Futility Closet An idler's miscellany of compendious amusements

Neither is authentic. No one in 51 B.C. could have foreseen the existence of Jesus of Nazareth, much less that he would arrive in exactly 51 years. And George would not have been called George I until it became necessary to distinguish him from George II (“Victoria” will not become “Victoria I” until we have a “Victoria II”).

Puzzling Invitation:

Just read it straight through:

Here stop and spend a social hour
In harmless mirth and fun.
Let friendship reign, be just and kind,
And evil speak of none.

From William T. Dobson, Literary Frivolities, Fancies, Follies and Frolics, 1880.

See also Requiescat In Pace.

All that is necessary is to tilt the barrel as in Fig. 1, and if the edge of the surface of the water exactly touches the lip a at the same time that it touches the edge of the bottom b, it will be just half full. To be more exact, if the bottom is an inch or so from the ground, then we can allow for that, and the thickness of the bottom, at the top. If when the surface of the water reached the lip a it had risen to the point c in Fig. 2, then it would be more than half full. If, as in Fig. 3, some portion of the bottom were visible and the level of the water fell to the point d, then it would be less than half full.

This method applies to all symmetrically constructed vessels.

1. A clock strikes six in 5 seconds. How long does it take to strike twelve?
Answer: Not 10 seconds, but 11 — one second per interval.

2. A bottle and its cork together cost $1.10. The bottle costs a dollar more than the cork. How much does the bottle cost?
Answer: Not $1.00, but $1.05.

3. A train leaves New York for Chicago at 90 mph. At the same time, a bus leaves Chicago for New York at 50 mph. Which is farther from New York when they meet?
Answer: Neither — when they meet they’ll be equally distant from anywhere, n’est-ce pas?

You’ll hear from my attorney.

The rolling wheel makes two revolutions. You can prove this to yourself using two coins with milled edges—hold one in place and roll the other around it.

1. Mrs. A and Mrs. C cross
2. Mrs. A returns
3. Mrs. A and Mrs. B cross
4. Mrs. B returns
5. Mr. A and Mr. C cross
6. Mr. A and Mrs. A return
7. Mr. A and Mr. B cross
8. Mrs. C returns
9. Mrs. A and Mrs. B cross
10. Mrs. A returns
11. Mrs. A and Mrs. C cross

Other solutions are possible, but all require at least 11 crossings. (The price of propriety!)

No, it can’t be done.

Each domino covers two squares, a light one and a dark one. Therefore a board can be tiled perfectly only if it has an equal number of light and dark squares.

The original board met this requirement, but we’ve removed two squares of the same color (diagonal corners). So the task is impossible.

The white queen travels clockwise until she captures the black knight next to the black king, giving check. Black’s only legal response is to capture the white knight. (The bishop can’t take the queen because it’s pinned by the rook at left.) Then the white rook at the top travels counterclockwise until it captures the black queen, giving mate.

(Thanks for Ryan for the tip, and to Lawrence for permission.)

1.f3 e5 2.Kf2 h5 3.Kg3 h4+ 4.Kg4 d5#

Ten rows altogether. Not bad for a chicken.

We know that the brakeman doesn’t live in Detroit (2) and that his nearest neighbor is a passenger (4). So that passenger can’t be Mr. Robinson, who lives in Detroit (1).

From (3) and (4) we know that the brakeman’s nearest neighbor also can’t be Mr. Jones (because Mr. Jones earns $20,000 per year, which is not evenly divisible by 3).

Thus, if the brakeman’s nearest neighbor is not Mr. Jones and not Mr. Robinson, it must be Mr. Smith.

Now, if Mr. Robinson lives in Detroit (1) and Mr. Smith lives halfway between Chicago and Detroit (as we’ve just deduced), then the third passenger, Mr. Jones, must be the one referred to in (6) and thus lives in Chicago.

That means, also from (6), that the brakeman’s name is Jones. And if Smith is not the fireman (5), then by elimination Smith must be the engineer.

“Though this problem was much discussed in the Daily Mail from 18th January to 7th February, 1905, when it appeared to create great public interest, it was actually first propounded by me in the Weekly Dispatch of 14th June, 1903.

“Imagine the room to be a cardboard box. Then the box may be cut in various different ways so that the cardboard may be laid flat on the table. I show four of these ways and indicate in every case the relative positions of the spider and the fly and the straightened course which the spider must take, without going off the cardboard. These are the four most favourable cases, and it will be found that the shortest route is in No. 4, for it is only 40 feet in length (add the square of 32 to the square of 24 and extract the square root). It will be seen that the spider actually passes along five of the six sides of the room! Having marked the route, fold the box up (removing the side the spider does not use), and the appearance of the shortest course is rather surprising. If the spider had taken what most persons will consider the obviously shortest route (that shown in No. 1), he would have gone 42 feet! Route No. 2 is 43.174 feet in length and route No. 3 is 40.718 feet. I will leave the reader to discover which are the shortest routes when the spider and fly are 2, 3, 4, 5, and 6 feet from the ceiling and floor respectively.”

“‘Some here have asked me,’ continued Sir Hugh, ‘how they may find the cell in the dungeon of the Death’s Head wherein the noble maiden was cast. Beshrew me! but ’tis easy withal when you do but know how to do it. In attempting to pass through every door once, and never more, you must take heed that every cell hath two doors or four, which be even numbers, except two cells, which have but three. Now, certes, you cannot go in and out of any place, passing through all the doors once and no more, if the number of doors be an odd number. But as there be but two such odd cells, yet may we, by beginning at the one and ending at the other, so make our journey in many ways with success. I pray you, albeit, to mark that only one of these odd cells lieth on the outside of the dungeon, so we must perforce start therefrom. Marry, then, my masters, the noble demoiselle must needs have been waiting in the other.’

“The drawing makes this quite clear to the reader. The two ‘odd cells’ are indicated by the stars, and one of the many routes that will solve the puzzle is shown by the dotted line. It is perfectly certain that you must start at the lower star and end at the upper one; therefore, the cell with the star situated over the left eye must be the one sought.”

The earlier game must have reached this position:

From here, White played e5+, and Blacked responded … f5 (the only move). Then White captured the pawn en passant, exf5 e.p., producing the final position.