Futility Closet An idler's miscellany of compendious amusements

Black’s king is in check, so White must just have moved his knight from b4 to a2. But what was Black’s move before that? He cannot have moved his king, as there’s no square from which it can have come. The only other possibility is that Black moved a piece that’s not now on the board — a piece that White’s a2 knight has just captured. What can that have been? Not a knight or a pawn, as there’s no square that either of these can have come from. Nor a queen or a rook, as those could only have moved from a1, where they would have been checking the White king at the end of White’s last move. The only remaining possibility is a bishop: Black has just moved a bishop from b1 to a2, and White has responded by capturing it with a knight from b4.

None. The same amount of water is necessary to raise any boat.

One way to make this intuitive is to imagine that a 2-meter “slab” of water is inserted at the bottom of the lock chamber. This will raise a boat of any size to the level of the higher lake.

Yes. The water imparts an upward force on your finger as you immerse it, so you must exert a downward force to overcome it. That force is transferred to the glass and then to the scale, which sinks.

From Yuri B. Chernyak, The Chicken From Minsk, 1995.

White could mate in three different ways if it were his turn, but it can’t be, as there’s no legal move that Black can just have made. So Black plays Bxf6#.

It seems at first that the chance must be 1/2, as we’ve simply added a white counter and then removed it again, returning the bag to its original state.

But this is an error. After the white counter is added, the bag contains either (a) 2 white counters or (b) 1 white counter and 1 black counter. These states are equally probable. The chance of drawing a white counter from bag (a) is 1; from bag (b) it’s 1/2. So when the white counter is drawn, the chance that the bag had contained (a) 2 white counters or (b) 1 white counter and 1 black counter are proportional to (a) 1/2 × 1 (b) 1/2 × 1/2, i.e., 2 to 1. So there’s a 2/3 chance that the remaining counter is white.

In other words, the white counter that we drew is twice as likely to have come from a white-white bag than from a white-black bag.

Both were wrong, and the lady was badly cheated. She only got half the quantity that would be contained in the large bundle, and therefore should have been charged half the original price. A circle with the circumference half that of another must have its area a quarter that of the other.

There’s no legal move by which Black could have removed one of his own pieces from the board, so he must have moved the king. It’s illegal for two kings ever to occupy adjacent squares, so the king must have moved to a8 from a7. White’s previous move must have placed him in check there. White cannot have accomplished this with a bishop move, as the white bishop cannot have moved to its present position from another diagonal. The only other possibility is a discovered check by a knight:

So, in this position, White moved the knight from b6 to a8, discovering check, and Black captured the knight with his king.

Once. You swing and miss each time you serve, and your opponent commits a double fault each time he serves. Eventually you find yourself serving at 6-5 or 7-6 in the tiebreak, and then you heroically serve an ace.

From Dick Hess, All-Star Mathlete Puzzles, 2009.

UPDATE: A reader points out that a superlatively lazy player can win a set without ever hitting the ball, thanks to the hindrance rule:

26. HINDRANCE
If a player is hindered in playing the point by a deliberate act of the opponent(s), the player shall win the point.

“On the crucial tie-break point, your opponent dashes across the net and tackles you, awarding you the point and the set.”

(Thanks, Scott.)

The question asks us to discover whatever strange factor could cause 5/2 to give the result 3, and to apply the same factor to 10/3. So:

5/2 : 3 = 10/3 : x

Solving for x gives the result 4.

Never. After the mother’s first four strides and the daughter’s first six, they will again step together with the right foot, and the cycle repeats. Nowhere in that interval do they step together with the left foot.

Turn the board around.

The shortest possible way is to move the articles in the following order: piano, bookcase, wardrobe, piano, cabinet, chest of drawers, piano, wardrobe, bookcase, cabinet, wardrobe, piano, chest of drawers, wardrobe, cabinet, bookcase, piano. Thus seventeen removals are necessary. The landlady could the move chest of drawers, wardrobe, and cabinet. Mr. Dobson did not mind the wardrobe and chest of drawers changing rooms so long as he secured the piano.

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).

It’s easy to get bogged down in calculating proportions, but there’s a simpler solution. If I continue until the bottle is empty, then eventually I will have drunk all the water that I’ve added to the bottle. I add 1 ounce on the first day, 2 on the second, and so on to the 15th day. (I drink off the entire bottle on the 16th day, so no water is added then.) So in all I’ll have drunk 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 = 120 ounces of water.

(Another solution is to calculate that in the end I’ve drunk 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 = 136 ounces in total, and then deduct the 16 ounces of wine.)

When I’m standing on the scale and my wife joins me, the difference in the readings will show us her true weight, 292 – 170 = 122 pounds. So the scale is set 130 – 122 = 8 pounds too high.