1. Be4! and the dark-squared bishop will mate.

From Skakbladet, July 1912.

None. Each village is converted by its namesake missionary, unless it has already been converted by someone else, so every village gets converted at some point. But then each becomes a deathtrap for the next missionary who visits it. As each missionary works his way around the island, he leaves a string of such traps behind him, the first of which will spring on him when he comes around again unless he’s killed even sooner by stumbling on a village converted by someone else. So all 26 missionaries must die. Since a murderous village reverts to a pagan state, over time the 26 deaths provide 26 unconverted villages, but these are not necessarily different. If we could show that once a village has killed a missionary it is never visited again, and is thus effectively removed from the story, then the 26 deaths leave behind 26 different unconverted villages.

So suppose village A has just murdered its first victim. That means it has been visited twice, once by the missionary who converted it from paganism and then by the missionary whom it murdered. What is the possibility that it will receive a third visit? Well, one of the two known visitors might be missionary A on his first visit to the island. But any other visitor to A must come from village Z. So if A were to get 3 visitors, then at least 2 of them must have come from Z; that is, at least 2 missionaries must have got into and out of village Z. But Z kills every other missionary who visits it. So if 2 missionaries left Z, then at least 3 must have entered it. Thus in order for A to receive a third visitor, Z must already have had at least 3 visitors. By the same reasoning, Y must have had at least 3 visitors, and so on around the island, until we reach the impossible condition that in order to have a third visitor, village A must already have had 3 visitors. This shows that A cannot receive a third visitor, and the conclusion follows.

(From Crux Mathematicorum, via Ross Honsberger’s More Mathematical Morsels, 1991.)

1.Qh1! and White will mate with 2. Qh8# or 2. c7#.

From Schweizerische Schachzeitung, 1907.

Consider the troublesome statement “If Allen is out, then Brown is in” (supposing that Carr is out). We had evaluated this as false. But saying “If Allen is out, then Brown is in” is just the same as saying “Either Allen is in or Brown is in.” And this is true, not false — it admits the possibility that Allen remains in the shop, as of course he must. So the problem disappears.

In simple terms, when Carr leaves the shop then Allen (and perhaps Brown) stays behind, as our common sense says he does.

(Lewis Carroll, “A Logical Paradox,” Mind 3:11 [July 1894], 436-438.)

In the figure, parallel lines meet at the horizon, which represents infinity. If all the ties are evenly spaced, then the railroad track is a series of identical rectangles, and the diagonals of these rectangles are parallel. If diagonal BC reaches the horizon at P, then the line through D that’s parallel to it is PD. That gives us E, the point where the next tie meets the left track, and all that remains is to draw EF parallel to the horizon.

(From Ross Honsberger, More Mathematical Morsels, 1991.)

Here’s the first move of each solution:

Rank:

1: 1. Bd4
2: 1. Qb4
3: 1. Bgd6
4: 1. Qc8+
5: 1. Ra3
6: 1. Kb7
7: 1. Qe5
8: 1. Rc7

File:

a: 1. Bc2
b: 1. Qd4
c: 1. Qxc6
d: 1. Rxd4
e: 1. Qe4
f: 1. Bd5+
g: 1. Qd7
h: 1. Qe4

In general this type of problem is known as a “patchwork.”

It’s a transposition cipher. Start with the letter S and skip backward three characters at a time through the string, counting spaces as characters and cycling back to the end as often as necessary:

SAY HELLO TO A GOOD BUY