has an equal or greater amount in his own cup before pouring, and thus gives D at least a/6 ounces. This would leave D with at least a ounces in his own cup just before pouring; the only way he can receive precisely a ounces, as we know he does, is if each other dwarf gives him precisely a/6 ounces and no more, i.e., that each cup, just before pouring, contains the same quantity of milk. So after any given pouring the cups contain a, 5/6a, 4/6a, 3/6a, 2/6a, 1/6a, and 0 ounces; if there are 42 ounces in total then this works out to 12, 10, 8, 6, 4, 2, and 0 ounces.

No. Number the squares as shown. Now each tile, however it’s placed, must cover a 1, a 2, and a 3. But the grid contains 19 1s, 18 2s, and 17 3s. So the task is impossible.

From Pierre Berloquin, The Garden of the Sphinx, 1981.

One pole was 78 inches high, and the other 91 inches. Multiply these two numbers together, and also add them together. Then divide the first result by the second, and you get 42 inches as the height above ground of the point where the two strings intersect, no matter how far the poles may be apart.

1. Bc7 and the queen will mate.

From J. Paul Taylor, Chess Chips, 1878.

Black can have moved only his knights and his rooks. The knights began on squares of different colors, and a moving knight lands on light and dark squares alternately, so between them the two knights have made an even number of moves (even if they’ve exchanged places). While the knights were abroad the rooks may have shifted back and forth a bit, but they’re on their home squares now and so have made an even number of moves themselves (possibly zero). So Black has made an even number of moves altogether.

If Black has made an even number of moves, then so has White. But this presents a problem: White’s knights and rooks have moved an even number of times, but the f-pawn’s position on f3 means that White has made an additional move, giving an odd total. In order to make the total even we must find an odd quantity of additional moves that White can have made, and they can’t have been made with pawns, knights, or rooks. The simplest solution is that the king has made seven moves, e.g., e1-f2-e3-f4-g4-g3-f2-e1.

So the answer is that each player has made at least eight moves: The shortest game that could produce the position above with White to play begins with 1. f3 and then the white king makes seven moves while Black marks time with his knights (and perhaps his rooks).