No. When colored like a checkerboard, the rectangle has an equal number of light and dark squares. So does each of the tetrominoes … with one exception. So the task is impossible.

1. Bg5! leads to two pretty mates:

1. … Kf5 2. Rc5#
2. … Kd6 2. Bf4#

No one in this country can ever claim to be a hunter — a noble who did so would be lying, and a hunter who did so would be telling the truth. This means that Ali must have been lying when he attributed that claim to Ahmed — which tells us that Ali is a hunter. And if Ali always lies, then his statement that Azab is a hunter is false, which tells us that Azab is a noble. That means that we can trust Azab when he tells us that Ahmed is a noble.

(From Goodman’s Problems and Projects, 1972.)

The prisoners agree that one of them (say, Alice) will try repeatedly to turn the light on, and each of the others will turn it off twice.

That is, if Alice finds the light off she’ll turn it on, and if she finds it on she’ll leave it on. Each of the others will turn it off the first two times he finds it on and otherwise leave its state unchanged.

Alice counts the number of times she finds the room dark after her first visit. When she’s returned to find the room dark 2n – 3 times, she can announce that everyone has visited the room.

Each time Alice returns to the room and finds it dark, that’s a sign that one of the other n – 1 prisoners has visited it. If a prisoner exists who still hasn’t visited the room, then the light can’t have been turned off more than 2(n – 2) = 2n – 4 times. So once Alice’s total exceeds that, reaching 2n – 3, she knows that everyone’s visited the room.

(Note that the light’s initial state doesn’t matter: If the first visitor isn’t Alice and he finds the light on, he’ll turn it off, and that sign of his presence won’t be evident to her. But she’ll have to turn the light on again before he can perform his second off-turning, and that ensures that she’ll register his second visit.)

(From Peter Winkler’s excellent Mathematical Puzzles: A Connoisseur’s Collection, 2003.)

1. Kc6 Ka7 2. c8=R! Ka6 3. Ra8#

From Arthur Ford Mackenzie, Chess Lyrics: A Collection of Chess Problems, 1905.

The solution appears in the May 1941 issue:

“Of the 128 people at Piccadilly, 104 either had used or were about to use the Bakerloo Line. Hence the answer required is 80.”

(The Eureka archive is published under a Creative Commons license.)

Put him on h5 and White can mate in two starting with 1. Qb3.

From Arthur Ford Mackenzie, Chess: Its Poetry and Its Prose, 1887.

Yes. The thing to notice is that we can’t have three routes in the pattern (A,B,C), (A,B,D), (A,C,D), because in that case it’s impossible to fulfill the third condition: If the remaining ports are E and F, then we’ve left ourselves only one permissible route (not two) that visits A and E. That’s (A, E, F). Any other three-port route that includes A and E would “overload” another segment.

This means that if two of the routes are (A,B,C) and (A,B,D), then the second one that covers A,C must be (without loss of generality) (A,C,E). With that in mind the other routes can be worked out uniquely: (A,D,F), (A,E,F), (B,D,E), (B,E,F), (B,C,F), (C,D,E), (C,D,F).

David’s solution:

If one die has X different numbers and the other die has Y different numbers, there are at most XY possible totals. Since we need 11 totals from 2 to 12, XY must be at least 11. In particular, at least one die must have at least four different numbers.

If the first die has six different numbers, then any number on the second die has at most a 1/6 probability of getting the right number on the first die to get a total of 7. Therefore, the probability of rolling a 7 is at most 1/6.

If the first die has five different numbers, with one appearing twice, the second die must have at least three different numbers. There is only one possible number on the second die which totals a 7 with the duplicated number on the first die. The second die has at most four copies of the number which totals a 7 with the duplicated number and has two ways to roll a 7; all other numbers on the second die have only one way to roll a 7. This is a maximum total of 10 rolls for a probability of 10/36.

If each die has four different numbers, they can be split 2,2,1,1 or 3,1,1,1. Each number on each die can pair with at most one number on the other die to make a 7. The largest possible sum of products, which gives the number of ways to roll a 7, is 3 × 3 + 1 × 1 + 1 × 1 + 1 × 1 = 12. The maximum probability is thus 12/36.

If the first die has four different numbers and the second has three, there are only 12 different pairs of values, and thus there must be at most two different value pairs which total 7 in order to have 11 possible totals.

Therefore, the largest possible sum of products is 3 × 4 + 1 × 1 = 13, not 3 × 4 + 1 × 1 + 1 × 1 = 14. And 13 is possible, with one die having 1,2,6,6,6,7 and the other having 1,1,1,1,3,5 (or 1,2,2,2,6,7 and 1,3,5,5,5,5); this gives a probability of 13/36.